Felix' Math Place http://math.fontein.de Focussed on, but not limited to Computational Number Theory Sat, 30 Jul 2011 12:35:49 +0000 en hourly 1 http://wordpress.org/?v=3.1 A Cute Identity. http://math.fontein.de/2011/07/30/a-cute-identity/ http://math.fontein.de/2011/07/30/a-cute-identity/#comments Sat, 30 Jul 2011 12:35:49 +0000 Felix Fontein https://math.fontein.de/?p=869 Recently, while doing some computations, I stumbled about a very interesting identity, which I do not want to withhold from you all:

Theorem.

Let K be a field and x_1, \dots, x_n \in K such that 1 + \sum_{j=1}^i x_i \neq 0 for all i = 1, \dots, n. Then \displaystyle 1 - \sum_{i=1}^n \frac{x_i}{\Bigl( 1 + \sum_{j=1}^{i-1} x_j \Bigr) \Bigl( 1 + \sum_{j=1}^i x_j \Bigr)} = \frac{1}{1 + \sum_{i=1}^n x_i}.

It is very easy to prove it by induction:

Proof.

For n = 1, the left-hand side equals \displaystyle 1 - \frac{x_1}{( 1 ) ( 1 + x_1 )} = \frac{1 + x_1 - x_1}{1 + x_1} = \frac{1}{1 + x_1}, which equals the right-hand side for n = 1. Hence, the statement is true for n = 1. Now assume that it holds for n. Then
& 1 - \sum_{i=1}^{n+1} \frac{x_i}{\Bigl( 1 + \sum_{j=1}^{i-1} x_j \Bigr) \Bigl( 1 + \sum_{j=1}^i x_j \Bigr)} \\ {}={} & 1 - \sum_{i=1}^n \frac{x_i}{\Bigl( 1 + \sum_{j=1}^{i-1} x_j \Bigr) \Bigl( 1 + \sum_{j=1}^i x_j \Bigr)} \\ {}-{} & \frac{x_{n+1}}{\Bigl( 1 + \sum_{j=1}^n x_j \Bigr) \Bigl( 1 + \sum_{j=1}^{n+1} x_j \Bigr)} \\ {}={} & \frac{1}{1 + \sum_{i=1}^n x_i} - \frac{x_{n+1}}{\Bigl( 1 + \sum_{j=1}^n x_j \Bigr) \Bigl( 1 + \sum_{j=1}^{n+1} x_j \Bigr)} \\ {}={} & \frac{1 + \sum_{j=1}^{n+1} x_j}{\Bigl(1 + \sum_{i=1}^n x_i\Bigr) \Bigl( 1 + \sum_{j=1}^{n+1} x_j \Bigr)} - \frac{x_{n+1}}{\Bigl( 1 + \sum_{j=1}^n x_j \Bigr) \Bigl( 1 + \sum_{j=1}^{n+1} x_j \Bigr)} \\ {}={} & \frac{1 + \sum_{j=1}^n x_j}{\Bigl(1 + \sum_{i=1}^n x_i\Bigr) \Bigl( 1 + \sum_{j=1}^{n+1} x_j \Bigr)} = \frac{1}{1 + \sum_{j=1}^{n+1} x_j}, what we had to show.
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Yet, I have no idea what this identity should tell me. The left-hand side looks so complicated, there is no indication it should simplify to something like the right-hand side. This identity miraculously appeared when I computed the Gram-Schmidt orthogonalization of the linearly independent system v_i = \lambda_i e_i + e_n, 1 \le i < n, where \lambda_i \in \R \setminus \{ 0 \} and e_1, \dots, e_n is the standard orthonormal base of \R^n. It turns out that one can explicitly describe the Gram-Schmidt orthogonalization, namely it is \hat{v}_1, \dots, \hat{v}_{n-1} with \displaystyle \hat{v}_i = \lambda_i e_i + \frac{1}{1 + \sum_{j=1}^{i-1} \lambda_j^{-2}} \biggl( -\sum_{j=1}^{i-1} \lambda_j^{-1} e_j + e_n \biggr), and the squared norm of \hat{v}_i is given by \displaystyle \langle \hat{v}_i, \hat{v}_i \rangle = \lambda_i^2 \cdot \frac{1 + \sum_{j=1}^i \lambda_j^{-2}}{1 + \sum_{j=1}^{i-1} \lambda_j^{-2}}. This one can also easily show by induction, using the above identity; it appears two times with x_i = \lambda_i^{-2}. Note that the system (v_1, \dots, v_{n-1}) already appeared once in this blog, namely when I tried to find the closest vector in its span to e_n; this was done in this post.

In case you have seen this identity before, let me know. I’m really curious if it has been used somewhere else.

]]> http://math.fontein.de/2011/07/30/a-cute-identity/feed/ 0 Solving Certain Linear Systems over the Integers. http://math.fontein.de/2011/06/17/solving-certain-linear-systems-over-the-integers/ http://math.fontein.de/2011/06/17/solving-certain-linear-systems-over-the-integers/#comments Fri, 17 Jun 2011 18:52:49 +0000 Felix Fontein https://math.fontein.de/?p=831 Assume you have a linear system of equations A x = b, where A \in \Z^{n \times n} and b \in \Z^n. Assume that \det A \neq 0, and that we know that a solution in \Z^n exists. One question is: how can we efficiently compute x? Clearly, any algorithm solving linear systems over the integers or rationals will do; for example, the algorithms from the Integer Matrix Library by Z. Chen, C. Fletcher and A. Storjohann will do. That library will find any solution x \in \Q^n, and also does not require that A is invertible (over the rationals) or that A is square. But for our purposes, using such a general solver is overkill.

Note that the below material is well-known among experts.

Let p be any prime not dividing \det A. Then A modulo p is invertible, and modulo p, the system A x = b has a unique solution. Moreover, for any integer e, the system A x = b has a unique solution modulo p^e: this is true since A is also invertible modulo p^e – for that, it suffices to check that the determinant of A is a unit, which it is since it is coprime to p^e. Moreover, if y \in \Z^n is a solution to A x = b over the integers, then y modulo p^e is the unique solution of A x = b modulo p^e. Hence, if we choose e such that \frac{1}{2} p^e bounds all coefficients of the solution y, we can recover a solution to A x = b over the integers from a solution to A x = b modulo p^e, by chosing the unique preimages in (-\tfrac{1}{2} p^e, \tfrac{1}{2} p^e].

This opens the question on how to solve A x = b modulo p^e. For that, a Hensel-like lifting technique can be used. (In fact, this follows from Bourbaki’s generalization since the Jacobian of the map f : (\Z/p^e\Z)^n \to (\Z/p^e\Z)^n, x \mapsto A x - b equals A.) Assume that we have an x \in \Z^n which satisfies A x \equiv b \pmod{p^{e-1}}. We want to find x' \in \Z^n with A x' \equiv b \pmod{p^e}. Write x' = x + p^{e-1} x'' with x'' \in \{ 0, \dots, p - 1 \}^n. As A x' = A x + p^{e-1} A x'', and as A x - b is divisible by p^{e-1}, we obtain the linear system A x'' \equiv \frac{A x - b}{p^{e-1}} \pmod{p}. Hence, it suffices to solve e linear systems over the prime field \F_p to solve A x = b over \Z/p^e\Z.

This yields the following algorithm:

  1. Choose p := 2.
  2. Solve A x = b modulo p.
  3. If a unique solution exists:
    1. Set e = 0 and lift x to the integers with coordinates in (\tfrac{1}{2} p, \tfrac{1}{2} p].
    2. Compute c := A*x - b.
    3. If c = 0, return x.
    4. Solve A y = c/p^e modulo p.
    5. Set x := x + y*p^e and e := e + 1.
    6. Adjust x modulo p^{e+1} such that all coefficients are in (\tfrac{1}{2} p^{e+1}, \tfrac{1}{2} p^{e+1}].
    7. Go back to Step 3.2.

    Else:

    1. Choose the next prime p and go back to Step 2.

The only subprogram we need is a linear systems solver for A x = b with square A over a finite field, which returns information on the number of solutions. (Note that \det A is not divisible by p if and only if there is a unique solution.) If more information is known on the matrix A, for example its determinant has been already computed, this information can be used as well.

Let us analyze the running time of this algorithm. Denote by NP(A) the smallest prime not dividing \det A, and by S(n, p) the time the linear system solver over \F_p needs to solve a system of size n. Let \|A\|_\infty (resp. \|x\|_\infty resp. \|b\|_\infty) denote the largest absolute value of an coefficient of A (resp. x resp. b).

Clearly, the number of iterations is in O(\log_{NP(A)} \|x\|_\infty) = O(\frac{\log \|x\|_\infty}{\log NP(A)}). In each iteration, one linear system over \F_p of size n has to be solved, and A x - b has to be evaluated. The former takes S(n, NP(A)) operations, and the latter involves n^2 multiplications and additions of integers of size O(\log \|A\|_\infty) and O(e \log NP(A)), and n substractions of integers of size O(\log \|A\|_\infty + e \log NP(A)) and O(\log \|b\|_\infty). For simplicity, assume that \log \|b\|_\infty = O(\log \|A\|_\infty). Finally, to compute x = x + y p^e, we need n multipliations of integers of size O(\log NP(A)) and O(e \log NP(A)), and n additions which can be neglected. Clearly, the n multiplications can also be neglected, since the evaluation of A x already is slower.

Let M(m) denote the time a multiplication of two numbers of size m needs. Then inside the main loop, we need \displaystyle O\bigl(S(n, NP(A)) + n^2 M(\max\{ \log \|A\|_\infty, e \log NP(A) \})\bigr) time units, and the main loop alltogether needs & O\Biggl(\sum_{e=1}^{\frac{\log \|x\|_\infty}{\log NP(A)}} \biggl( S(n, NP(A)) + n^2 M(\max\{ \log \|A\|_\infty, e \log NP(A) \}) \biggr) \Biggr) \\ {}={} & O\Biggl(\frac{\log \|x\|_\infty}{\log NP(A)} \bigl( S(n, NP(A)) + n^2 M(\max\{ \log \|A\|_\infty, \log \|x\|_\infty \}) \biggr) \Biggr) time units. Finding p needs \displaystyle O\Biggl(\frac{NP(A)}{\log NP(A)} S(n, NP(A)) \Biggr) time units (using the Prime Number Theorem).

Assuming that we use a naive Gaussian algorithm as well as naive multiplication, i.e. S(n, p) = n^3 (\log p)^2 and M(m) = m^2, we obtain a total running time of O\Biggl( & n^3 \bigl( \log \|x\|_\infty + NP(A) \bigr) \log NP(A) \\ & {}+ n^2 \max\biggl\{ \frac{(\log \|A\|_\infty)^2 \log \|x\|_\infty}{\log NP(A)}, \frac{(\log \|x\|_\infty)^3}{\log NP(A)} \biggr\} \Biggr). Using fast multiplication, i.e. M(m) = m^{1 + \varepsilon} (using FFT methods), and fast linear system solving, i.e. S(n, p) = O(n^\omega (\log p)^{1 + \varepsilon}), where \omega \le 2.376, we obtain a total running time of O\Biggl( & (NP(A) + \log \|x\|_\infty) n^\omega (\log NP(A))^{\varepsilon} \\ & {}+ n^2 \max\biggl\{ \frac{\log \|x\|_\infty (\log \|A\|_\infty)^{1+\varepsilon}}{\log NP(A)}, \frac{(\log \|x\|_\infty)^{2 + \varepsilon}}{\log NP(A)} \biggr\} \Biggr)

Now let us try to eliminate NP(A) from this expression. Clearly, the the second part, we can use that NP(A) \ge 2. To eliminate NP(A) from the first part, we need to find an upper bound. For that, let us first stick to NP(t), the smallest prime not dividing the integer t. (Letting A be a 1 \times 1-matrix A yields NP(t) = NP(A); in general, NP(A) = NP(\det A) using this notation.) Now t is divisible by \prod_{p < NP(t)} p, whence for t < \prod_{p < x} p we have NP(t) < x. Note that for integral x, \log \bigl( \prod_{p < x} p \bigr) = \vartheta(x - 1) \le \vartheta(x) \sim x, where \vartheta denotes the Chebyshev function. Using known bounds on \vartheta(x), we get \displaystyle \prod_{p < x} p = \exp(x + O(x/\log x)) = \exp((1 + o(1)) x). Therefore, \prod_{p < x} p > \exp((1 - \varepsilon) x) becomes true for x \to \infty for every \varepsilon. This shows that NP(t) < \frac{\log t}{1 - \varepsilon} eventually holds for t \to \infty, yielding NP(t) = O(\log t) and, thus, NP(A) = O(\log \det A). Using the Leibniz formula, \log \det A = O(n \log n + n \log \|A\|_\infty).

Finally, we can use some linear algebra to bound \|x\|_\infty in terms of A and b. First note that A A^\# = (\det A) I_n, where I_n denotes the n \times n identity matrix and A^\# denotes the adjungate matrix of A. As x = A^{-1} b = \frac{1}{\det A} A^\# b, we see that it suffices to bound \|A^\#\|_\infty. Now the coefficients of A^\# are determinants of (n - 1) \times (n - 1) matrices with coefficients coming from A, whence \|A^\#\|_\infty \le (n - 1)! \|A\|_\infty^n. Therefore, \displaystyle \log \|x\|_\infty \le n \log n + n \log \|A\|_\infty + \log \|b\|_\infty = O(n \log \|A\|_\infty) when assuming that \log n, \log \|b\|_\infty = O(\log \|A\|_\infty).

This can be combined into the following theorem:

Theorem.

Assuming that \log n = O(\log \|A\|_\infty) and \log \|b\|_\infty = O(\log \|A\|_\infty), the above algorithm needs \displaystyle O\bigl( n^5 (\log \|A\|_\infty)^3 \bigr) time units to compute the unique solution of A x = b using naive arithmetic in \Z, \F_p, and naive Gaussian elimination to solve linear systems over \F_p. Using fast linear algebra and fast multiplication, we only need \displaystyle O\bigl( n^{4 + \varepsilon} (\log \|A\|_\infty)^{2 + \varepsilon} \bigr) time units for any \varepsilon > 0.
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]]> http://math.fontein.de/2011/06/17/solving-certain-linear-systems-over-the-integers/feed/ 0 On a Certain Determinant. http://math.fontein.de/2011/03/25/on-a-certain-determinant/ http://math.fontein.de/2011/03/25/on-a-certain-determinant/#comments Fri, 25 Mar 2011 09:55:14 +0000 Felix Fontein https://math.fontein.de/?p=814 Yesterday, I managed to compute a determinant of a certain matrix. This matrix appears in the work I’m currently doing, and having an explicit formula for it allowed me to explicitly solve a system of linear equations using . In this post, I want to present the result with two different proofs by induction.

I wouldn’t be surprised if this is already well-known, but at least I didn’t see it before. Anyway, if you have seen it before, please feel free to tell me about it.

Edit: in fact, the theorem follows from the more general result stated here. Also see the comments.

Theorem.

Let K be a field and x_1, \dots, x_n \in K. Then the matrix \displaystyle M(x_1, \dots, x_n) := \Matrix{ 1 + x_1 & 1 & \cdots & 1 \\ 1 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 1 \\ 1 & \cdots & 1 & 1 + x_n } \in K^{n \times n} has determinant \displaystyle \prod_{i=1}^n x_i + \sum_{j=1}^n \prod_{i \neq j} x_i.

Before I present the proofs, I want to say a few words on how this matrix comes up. Consider the linear system A x = b, where \displaystyle A = \Matrix{ y_1 & & 0 \\ & \ddots & \\ 0 & & y_n \\ 1 & \cdots & 1 } \in \R^{(n + 1) \times n} \quad \text{and} \quad b = \Matrix{ 0 \\ \vdots \\ 0 \\ 1 } \in \R^{n+1}. Assuming that all y_i \neq 0, one sees that this system has no solution. Instead, one can try to find a vector x which minimizes \| A x - b \|_2. A well-known fact in Linear Algebra says that this is the case for a unique x, and that x is the solution to the system A^T A x = A^T b. Now A^T A = M(y_1^2, \dots, y_n^2) and A^T b = (1, \dots, 1)^T, whence we can describe the unique solution x = (x_1, \dots, x_n)^T using Cramer’s rule by \displaystyle x_i = \frac{\det M(y_1^2, \dots, y_{i-1}^2, 0, y_{i+1}^2, \dots, y_n^2)}{\det M(y_1^2, \dots, y_n^2)}. Now the above theorem gives an easy to evaluate formula for the determinants, namely \displaystyle x_i = \frac{\prod_{j \neq i} y_j^2}{\prod_{j=1}^n y_j^2 + \sum_{k=1}^n \prod_{j \neq k} y_j^2}. The solution vector x can be evaluated in O(n) operations, and in case the y_i are rational numbers, one can hence efficiently compute an exact (i.e. rational) solution.

Proof (First proof).

For n = 1, 2 this is easy to verify. Hence, assume that n \ge 3 and that the statement is true for n - 1. Using the multilinearity of the determinant and Lagrange expansion, both for the first row, we see that \displaystyle \det M(x_1, \dots, x_n) = x_1 \det M(x_2, \dots, x_n) + \det M(0, x_2, \dots, x_n). The same argument applied to the second row of the second determinant, we obtain that the second determinant \displaystyle x_2 \det M(0, x_3, \dots, x_n) + \det M(0, 0, x_3, \dots, x_n). Since M(0, 0, x_3, \dots, x_n) has two identical rows – namely the first two –, its determinant is 0. Moreover, by induction hypothesis, \det M(0, x_3, \dots, x_n) = \prod_{i=3}^n x_i and \displaystyle \det M(x_2, \dots, x_n) = \prod_{i=2}^n x_i + \sum_{j=2}^n \prod_{i=2 \atop i \neq j}^n x_i. Plugging this in yields \displaystyle \det M(x_1, \dots, x_n) = x_1 \prod_{i=2}^n x_i + x_1 \sum_{j=2}^n \prod_{i=2 \atop i \neq j}^n x_i + \prod_{i=2}^n x_i, which shows the claim.
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Proof (Second proof).

For n = 1 this is clear. Hence, assume that n > 1 and that the statement is true for n - 1. We do a Lagrange expansion by the last column. This yields a term (-1)^{n + n} (1 + x_n) \det M(x_1, \dots, x_{n-1}), which by induction hypothesis equals \displaystyle \prod_{i=1}^{n-1} x_i + \sum_{j=1}^{n-1} \prod_{i=1 \atop i \neq j}^{n-1} x_i + \prod_{i=1}^n x_i + \sum_{j=1}^{n-1} \prod_{i=1 \atop i \neq j}^n x_i. The other terms are of the form (-1)^{n + i} \det M_i, where M_i is M(x_1, \dots, x_n) with the last column and the i-th row removed. Note that by cyclically permuting rows i to n - 1 of M_i, we obtain the matrix M(x_1, \dots, x_{i-1}, 0, x_{i+1}, \dots, x_{n-1}). The permutation has sign (-1)^{n - i + 1}, whence we see that \displaystyle (-1)^{n+i} \det M_i = -\det M(x_1, \dots, x_{i-1}, x_{i+1}, \dots, x_{n-1}) = -\prod_{j=1 \atop j \neq i}^{n-1} x_j. Combining everything so far, we see that & \det M(x_1, \dots, x_n) \\ {}={} & \prod_{i=1}^{n-1} x_i + \sum_{j=1}^{n-1} \prod_{i=1 \atop i \neq j}^{n-1} x_i + \prod_{i=1}^n x_i + \sum_{j=1}^{n-1} \prod_{i=1 \atop i \neq j}^n x_i - \sum_{i=1}^{n-1} \prod_{j=1 \atop j \neq i}^{n-1} x_j \\ {}={} & \prod_{i=1}^n x_i + \sum_{j=1}^n \prod_{i=1 \atop i \neq j}^n x_i, what we wanted to show.
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]]> http://math.fontein.de/2011/03/25/on-a-certain-determinant/feed/ 2 A Strange Inequality. http://math.fontein.de/2010/12/09/a-strange-inequality/ http://math.fontein.de/2010/12/09/a-strange-inequality/#comments Thu, 09 Dec 2010 07:49:09 +0000 Felix Fontein http://math.fontein.de/?p=795 Today, while trying to prove a result for a preprint I’m working on, I got so frustrated that I played around with something else from that paper. I got an idea to get rid of something non-nice, namely I had an asymptotic bound O(g/n + 1) for g, n \to \infty, and wanted to drop the +1 if possible. Of course, this is only possibe if g > 0 and if g \ge C n for some constant n.

So I began working this out to see how far I could get. It is rather easy to translate the whole problem into a question on some integers. Namely, assume that n > 1 is an integer, and we are given s integers 1 \le n_1, \dots, n_s < n with \gcd(n_1, \dots, n_s, n) = 1. Define the quantity g(n_1, \dots, n_s) as \displaystyle \frac{\biggl(n - \gcd\biggl(n, \sum\limits_{i=1}^s n_i\biggr)\biggr) + \sum\limits_{i=1}^s (n - \gcd(n, n_i)) - 2 (n - 1)}{2}. One can show that this is always a non-negative integer. Now I claim that g(n_1, \dots, n_s) is either 0, or \ge \frac{1}{6} n.

Where does this strange expression comes from? Consider the function field K = \C(x, y) over \C defined by \displaystyle y^n = \prod_{i=1}^s (x - i)^{n_i}. (In fact, one can do this over any field whose characteristic is coprime to n, and which has at least s elements. Moreover, over an algebraically closed field k, any function field extension K / k(x) which is cyclic of order n, where n is coprime to the characteristic of k, can be written in this form, since this is a Kummer extension.) One can show that this equation is irreducibe if and only if \gcd(n_1, \dots, n_s, n) = 1, and that the genus of this function field is given by g(n_1, \dots, n_s). This also explains why we must have that g(n_1, \dots, n_s) \ge 0, since the genus is always a nonnegative integer.

Now we have a struggle: is it easier to show the claim using some Elementary Number Theory, or using some (advanced?) Algebraic Geometry (considering the geometric side) or Algebraic Number Theory (considering the function field side)? I don’t have any idea how to use the latter two to prove this, but I found an elementary proof.

First, note that if s \ge 5, or in case n \ge 4 and n \nmid n_1 + \dots + n_s, at least five of the n - \gcd(n, \bullet) terms must be \ge \frac{n}{2} since \bullet is not divisible by n. Therefore, g(n_1, \dots, n_s) \ge \frac{5 \cdot \tfrac{1}{2} n - 2 (n - 1)}{2} \ge \tfrac{1}{2} n.

Moreover, if n \mid n_1 + \dots + n_s, we have n_s \equiv -(n_1 + \dots + n_{s-1}) \pmod{n}, and therefore 1 = \gcd(n_1, \dots, n_s, n) = \gcd(n_1, \dots, n_{s-1}, n) and \gcd(n, n_s) = \gcd(n, n_1 + \dots + n_{s-1}). Hence, \displaystyle g(n_1, \dots, n_s) = g(n_1, \dots, n_{s-1}). Therefore, it suffices to consider the case n \nmid n_1 + \dots + n_s and s \le 3.

For s = 1, note that \gcd(n, n_1) = 1 implies g(n_1) = 0. Hence, there is nothing to show.

For s = 2, let me consider three cases.

  1. \gcd(n, n_1) = n/p for some prime p;
  2. \gcd(n, n_1) = n/p^2 for some prime p;
  3. \gcd(n, n_1) = n/(p q) for two distinct prims p, q.

In all three cases, one can show by analyzing several cases that the claim is true. Thus, we are only interested in the cases where no \gcd(n, n_i) is of the form n/p, n/p^2, n/(pq). Therefore, \gcd(n, n_i) \le \frac{1}{12} for all i. Since not all \gcd(n, n_i)‘s cannot be the same – this would contradict \displaystyle 1 = \gcd(n_1, \dots, n_s, n) = \gcd(\gcd(n_1, n), \dots, \gcd(n_s, n)) – we must have & (n - \gcd(n, n_1 + n_2)) + \sum_{i=1}^2 (n - \gcd(n, n_i)) - 2 (n - 1) \\ {}\ge{} & n - (\tfrac{1}{2} + \tfrac{1}{12} + \tfrac{1}{13}) n + 2 \ge \tfrac{1}{3} n + 2.

Let me demonstrate how to do \gcd(n, n_1) = n/p. (In fact, this case suffices if one does not wants the constant \frac{1}{6}, but one is happy with the constant \frac{1}{24}, since 1 - \frac{1}{4} - \frac{1}{6} - \frac{1}{2} = \frac{1}{12}.) Assume that \gcd(n, n_1) = n/p for some prime p. Since \displaystyle 1 = \gcd(n_1, n_2, n) = \gcd(\gcd(n, n_1), \gcd(n, n_2) = \gcd(n/p, \gcd(n, n_2)), we must have \gcd(n, n_2) \mid p, with p^2 \nmid n in case \gcd(n, n_2) = p. In both cases, we have \gcd(n/p, n_1 + n_2) = 1.

In case \gcd(n, n_2) = 1, \gcd(n/p, n_1 + n_2) = 1 implies \gcd(n, n_1 + n_2) \mid p. Therefore, \gcd(n, n_1) + \gcd(n, n_2) + \gcd(n, n_1 + n_2) \le n/p + 1 + p. In case \gcd(n, n_2) = p, \gcd(n/p, n_1 + n_2) = 1 implies \gcd(n, n_1 + n_2) = 1. Therefore, we also have \displaystyle \gcd(n, n_1) + \gcd(n, n_2) + \gcd(n, n_1 + n_2) \le n/p + p + 1. This yields & (n - \gcd(n, n_1 + n_2)) + \sum_{i=1}^2 (n - \gcd(n, n_i)) - 2 (n - 1) \\ {}\ge{} & n - (n/p + p + 1) + 2 = n \tfrac{p - 1}{p} - p + 1 \overset{!}{\ge} \tfrac{1}{3} n. The latter inequality is true if and only if \displaystyle n \ge \frac{3 p (p - 1)}{2 p - 3}. If we write n = p k, this translates to \displaystyle p \ge 3 \frac{k - 1}{2 k - 3}; if k \ge 2, 3 \frac{k - 1}{2 k - 3} \le 3, and if k \ge 3, 3 \frac{k - 1}{2 k - 3} \le 2. Hence, the only cases were the above argument does not work are n = p (k = 1) and n = 4 (k = 2, p = 2).

In case n = 4, we must have n_1 = 2 and n_2 \in \{ 1, 3 \}. In that case, \gcd(n, n_1) + \gcd(n, n_2) + \gcd(n, n_1 + n_2) = 2 + 1 + 1 = 4, whence & (n - \gcd(n, n_1 + n_2)) + \sum_{i=1}^2 (n - \gcd(n, n_i)) - 2 (n - 1) \\ {}={} & (4 - 1) + (4 - 2) + (4 - 1) - 2 (4 - 1) = 2 \ge \tfrac{1}{3} \cdot 4. In case n = p, since we do by assumption p \nmid n_1 + n_2, we obtain \gcd(n, n_1) + \gcd(n, n_2) + \gcd(n, n_1 + n_2) = 1 + 1 + 1 = 3, whence \displaystyle (n - \gcd(n, n_1 + n_2)) + \sum_{i=1}^2 (n - \gcd(n, n_i)) - 2 (n - 1) = n - 1 \ge \tfrac{1}{3} n since n \ge 3/2.

The cases \gcd(n, n_1) = n / p^2 and \gcd(n, n_1) = n / (p q) are proven analogously, with a few more case distinctions.

So we are left with the case s = 3. Here, one can proceed in a similar, painful way. Or one increases the constant to \frac{1}{12}, since we know that not all \gcd(n, n_i)‘s can be n/2, whence one is at least \le n/3. Hence, & (n - \gcd(n, n_1 + n_2)) + \sum_{i=1}^3 (n - \gcd(n, n_i)) - 2 (n - 1) \\ {}\ge{} & 4 n - 3 \cdot \tfrac{1}{2} n - \tfrac{1}{3} n - 2 n + 2 = \tfrac{1}{6} n + 2, which yields the claim.

To sum everything up, we showed the following theorem:

Theorem.

Let n > 1 be an integer, s \ge 1, and n_1, \dots, n_s \in \{ 1, \dots, n - 1 \} satisfy \gcd(n_1, \dots, n_s, n) = 1. Then g(n_1, \dots, n_s), defined as \displaystyle \frac{\biggl(n - \gcd\biggl(n, \sum\limits_{i=1}^s n_i\biggr)\biggr) + \sum\limits_{i=1}^s (n - \gcd(n, n_i)) - 2 (n - 1)}{2}, satisfies g(n_1, \dots, n_s) \ge 0, and if it is strictly larger than zero, \displaystyle g(n_1, \dots, n_s) \ge \frac{1}{24} n.
□

As mentioned, if one invests more work, one can actually show g(n_1, \dots, n_s) \ge \frac{1}{6} n. For my preprint though, this is not worth the trouble.

]]> http://math.fontein.de/2010/12/09/a-strange-inequality/feed/ 0 Multiplicity of the Determinant. http://math.fontein.de/2010/11/10/multiplicity-of-the-determinant/ http://math.fontein.de/2010/11/10/multiplicity-of-the-determinant/#comments Wed, 10 Nov 2010 00:55:15 +0000 Felix Fontein http://math.fontein.de/?p=790 I just learned about a nice trick to show that \det(A B) = \det A \cdot \det B for matrices A, B \in K^{n \times n} from my colleague Francesco Sica, who attributed it to F. Catanese.

Assuming that it is known that there is, up to scale, only one alternating n-linear form K^{n \times n} \to K (i.e. that \dim_K \bigwedge^n K^n = 1), one can proceed as follows. Given A \in K^{n \times n}, consider the map f_A : K^{n \times n} \to K, B \mapsto \det(A B). This is clearly n-linear and alternating, whence there exists some \lambda \in K such that f_A = \lambda \cdot \det. Evaluating f_A at the identity matrix I gives \lambda = \lambda \det(I) = f_A(I) = \det(A I) = \det A. Evaluating f_A at B gives \det(A B) = f_A(B) = \lambda \det B = \det A \cdot \det B.

Of course, using the trick similar to the first lemma here, it suffices to show this for K = \C to obtain it for any unitary commutative ring, after showing that the determinant is in fact a polynomial equation with integer coefficients (for example, by showing the Leibniz formula).

]]> http://math.fontein.de/2010/11/10/multiplicity-of-the-determinant/feed/ 2 Rigorous Arithmetic in the Arakelov Divisor Class Group of a Number Field. http://math.fontein.de/2010/07/27/rigorous-arithmetic-in-the-arakelov-divisor-class-group-of-a-number-field/ http://math.fontein.de/2010/07/27/rigorous-arithmetic-in-the-arakelov-divisor-class-group-of-a-number-field/#comments Tue, 27 Jul 2010 09:50:37 +0000 Felix Fontein http://math.fontein.de/?p=778 This year at the IX. Algorithmic Number Theory Symphosium, held in Nancy, I had a poster in the poster session. You can see it here (click to see a larger version):

You can also get a PDF version here (9.1 MB).
The poster discusses how to effectively compute in the Arakelov divisor class group \Pic^0(K) of a number field K, which is assumed to be totally real in the current implementation described in the poster, but the same method works as long as there is at least one real embedding of K. In case K is totally imaginary, the only thing which gets more complicated is doing comparisms. The arithmetic uses f-representations as the main tool, i.e. it allows to compute in the infrastructure of K.

]]> http://math.fontein.de/2010/07/27/rigorous-arithmetic-in-the-arakelov-divisor-class-group-of-a-number-field/feed/ 1 Inequalities. http://math.fontein.de/2010/02/09/inequalities/ http://math.fontein.de/2010/02/09/inequalities/#comments Tue, 09 Feb 2010 06:29:04 +0000 Felix Fontein http://math.fontein.de/?p=732 Inspired by a tweet, I decided to play a bit with Graphviz to create a graph on which inequalities imply which ones. Here’s the result:

Click here to obtain a larger version. Click any box on the image to obtain the Wikipedia page on the inequality.

]]> http://math.fontein.de/2010/02/09/inequalities/feed/ 3 How to Compute the 5-adic Expansion of 1/2; or: Hensel’s Lemma and (Non-Analytic) Newton Iteration. http://math.fontein.de/2010/02/06/how-to-compute-the-5-adic-expansion-of-12-or-hensels-lemma-and-non-analytic-newton-iteration/ http://math.fontein.de/2010/02/06/how-to-compute-the-5-adic-expansion-of-12-or-hensels-lemma-and-non-analytic-newton-iteration/#comments Sat, 06 Feb 2010 22:03:24 +0000 Felix Fontein http://math.fontein.de/?p=690 Today, I want to discuss three topics: p-adic integers and numbers, Hensel’s Lemma as well as Newton iteration. Three topics which, on the first glance, seem to have nothing in common. But nonetheless, there is a tight relation between them.

The p-adic numbers.

Let us begin with the p-adic numbers. They are a non-achrimedean analogue to the real numbers, whence we want to discuss the real numbers first. There are many different constructions of the real numbers. We pick a certain one, namely construction as a completion of the rational numbers. For that, consider the archimedean distance d_\infty(x, y) = |x - y| respectively the archimedean absolute value \displaystyle |x| = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0. \end{cases} Recall that a sequence (a_n)_{n\in\N} is said to be convergent with limit a \in \Q if for every \varepsilon > 0, there exists some n_0 \in \N with d_\infty(a_n, a) < \varepsilon for all n \ge n_0. Note that every convergent series is a Cauchy sequence, i.e. it satisfies \displaystyle \forall \varepsilon > 0 \; \exists N \in \N \; \forall n, m \ge N : d_\infty(a_n, a_m) < \varepsilon. But not every Cauchy sequence in \Q converges. One reason to use the real numbers is to add limits of Cauchy sequences, so that every Cauchy sequence (with coefficients in \Q) converges. More precisely, consider the set of Cauchy sequences, C; this is an \Q-subspace of \Q^\N, the space of all functions \N \to \Q (i.e. all sequences). Consider the subspace c of sequences converging to 0 \in \Q; note that c \subseteq C. Therefore, we can consider the quotient \R := C / c; \Q embeds via the diagonal embedding, i.e. q \in \Q maps to (n \mapsto q) + c \in C / c = \R. One quickly checks that \R is a ring, and that every non-zero element is in fact invertible, i.e. it is a field. Moreover, one quickly checks that the canonical order \le on \Q extends to \R; this allows to define d_\infty(x, y) and \abs{x} for x, y \in \R in the same manner as for rational numbers. Moreover, one sees that all Cauchy sequences in \R actually converge.

The p-adic numbers can be constructed in a very similar way. Fix a prime number p, and condider the p-adic valuation on \Q, defined by \displaystyle \nu_p : \Q^* \to \Z, \qquad p^t \frac{a}{b} \mapsto t if a, b \in \Z are not divisible by p. Moreover, set \nu_p(0) := \infty. Then, we can define the p-adic absolute value \abs{\bullet}_p : \Q \to \Q, z \mapsto p^{-\nu_p(z)}; then \abs{z}_p = 0 if, and only if, z = 0; we have that the strict triangle inequality \displaystyle \abs{x + y}_p \le \max\{ \abs{x}_p, \abs{y}_p \} \le \abs{x}_p + \abs{y}_p is satisfied; and we have that \abs{x y}_p = \abs{x}_p \abs{y}_p; here, x,y, z \in \Q are arbitrary. Such absolute values which satisfy the strict triangle inequality \abs{x + y}_p \le \max\{ \abs{x}_p, \abs{y}_p \} are called non-archimedean absolute values.

Remark.
In fact, one can show that these p-adic absolute values together with the above Archimedean absolute value are all absolute values on \Q up to equivalence; here, we say that two absolute values \abs{\bullet}_i, \abs{\bullet}_j are equivalent if there exists some number t > 0 with \abs{z}_i = \abs{z}_j^t for all z \in \Q.

Now let us continue with the p-adic numbers. We can define Cauchy sequences and the notion of convergence as above, by replacing d_\infty(x, y) by d_p(x, y) := \abs{x - y}_p. As above, we obtain that \Q has a completion with respect to d_p which forms a field, denoted by \Q_p. This is the field of p-adic numbers. Also, d_p and \abs{\bullet}_p extend onto \Q_p. As opposed to the case of the real numbers, the image of d_p resp. \abs{\bullet}_p do not change; the reason is that \nu_p is a discrete valuation, i.e. attains only integers. Actually, this field \Q_p has several interesting properties, which we want to collect.

Theorem.
  1. For every x, y \in \Q_p, \abs{x + y}_p \le \max\{ \abs{x}_p, \abs{y}_p \} \le \abs{x}_p + \abs{y}_p.
  2. The set \{ x \in \Q_p \mid \abs{x}_p \le 1 \} is a subring of \Q_p; denote this subring by \Z_p.
  3. For any B \in \R_{>0}, we have that \{ x \in \Q_p \mid \abs{x} \le B \} and \{ x \in \Q_p \mid \abs{x} < B \} are both open and closed in \Q_p. In particular, \Z_p is both open and closed.
  4. We have that \displaystyle \Z_p = \{ z \mid \exists (z_n)_n \text{ sequence in } \Z : \lim z_n = z \}.
  5. The ring \Z_p is local with maximal ideal \frakm_p := \{ x \in \Q_p \mid \abs{x}_p < 1 \}, and \Z_p / \frakm_p \cong \Z/p\Z. In fact, \Z_p is a discrete valuation ring.
  6. The series \sum_{n=0}^\infty a_n with a_n \in \Q_p converges if, and only if, \lim_{n\to\infty} a_n = 0.
  7. Every non-zero element z \in \Q_p^* can be written uniquely in the form z = \sum_{n=k}^\infty a_n p^n, where a_n \in \{ 0, \dots, p - 1 \} for all n, k = \nu_p(z) and a_k \neq 0.
  8. For every n \in \N, \Z_p / \frakm_p^n \cong \Z/p^n\Z.
  9. If \Q_p is equipped with the topology induced by the metric d_p, then \Z_p is the maximal compact subring of \Q_p.
Proof.
  1. Let (x_n)_n, (y_n)_n sequences in \Z with \lim x_n = x, \lim y_n = y. Then, for every n, \max\{ \abs{x_n}_p, \abs{y_n}_p \} - \abs{x_n + y_n}_p \ge 0. Now \Q_p is a topological ring, and \abs{\bullet}_p is continuous, whence the result follows from applying \lim_{n\to\infty}.
  2. From (a), we see that this set is closed under addition. That it is closed under multiplication is clear, and 0, 1 are contained in it as well.
  3. Write B = p^{-t + \varepsilon} with t \in \Z and 0 \le \varepsilon < 1. Now \abs{x}_p \le B \Leftrightarrow \abs{x}_p \le p^{-t} \Leftrightarrow \abs{x}_p < p^{-t + 1}, and \displaystyle \abs{x}_p < B \Longleftrightarrow \begin{cases} \abs{x}_p \le B & \text{if } \varepsilon > 0, \\ \abs{x} \le p^{-t - 1} & \text{if } \varepsilon = 0. \end{cases}
  4. Let (z_n)_n be a sequence of elements in \Z which converges with respect to d_p. Then \abs{z_n}_p \le 1 for all n \in \N as \nu_p(z_n) \ge 0; as \abs{\bullet}_p is continuous, \abs{\lim z_n}_p = \lim \abs{z_n}_p \le 1.
    Conversely, let z \in \Z_p and let (z_n)_n be a sequence of elements in \Q with \lim z_n = z. Now \Z_p is open; therefore, there exists some n_0 with \abs{z_n}_p \le 1 for all n \ge n_0. Without loss of generality, we can assume that n_0 = 0, i.e. all z_n lie in \Z_p \cap \Q. Moreover, we can assume that z_n \neq 0 for all n. We want to construct a sequence (a_n)_n in \Z with \lim a_n = \lim b_n. For that, write z_n = \frac{x_n}{y_n} with x_n, y_n \in \Z, where y_n is not divisible by p. Let \tilde{y}_n \in \{ 0, \dots, p^n - 1 \} be such that y_n \tilde{y}_n \equiv 1 \pmod{p^n} and set b_n := x_n \tilde{y}_n; moreover, write y_n \tilde{y}_n = 1 + c_n p^n with c_n \in \Z. Then \displaystyle b_n - a_n = b_n (1 - \tilde{y}_n y_n) = -b_n c_n p^n, whence Formula does not parse: \bas{b_n - a_n} \le p^{-n}. Therefore, one obtains that \lim a_n = \lim b_n = z.
  5. One quickly checks using 1. that \frakm_p is closed under addition. It is clearly also closed under multiplication by elements of \Z_p, and contains 0; therefore, it is an ideal. Now consider the map \Z \to \Z_p / \frakm_p; clearly, p \in \frakm_p, whence p \Z is contained in the kernel of this map. But 1 is not contained in the kernel, as 1 \not\in \frakm_p; therefore, \Z_p / \frakm_p contains a copy of \Z/p\Z. To see that this is everything, let x + \frakm_p \in \Z_p / \frakm_p; let (x_n)_n be a sequence in \Z with \lim x_n = x. Let n_0 be such that for all n \ge n_0, d_p(x_n, x) < 1; then x_n - x \in \frakm_p, whence x + \frakm_p = x_n + \frakm_p. But x_n + \frakm_p is contained in the copy of \Z/p\Z.
  6. Clearly, if the series converges, we must have that \lim a_n = 0.
    Now, conversely, assume that \lim a_n = 0. We show that (b_n)_n with b_n = \sum_{k=0}^n a_k is a Cauchy series. For that, let \varepsilon > 0 be given. Choose N such that for all n \ge N, \abs{a_n}_p < \varepsilon. Now, if n, m \ge N with n \ge m, we have that \displaystyle \abs{b_n - b_m}_p = \abs{\sum_{k=m}^n a_k}_p \le \max\{ \abs{a_m}_p, \dots, \abs{a_n}_p \} < \varepsilon by 1., what we had to show.
  7. First, for any choice of the k‘s and a_n‘s, we obtain an element z = \sum_{n=k}^\infty a_n p^n in \Q_p. Now assume that \sum_{n=k}^\infty a_n p^n = \sum_{n=k}^\infty b_n p^n with a_n, b_n \in \{ 0, \dots, p - 1 \}. Assume that there exists some t with a_t \neq b_t, and further assume that t is chosen to be minimal under this condition, i.e. a_n = b_n for all n < t. Then 0 = \sum_{n=t}^\infty (a_n - b_n) p^n. Multiplying with p^{-t} gives 0 = z := \sum_{n=0}^\infty (a_{n+t} - b_{n+t}) p^n with a_t - b_t \in \{ -p + 1, \dots, -2, -1, 1, 2, \dots, p - 1 \}. Moreover, z \in \Z_p, and z + \frakm_p = (a_t - b_t) + \frakm_p. But now a_t - b_t \not\in \frakm_p, whence z + \frakm_p \neq 0 + \frakm_p. But by construction z = 0, a contradiction. Therefore, the representations \sum_{n=k}^\infty a_n p^n are unique.
    We have to show that every element in \Q_p can be written in this way. For that, let z \in \Q_p^*; now z' := z p^{\nu_p(z)} satisfies \abs{z'}_p = 1; in particular, z' \in \Z_p \setminus \{ 0 \}. We have to show that we can write z' = \sum_{n=0}^\infty a_n p^n with a_0 \neq 0. For that, let (z_n)_n be a sequence in \Z with \lim z_n = z; without loss of generality, we can assume that \abs{z - z_n}_p \le p^{-n-1} for every n. Then we also have \abs{z_n - z_m}_p \le p^{-n-1} for every m \ge n, whence z_n \equiv z_m \pmod{p^{n+1}}. Therefore, we can choose a_n \in \{ 0, \dots, p - 1 \} inductively such that \sum_{t=0}^n a_t p^t \equiv z_n \pmod{p^{n+1}}, and we obtain that z = \lim z_n = \lim \sum_{t=0}^n a_t p^t = \sum_{n=0}^\infty a_n p^n. Finally, since 0 = \nu_p(\sum_{n=0}^\infty a_n p^n) = \min\{ n \mid a_n \neq 0 \} (which follows from the strict triangle inequality), it follows that a_0 \neq 0.
  8. Clearly, \frakm_p^n = \{ x \in \Q_p \mid \nu_p(x) \ge n \}; therefore, using 7., we see that every residue class in \Z_p / \frakm_p^n is uniquely described by \sum_{t=0}^{n-1} a_n p^n + \frakm_p, where a_n \in \{ 0, \dots, p - 1 \}. Hence, \abs{\Z_p / \frakm_p^n} = p^n. Now, as in the proof of 5., we see that \Z/p^n\Z injects into \Z_p / \frakm_p^n, whence this injection is in fact a bijection. Thus \Z_p / \frakm_p^n \cong \Z/p^n\Z.
  9. Let R be any compact subring of \Q_p and x \in R. If \abs{x}_p > 1, then \abs{x^n}_p \to \infty for n \to \infty. Hence, R is unbounded, a contradiction.
    Hence, it is left to show that \Z_p is compact. For that, it suffices to show that any sequence in \Z_p has at least one accumulation point. Let (x_n)_n be any sequence in \Z_p. We claim that for any n, there exists some z_n \in \Z such that z_n + \frakm_p^n contains infinitely many elements of the sequence. For n = 0 this is clear for any choice of z_n; hence, assume that this is the case for some n. Now z_n + \frakm_p^n = \bigcup_{i=0}^{p-1} (z_n + i p^n) + \frakm_p^{n+1}; now there must be some i \in \{ 0, \dots, p - 1 \} such that infinitely many of the x_n‘s lie in (z_n + i p^n) + \frakm_p^{n+1} (otherwise, only finitely many can lie in z_n + \frakm_p^n); set z_{n+1} := z_n + i p^n. We see that (z_n)_n is a Cauchy sequence, whence z = \lim z_n \in \Z_p exists. Now by construction, z is an accumulation point of (x_n)_n. Therefore, \Z_p is compact.
□

Before we continue, we want to explore another construction of \Z_p which is completely algebraic. For that, we need the projective limit in the category of rings:

Definition.
Let (I, \le) be a directed set and for every i \in I, let R_i be a ring. Assume that for every i, j \in I with i \le j there exists a homomorphism \phi_{ij} : R_j \to R_i such that for all i, j, k with i \le j \le k, we have \phi_{ik} = \phi_{ij} \circ \phi_{jk}, and we have \phi_{ii} = \id_{R_i}. Such a tuple ((I, \le), (R_i)_i, (\phi_{ij})_{ij}) is called a projective system. \displaystyle \xymatrix{ R_k \ar[dd]_{\phi_{ik}} \ar[dr]^{\phi_{jk}} & \\ & R_j \ar[dl]^{\phi_{ij}} \\ R_i & } A projective limit of ((I, \le), (R_i)_i, (\phi_{ij})_{ij}) is a ring R together with homomorphisms \pi_i : R \to R_i such that \pi_j = \pi_i \circ \phi_{ij} if i \le j, which satisfies the following universal property:
If R' is any other ring and \pi'_i : R \to R_i any other family of ring homomorphisms with \pi_j = \pi_i \circ \phi_{ij} whenever i \le j, there exists a unique homomorphism \psi : R' \to R with \pi_i \circ \psi = \pi'_i for all i \in I. \displaystyle \xymatrix{ R' \ar@{->}[r]^{\exists! \psi} \ar[dr]_{\pi'_i} & R \ar[d]^{\pi_i} \\ & R_i }

We have the following, classical result:

Theorem.
For every projective system ((I, \le), (R_i)_i, (\phi_{ij})_{ij}) of rings, there exists a projective limit which is unique up to unique isomorphism; i.e., for any two projective limits (R, (\pi_i)_i) and (R', (\pi'_i)_i) there exists a unique isomorphism \psi : R' \to R such that \pi_i \circ \psi = \pi'_i for all i.
Moreover, a projective limit can be constructed as \displaystyle R := \biggl\{ (r_i)_i \in \prod_{i \in I} R_i \;\biggm|\; \forall (i, j) \in {\le} : \psi_{ij}(r_j) = r_i \biggr\}, where \pi_i is the restriction of the canonical projection \prod_{j \in I} R_j \to R_i to the subset R.
□
Definition.
Let ((I, \le), (R_i)_i, (\phi_{ij})_{ij}) be a projective system. Define the projective limit of it as any projective limit, and denote it by \varprojlim_{i \in I} R_i.
We choose I = \N with the usual order, and let R_n := \Z/p^n\Z. Then, if n \le m, one has the projection \phi_{nm} : \Z/p^m\Z \to \Z/p^n\Z, x + p^m\Z \mapsto x + p^n\Z. Hence, we can define \hat{\Z}_p := \varprojlim_{n\in\N} R_n. Now this definition coincides with the old one; this can be seen using \displaystyle \hat{\Z}_p = \{ (a_n)_n \mid a_n \in \{ 0, \dots, p^n - 1 \}, \; a_{n+1} \equiv a_n \pmod{p^n} \}; then, the map \displaystyle \Z_p \to \hat{\Z}_p, \qquad \sum_{n=0}^\infty a_n p^n \mapsto \biggl( \sum_{t=0}^{n-1} a_t p^t \biggr)_n is obviously an isomorphism.

Hensel’s Lemma.

Hensel’s Lemma can be formulated in a very algebraic way. All rings in this section are commutative and unitary.

Definition.
Let R be a ring and \fraka an ideal of R. We say that \fraka is nilpotent if \fraka^n = 0 for some n \in \N. We say that \fraka is a nilideal if every x \in \fraka is nilpotent, i.e. if for every x \in \fraka there is some n_x \in \N with x^{n_x} = 0.

Note that in Noetherian rings, nilideals are already nilpotent, since ideals generated by finitely many nilpotent elements are always nilpotent. (Note that our rings are commutative. Otherwise it won’t work.)

Proposition (Hensel's Lemma).
Let R be a ring and \fraka a nilideal in R. Let f \in R[x] and a \in R such that f(a) \in \fraka and f'(a) + \fraka \in (R / \fraka)^*. Then there exists a unique b \in R with a - b \in \fraka and f(b) = 0.
Proof (Part I).
First, assume that both b and b' satisfy a - b, a - b' \in \fraka and f(b) = 0 = f(b'). Using t := b' - b, we see that 0 ={} & f(b') = f(b + t) = f(b) + f'(b) t + e t^2 \\ {}={} & (b' - b) ( f'(b) + e (b' - b) ) for some e \in R using Taylor expansion. Since f'(b) + e (b' - b) + \fraka = f'(b) + \fraka = f'(a) + \fraka is a unit in R / \fraka, there exist some a \in R, c \in \fraka with a (f'(b) + e (b' - b)) = 1 + c. How c is nilpotent, whence (1 + c) \sum_{n=0}^\infty (-c)^n = 1, i.e. 1 + c \in R^*. But this means that f'(b) + e (b' - b) \in R^*, whence b' - b = 0. This shows uniqueness.
Moreover, we want to show that it suffices to require that \fraka is nilpotent. Cosider the subring R' of R which is the smallest (unitary) subring containing the coefficients of f and a, and some fixed b \in R with f'(a) b - 1 \in \fraka. This ring is clearly finitely generated, whence Noetherian, and \fraka' := \fraka \cap R' is a nilideal in R' as well. Since R' is Noetherian, \fraka' is in fact nilpotent. Moreover, f'(a) + \fraka' \in (R' / \fraka')^* since f'(a) b - 1 \in \fraka'. Hence, it suffices to prove the lemma for (R', \fraka') instead of (R, \fraka).
□

We will complete the proof later. First, let us discuss some implications of this result. Consider the ring R = \Z/p^n\Z and \fraka = p^m R, where m > 0. Then clearly \fraka is nilpotent in R, whence we can apply Hensel’s lemma to this situation. Assume that f \in \Z[x] is a polynomial and a \in \Z with f(a) \equiv 0 \pmod{p^m}; if then f'(a) \not\equiv 0 \pmod{p}, there exists a unique b \in R with f(b) = 0 and b \equiv a \pmod{p^m}. Doing this construction for m = 1 and all n \in \N, we obtain a sequence (b_n)_n of elements with b_n \in \{ 0, \dots, p^n - 1 \} and b_{n+1} \equiv b_n \pmod{p^n}, i.e. we obtain an element of \hat{\Z}_p = \Z_p!

In particular, let a \in \Z be any element with p \nmid a. Consider f := a x - 1 \in \Z[x]. Now there exists some b, c \in \Z with a b + c p = 1; therefore, f(b) \equiv 0 \pmod{p}. But this implies that there exists a unique z \in \Z_p with f(z) = 0 and z \equiv b \pmod{p}. (Since b is unique modulo p, it follows that z is uniquely defined by f(z) = 0, i.e. by a z = 1.) Hence, we have shown that any integer a is invertible in \Z_p if p \nmid a. But how to compute z? We will discuss this later; for now, note that we can use the Extended Euclidean Algorithm to find b_n, c_n \in \Z with a b_n + c_n p^n = 1; then b_n \equiv z \pmod{p^n}, i.e. we can approximate z up to arbitrary precision.

What about p \mid a? In that case, \frac{1}{a} \not\in \Z_p: if a would have an inverse in \Z_p, say b \in \Z_p, then we would have 0 \equiv a \cdot b \equiv 1 \pmod{p}, a contradiction. Therefore, we get:

Proposition.
We have that \displaystyle \Z_p \cap \Q = \biggl\{ \frac{a}{b} \in \Q \;\biggm|\; a, b \in \Z, \; p \nmid b \biggr\}.
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Newton Iteration.

In the last section, we ignored two points: first, how to prove existence in Hensel’s lemma, and second, how to compute this element in a constructive way. We want to fix this now. For this, we describe how Newton’s method works in arbitrary rings!

Let f \in R[x] be a polynomial, \fraka \subseteq R a nilpotent ideal, and a \in R such that f(a) \in \fraka and f'(a) + \fraka \in (R / \fraka)^*. Fix some \hat{a} \in R with f'(a) \hat{a} - 1 \in \fraka, and consider the sequence \displaystyle x_0 := a, \qquad x_{n+1} := x_n - \hat{a} f(x_n), \quad n \in \N. We claim that f(x_n) \in \fraka^{n+1} for all n \in \N; since \fraka is nilpotent, this means that the sequence becomes stationary eventually and gives a root of f. Moreover, we claim that x_n - x_{n-1} \in \fraka^n, whence x_n - a \in \fraka for all n.

Assume that this is true for some n, i.e. we have f(x_n) \in \fraka^{n+1}. Then x_{n+1} = x_n - \hat{a} f(x_n). Now f(x_n) \in \fraka^{n+1}, whence x_{n+1} - x_n \in \fraka^{n+1}. To show that f(x_{n+1}) \in \fraka^{n+2}, we again use the Taylor expansion; by it, f(x_{n+1}) ={} & f(x_n - \hat{a} f(x_n)) \\ {}={} & f(x_n) - f'(x_n) (\hat{a} f(x_n)) + e \hat{a}^2 f(x_n)^2 \\ {}={} & f(x_n) (1 - f'(x_n) \hat{a}) + e \hat{a}^2 f(x_n)^2 for some e \in R. Now 1 - f'(x_n) \hat{a} \in \fraka, whence f(x_n) (1 - f'(x_n) \hat{a}) \in \fraka^{n+2}. Moreover, f(x_n)^2 \in (\fraka^{n+1})^2 \subseteq \fraka^{n + 2}. Combining this gives f(x_{n+1}) \in \fraka^{n+2}.

Therefore, the proof of Hensel’s lemma is completed. Moreover, we obtained an algorithm to refine an approximation of \frac{1}{a} \in \Z_p without using the Extended Euclidean Algorithm: as soon as \hat{a} \in \Z with a \hat{a} \equiv 1 \pmod{p} is known, we can compute \frac{1}{a} modulo \frakm_p^n by applying the Newton iteration x \mapsto x - \hat{a} f(x) = x - \hat{a} (a x - 1) = x (1 - \hat{a} a) + \hat{a} a only n - 1 times. Moreover, we can start with some intermediate result, say \frac{1}{a} modulo p^m, to compute \frac{1}{a} modulo p^n (with n > m) in at most n - m iterations (which each need one multiplication and one addition modulo p^n). This is considerably faster than applying the Extended Euclidean Algorithm for a and p^n.

So, What About \frac{a}{b} in \Z_p?

Assume that p \nmid b; then we know that \frac{a}{b} \in \Z_p. Consider the polynomial f := b x - a \in \Z[x]; clearly, f(\frac{a}{b}) = 0 in \Z_p, and f'(\frac{a}{b}) = b is a unit in \Z_p. Therefore, we can use the methods from above to compute \frac{a}{b} \in \Z_p.

First, we need an approximation of \frac{a}{b} modulo p. For that, use the Extended Euclidean Algorithm to compute c, c' \in \Z with b c + p c' = 1; then a c satisfies (a c) b \equiv a \pmod{p}, i.e. \frac{a}{b} + \frakm_p = a c + \frakm_p.

Set a_0 := a c \mod p and \displaystyle a_{n+1} := a_n - c (b a_n - a) \mod p^{n+2} = a_n (1 - c b) + c a \mod p^{n+2}, n \ge 0. Then, by the above, a_n + \frakm_p^{n+1} = \frac{a}{b} + \frakm_p^{n+1}. Hence, this allows to approximate \frac{a}{b} up to an error of \frakm_p^n in n - 1 iterations; we only need to perform the Extended Euclidean Algorithm once (to get a starting value), and from that, we can refine the approximation by applying a linear map.

As an example, let us consider \frac{1}{2} in \Z_5, i.e. a = 1, b = 2 and p = 5. Clearly, 3 \cdot 2 - 1 \cdot 5 = 1, whence we get c = 3. Hence, a_0 = 3. Now 1 - c b = -5 and c a = 3, whence a_{n+1} = 3 - 5 a_n \mod 5^{n+2}, n \in \N. We rapidly get a_0 ={} & 3, \\ a_1 ={} & 13 = 2 \cdot 5 + 3, \\ a_2 ={} & 63 = 2 \cdot 5^2 + 2 \cdot 5 + 3, \\ a_3 ={} & 313 = 2 \cdot 5^3 + 2 \cdot 5^2 + 2 \cdot 5 + 3, \\ a_4 ={} & 1563 = 3 + 2 \cdot \sum_{n=1}^4 5^n, \\ a_5 ={} & 7813 = 3 + 2 \cdot \sum_{n=1}^5 5^n, \\ a_6 ={} & 39063 = 3 + 2 \cdot \sum_{n=1}^6 5^n, \\ a_7 ={} & 195313 = 3 + 2 \cdot \sum_{n=1}^7 5^n, \\ a_8 ={} & 976563 = 3 + 2 \cdot \sum_{n=1}^8 5^n, \\ a_9 ={} & 4882813 = 3 + 2 \cdot \sum_{n=1}^9 5^n, \\ a_{10} ={} & 24414063 = 3 + 2 \cdot \sum_{n=1}^{10} 5^n, \\ \vdots\;\; & Hence, it seems that \displaystyle a_n = 3 + 2 \cdot \sum_{i=1}^n 5^i = 3 + 2 \cdot 5 \cdot \sum_{i=0}^{n-1} 5^i = 3 + \tfrac{5}{2} (5^n - 1), i.e. we have \displaystyle \frac{1}{2} = 3 + \sum_{n=1}^\infty 2 \cdot 5^n \in \Z_5. And indeed: \displaystyle 2 \cdot \biggl( 3 + \sum_{n=1}^\infty 2 \cdot 5^n \biggr) = 2 + 4 \cdot \sum_{n=0}^\infty 5^n = 1 since \displaystyle -1 = (p - 1) \sum_{n=0}^\infty p^n \in \Z_p. Note that -1 = (p - 1) \sum_{n=0}^\infty p^n follows from the fact that \nu_p(p) < 0, whence \sum_{n=0}^\infty p^n converges in \Z_p (see the theorem); this is a geometric series, whence its value is \frac{1}{1 - p}. Hence, if we multiply by p - 1, we obtain -1.

Finally, let us consider another example, namely \frac{432}{1234} in \Z_{17}, i.e. a = 432, b = 1234, p = 17; then (-5) b + 363 p = 1, i.e. c = -5. Hence, we obtain a_0 := a c \mod p = 16 and \displaystyle a_{n+1} = (1 - c b) a_n + c a \mod 17^{n+2} = 6171 a_n - 2160 \mod 17^{n+2}, n \in \N. We then obtain a_0 ={} & 16, \\ a_1 ={} & 50 = 16 + 2 \cdot 17, \\ a_2 ={} & 1784 = 16 + 2 \cdot 17 + 6 \cdot 17^2, \\ a_3 ={} & 65653 = 16 + 2 \cdot 17 + 6 \cdot 17^2 + 13 \cdot 17^3, \\ a_4 ={} & 483258 = 16 + 2 \cdot 17 + 6 \cdot 17^2 + 13 \cdot 17^3 + 5 \cdot 17^4, \\ a_5 ={} & 13261971 = 16 + 2 \cdot 17 + 6 \cdot 17^2 + 13 \cdot 17^3 + 5 \cdot 17^4 \\ {}+{} & 9 \cdot 17^5, \\ a_6 ={} & 182224954 = 16 + 2 \cdot 17 + 6 \cdot 17^2 + 13 \cdot 17^3 + 5 \cdot 17^4 \\ {}+{} & 9 \cdot 17^5 + 7 \cdot 17^6, \\ a_7 ={} & 1413240973 = 16 + 2 \cdot 17 + 6 \cdot 17^2 + 13 \cdot 17^3 + 5 \cdot 17^4 \\ {}+{} & 9 \cdot 17^5 + 7 \cdot 17^6 + 3 \cdot 17^7, \\ a_8 ={} & 64195057942 = 16 + 2 \cdot 17 + 6 \cdot 17^2 + 13 \cdot 17^3 + 5 \cdot 17^4 \\ {}+{} & 9 \cdot 17^5 + 7 \cdot 17^6 + 3 \cdot 17^7 + 9 \cdot 17^8, \\ a_9 ={} & 1012898069918 = 16 + 2 \cdot 17 + 6 \cdot 17^2 + 13 \cdot 17^3 + 5 \cdot 17^4 \\ {}+{} & 9 \cdot 17^5 + 7 \cdot 17^6 + 3 \cdot 17^7 + 9 \cdot 17^8 + 8 \cdot 17^9, \\ a_{10} ={} & 13108861472612 = 16 + 2 \cdot 17 + 6 \cdot 17^2 + 13 \cdot 17^3 + 5 \cdot 17^4 \\ {}+{} & 9 \cdot 17^5 + 7 \cdot 17^6 + 3 \cdot 17^7 + 9 \cdot 17^8 + 8 \cdot 17^9 + 6 \cdot 17^{10}, \\ \vdots\;\; & This can be continued a long time, without seeing any pattern.

One would expect that the sequence of digits evenutally gets (eventually) periodic, as it happens with the decimal expansion of rational numbers. For that, assume that we know \frac{a}{b} = p^t \frac{a'}{b'}, where a', b' are coprime and not divisible by p, and b' > 0. Now let \ell be the order of p in the multiplicative group \Z/b'\Z; hence, it is the smallest non-negative rational number with b' \mid (p^\ell - 1). Then \frac{a}{b} = p^t a' \frac{p^\ell - 1}{b'} \frac{1}{p^\ell - 1}. Now, \displaystyle \frac{1}{p^\ell - 1} = -\sum_{n=0}^\infty p^{\ell n} by the geometric series (as above), whence \displaystyle \frac{a}{b} = a' \frac{p^\ell - 1}{b'} \cdot \sum_{n=0}^\infty p^{\ell n + t}. Moreover, write \frac{a'}{b'} = \frac{a''}{b'} + a''' with a''' \in \Z such that 0 \le a'' < b'. Then a' \frac{p^\ell - 1}{b'} = \frac{a'' (p^\ell - 1)}{b'} + a''' (p^\ell - 1), and 0 \le \frac{a'' (p^\ell - 1)}{b'} < p^\ell. Set x := \frac{a'' (p^\ell - 1)}{b'}; then, if we write x = \sum_{i=0}^{\ell - 1} x_i p^i, we see that \displaystyle \frac{a}{b} = a''' + \sum_{n=0}^\infty \sum_{i=0}^{\ell - 1} a_i p^{\ell n + i + t} = a''' + \sum_{n=0}^\infty a_{\ell \mod n} p^{\ell + t}. We are left to consider the a''' part. In case a''' > 0, the p-adic expansion of a''' has finite length, whence adding it does not change the periodicity of \frac{a}{b}. But what if a''' < 0?

Lemma.
Let x = \sum_{n=0}^\infty a_n p^n \in \Z_p, where 0 \le a_n < p. Then \displaystyle -x = 1 + \sum_{n=0}^\infty (p - 1 - a_n) p^n.
Proof.
Clearly, \displaystyle \sum_{n=0}^\infty a_n p^n + \sum_{n=0}^\infty (p - 1 - a_n) p^n = (p - 1) \sum_{n=0}^\infty p^n = -1 \in \Z_p, whence adding 1 results in the statement.
□

Therefore, the p-adic expansion of a''' is periodic if a''' < 0, with almost all p-adic digits being p - 1. Now, we can conclude with the fact that the sum of two periodic p-adic expansions is periodic.

Conversely, assume that x = x' + \sum_{n=0}^\infty a_{n \mod m} p^n \in \Z_p with x' \in \Z and a_0, \dots, a_{m-1} \in \{ 0, \dots, p - 1 \}; we claim that x \in \Q. Clearly, without loss of generality, we can assume that x' = 0. But then, if we set x'' := \sum_{i=0}^{m-1} a_i p^i \in \Z, \displaystyle x = \sum_{n=0}^\infty \sum_{i=0}^{m-1} a_i p^{m n + i} = x'' \cdot \sum_{n=0}^\infty p^{m n} = \frac{x''}{1 - p^m} \in \Q. Hence, we proved:

Proposition.
An element x = \sum_{n=t}^\infty a_n p^n \in \Q_p with 0 \le a_n < p lies in \Q if, and only if, there exists some m, m' \in \N such that for all i \in \N, a_{m' + i} = a_{m' + i + m}.
□

Finally, note that the order of 17 in \Z/1234\Z is \phi(1234) = 616; hence, the period length of \frac{432}{1234} is probably 616. We would have had to compute a high amount of p-adic digits of \frac{432}{1234} to see this.

]]> http://math.fontein.de/2010/02/06/how-to-compute-the-5-adic-expansion-of-12-or-hensels-lemma-and-non-analytic-newton-iteration/feed/ 2 Finding Lattice Points, Finite Abelian Groups, and Explaining Algorithms. http://math.fontein.de/2010/01/29/finding-lattice-points-finite-abelian-groups-and-explaining-algorithms/ http://math.fontein.de/2010/01/29/finding-lattice-points-finite-abelian-groups-and-explaining-algorithms/#comments Fri, 29 Jan 2010 10:20:34 +0000 Felix Fontein http://math.fontein.de/?p=627 In this article, I want to discuss three questions, which turn out to be closely related. The first question is,

“Given a lattice \Lambda \subseteq \R^n. How do I find a basis of this lattice?”

(Note that this question is far from being well-posed.) The second question is,

“If G is a finite abelian group and g_1, \dots, g_n \in G, how do I compute the structure of \langle g_1, \dots, g_n \rangle, the subgroup generated by these elements?”

The third question comes up in the description of algorithms, for example of the baby-step giant-step algorithm by D. Shanks, an optimization by D. Terr, and more general algorithms, like the Buchmann-Schmidt algorithm. These algorithms can be described in terms of the first question, making them easier to understand.

We begin with sketching the relation between lattices and finite abelian groups. If G is a finite abelian group, and g_1, \dots, g_n \in G some elements, for example, a set of generators, consider the map \displaystyle \Psi_{(g_1, \dots, g_n)} : \Z^n \to G, \qquad (\lambda_1, \dots, \lambda_n) \mapsto \sum_{i=1}^n \lambda_i g_i. This turns out to be a group homomorphism onto \langle g_1, \dots, g_n \rangle. The kernel of \Psi_{(g_1, \dots, g_n)}, which we will denote by \Lambda_{(g_1, \dots, g_n)}, is called the relation lattice. This is in fact a lattice in \R^n of volume \displaystyle \det \Lambda_{(g_1, \dots, g_n)} = \abs{\Z^n / \Lambda_{(g_1, \dots, g_n)}} = \abs{\langle g_1, \dots, g_n\rangle}; note that by the Homomorphism Theorem, \langle g_1, \dots, g_n \rangle \cong \Z^n / \Lambda_{(g_1, \dots, g_n)}. On the other hand, if \Lambda \subseteq \Z^n is a lattice, then G := \Z^n / \Lambda is a finite abelian group of \det G elements. Moreover, if the residue class of e_i, the vector consisting of zeroes except a one at the i-th position, in G is denoted by g_i, then \Lambda = \Lambda_{(g_1, \dots, g_n)}. Therefore, lattices in \Z^n and finite abelian groups with n generators are essentially the same thing.

How is this related to group structure computations? Recall the Smith normal form; this allows to convert a basis (v_1, \dots, v_n) of a lattice \Lambda into another basis, such that if one applies an invertible linear transformation to \Z^n, this basis is sent to \lambda_1 e_1, \dots, \lambda_n e_n with \lambda_i \in \N_{>0} such that \lambda_i divides \lambda_{i+1}, 1 \le i < n; then, \Z^n / \Lambda \cong \prod_{i=1}^n \Z/\lambda_i \Z. Hence, computing the structure of a finite abelian group generated by g_1, \dots, g_n can be split up in the two parts, (a) computation of a basis of the relation lattice \Lambda_{(g_1, \dots, g_n)} and (b) computation of a Smith normal form of this basis. There exist a lot of algorithms for computation of Smith normal forms; usually, the bottleneck is finding a basis of \Lambda_{(g_1, \dots, g_n)}.

Often, the process of finding a lattice equals determining the relation lattice of a finite abelian group, or something similar. One often has a way to test whether v + \Lambda = w + \Lambda, by computing a unique representation of the residue class v + \Lambda; then, one tries to find two ways v + \Lambda = w + \Lambda of writing the same residue class, but with v \neq w: then v - w is a non-trivial element of \Lambda. If one sets G := \Z^n / \Lambda, then this means that one seeks for pairs (v, g), (w, h), where g = v + \Lambda and Formula does not parse: h = w + \Lamda, such that g = h and v \neq w.

More generally, assume that X is a finite set and \pi : \Z^n \to X is a map such that \pi(x + \lambda) = \pi(x) for all \lambda \in \Lambda. Our above example, G = \Z^n / \Lambda and \pi(x) = x + \Lambda satisfies this. Moreover, assume that \pi : \Z^n / \Lambda \to X is “mostly injective”, i.e. that \pi(v) = \pi(w) implies that one can find \tilde{v}, \tilde{w} \in \Z^n such that \tilde{v} \approx v, \tilde{w} \approx w, \tilde{v} - \tilde{w} \in \Lambda with very little effort. Then one can work with elements (x, y) with y = \pi(x), and try to find two such pairs (x, y), (x', y') with y = y'; then this gives rise to an element of \Lambda near to x - x'.

Which brings us to the subject of explaining algorithms. Consider, for example, Shanks’ baby-step giant-step algorithm. You are given a group element of finite order g together with a bound B > 0. Then, for the algorithm, one computes g^0, g^1, \dots, g^{B-1}, as well as g^B, g^{2 B}, g^{3 B}, \dots, and compares this elements with the first B elements. Any match will result in a multiple of the order of g. But why is this the case? One can of course try to prove this; it is actually not very hard, in fact one just needs Fermat’s Little Theorem as well as long division. But one can do better, by visualizing the algorithm in a way which makes the solution looking obvious, and which allows even people who have no understanding of formal mathematics or computer science to immediately realize that the algorithm produces the correct result.

Namely, you might have guessed it, the order of g generates the relation lattice \Lambda_{(g)}. Hence finding the order is equivalent to finding a the smallest positive element in \Lambda_{(g)} \subseteq \Z. For this correspondence, we use the pairs (n, g^n) with n \in \Z as above. Two pairs (n, g^n), (m, g^m) with g^n = g^m gives a multiple n - m of the order of g. Now, the algorithm can be interpreted as translating the set of elements X_B := \{ -B+1, -B+2, \dots, -2, -1, 0 \} by B, 2 B, 3 B, and checking if a lattice element is contained in any of these translates. If one visualizes this, one immediately sees that this method will eventually find the smallest non-zero element of the lattice. First, this depicts the lattice \Z (gray dots) with its sublattice \Lambda_{(g)} (black dots), with the set X_B = \{ -B+1, \dots, 0 \} drawn in for B = 8:
\fbox{\begin{tikzpicture}[scale=0.3, node distance=0mm] \tikzstyle{gelt} = [draw, shape = circle, fill=black, inner sep=0pt, minimum size = 0.2cm]; \tikzstyle{empt} = [draw, shape = circle, inner sep=0pt, minimum size = 0.2cm]; \filldraw[black!67, fill=black!20] (-7.4,-0.4) to (-7.4,0.4) to (0.4,0.4) to (0.4,-0.4) to (-7.4,-0.4); \draw (-5,0) to (-5,-0.75); \node[empt] (gm5) at (-5,0) [label=below: \footnotesize \( g^{-5} \)] {}; \draw (0,0) to (0,-0.75); \node[gelt] (g0) at (0,0) [label=below: \footnotesize \( g^0 \)] {}; \draw (5,0) to (5,-0.75); \node[empt] (g5) at (5,0) [label=below: \footnotesize \( g^5 \)] {}; \draw (10,0) to (10,-0.75); \node[empt] (g10) at (10,0) [label=below: \footnotesize \( g^{10} \)] {}; \draw (23,0) to (23,-0.75); \node[gelt] (g23) at (23,0) [label=below: \footnotesize \( g^{23} \)] {}; \foreach \i in {-9,-8,-7,-6,-4,-3,-2,-1} \node[empt] at (\i,0) {}; \foreach \i in {1,2,3,4,6,7,8,9,11,12,13,14,15,16,17,18,19,20,21,22} \node[empt] at (\i,0) {}; \foreach \i in {24,25,26,27,28,29} \node[empt] at (\i,0) {}; \end{tikzpicture}}
The next figure depicts the translates of X_B by B, 2 B, 3 B, \dots:
\fbox{\begin{tikzpicture}[scale=0.3, node distance=0mm] \tikzstyle{gelt} = [draw, shape = circle, fill=black, inner sep=0pt, minimum size = 0.2cm]; \tikzstyle{empt} = [draw, shape = circle, inner sep=0pt, minimum size = 0.2cm]; \filldraw[black!67, fill=black!20] (0.6,-0.4) to (0.6,0.4) to (8.4,0.4) to (8.4,-0.4) to (0.6,-0.4); \filldraw[black!67, fill=black!20] (8.6,-0.4) to (8.6,0.4) to (16.4,0.4) to (16.4,-0.4) to (8.6,-0.4); \filldraw[black!67, fill=black!20] (16.6,-0.4) to (16.6,0.4) to (24.4,0.4) to (24.4,-0.4) to (16.6,-0.4); \filldraw[black!67, fill=black!20] (29.4,-0.4) to (24.6,-0.4) to (24.6,0.4) to (29.4,0.4); \draw (-5,0) to (-5,-0.75); \node[empt] (gm5) at (-5,0) [label=below: \footnotesize \( g^{-5} \)] {}; \draw (0,0) to (0,-0.75); \node[gelt] (g0) at (0,0) [label=below: \footnotesize \( g^0 \)] {}; \draw (5,0) to (5,-0.75); \node[empt] (g5) at (5,0) [label=below: \footnotesize \( g^5 \)] {}; \draw (10,0) to (10,-0.75); \node[empt] (g10) at (10,0) [label=below: \footnotesize \( g^{10} \)] {}; \draw (23,0) to (23,-0.75); \node[gelt] (g23) at (23,0) [label=below: \footnotesize \( g^{23} \)] {}; \foreach \i in {-9,-8,-7,-6,-4,-3,-2,-1} \node[empt] at (\i,0) {}; \foreach \i in {1,2,3,4,6,7,8,9,11,12,13,14,15,16,17,18,19,20,21,22} \node[empt] at (\i,0) {}; \foreach \i in {24,25,26,27,28,29} \node[empt] at (\i,0) {}; \end{tikzpicture}}
One directly sees that the translates X_B + k B, k \in \N_{>0} cover all positive integers, and that every positive integer is contained in exactly one translate. Moreover, one sees that if the first translate contains at most one lattice element, the first translate which contains a lattice element uniquely determines the smallest positive integer. Note that in case n is the order of g, then one can minimize the number of operations by chosing B \approx \sqrt{n}.

Now let us consider Terr’s modification of the baby-step giant-step algorithm; there, the situation is a bit more complicated. In Terr’s algorithm, the bound B from above constantly changes, starting with B = 1. Written as pseudo-code, the algorithm looks like this:

  1. Let a := g^1 and b := g^1, and let B := 1 and X_B := \{ (g^0, 0) \}.
  2. If (b, n) \in X_B for some n \in \N, return \frac{B (B + 1)}{2} - n.
  3. Set X_B := X_B \cup \{ (a, B) \}, and set B := B + 1.
  4. Compute a := a \cdot g and b := b \cdot a.
  5. Go back to Step 2.

There is no obvious reason why this should work. Note that the test whether (b, n) \in X_B means that one translates \{ -B + 1, \dots, 0 \} by the exponent c of b = g^c, which turns out to be \frac{B (B + 1)}{2}. One immediately gets the idea if one draws the first few translates:
\fbox{\begin{tikzpicture}[scale=0.3, node distance=0mm] \tikzstyle{gelt} = [draw, shape = circle, fill=black, inner sep=0pt, minimum size = 0.2cm]; \tikzstyle{empt} = [draw, shape = circle, inner sep=0pt, minimum size = 0.2cm]; \filldraw[black!67, fill=black!20] (0.6,-0.4) to (0.6,0.4) to (1.4,0.4) to (1.4,-0.4) to (0.6,-0.4); \filldraw[black!67, fill=black!20] (1.6,-0.4) to (1.6,0.4) to (3.4,0.4) to (3.4,-0.4) to (1.6,-0.4); \filldraw[black!67, fill=black!20] (3.6,-0.4) to (3.6,0.4) to (6.4,0.4) to (6.4,-0.4) to (3.6,-0.4); \filldraw[black!67, fill=black!20] (6.6,-0.4) to (6.6,0.4) to (10.4,0.4) to (10.4,-0.4) to (6.6,-0.4); \filldraw[black!67, fill=black!20] (10.6,-0.4) to (10.6,0.4) to (15.4,0.4) to (15.4,-0.4) to (10.6,-0.4); \filldraw[black!67, fill=black!20] (15.6,-0.4) to (15.6,0.4) to (21.4,0.4) to (21.4,-0.4) to (15.6,-0.4); \filldraw[black!67, fill=black!20] (21.6,-0.4) to (21.6,0.4) to (28.4,0.4) to (28.4,-0.4) to (21.6,-0.4); \filldraw[black!67, fill=black!20] (29.4,-0.4) to (28.6,-0.4) to (28.6,0.4) to (29.4,0.4); \draw (-5,0) to (-5,-0.75); \node[empt] (gm5) at (-5,0) [label=below: \footnotesize \( g^{-5} \)] {}; \draw (0,0) to (0,-0.75); \node[gelt] (g0) at (0,0) [label=below: \footnotesize \( g^0 \)] {}; \draw (5,0) to (5,-0.75); \node[empt] (g5) at (5,0) [label=below: \footnotesize \( g^5 \)] {}; \draw (10,0) to (10,-0.75); \node[empt] (g10) at (10,0) [label=below: \footnotesize \( g^{10} \)] {}; \draw (23,0) to (23,-0.75); \node[gelt] (g23) at (23,0) [label=below: \footnotesize \( g^{23} \)] {}; \foreach \i in {-9,-8,-7,-6,-4,-3,-2,-1} \node[empt] at (\i,0) {}; \foreach \i in {1,2,3,4,6,7,8,9,11,12,13,14,15,16,17,18,19,20,21,22} \node[empt] at (\i,0) {}; \foreach \i in {24,25,26,27,28,29} \node[empt] at (\i,0) {}; \end{tikzpicture}}
This is another tiling of \N_{>0}, where every positive integer is contained in exactly one translate. This tiling has the property that the first time (b, n) \in X_B occurs for some n \in \N is when the order of g is found, and one no longer has to fix a bound B before. The asymptotic complexity of this method is the same as the previous method in case B \approx \sqrt{n}, but in case B is chosen the wrong way, the first algorithm will perform worse than the second. Another way to visualize the second agorithm is to depict the set \{ -B+1, \dots, 0 \} together with \frac{B (B + 1)}{2}, where a = g^B and b = g^{B (B + 1)/2}, for B = 1, 2, \dots, 7:
\fbox{\begin{tikzpicture}[scale=0.3, node distance=0mm] \tikzstyle{gelt} = [draw, shape = circle, fill=black, inner sep=0pt, minimum size = 0.2cm]; \tikzstyle{empt} = [draw, shape = circle, inner sep=0pt, minimum size = 0.2cm]; \foreach \B in { 1, 2, 3, 4, 5, 6, 7 } { \filldraw[black!20, fill=black!67] (-\B+0.6,-\B-0.4) to (-\B+0.6,-\B+0.4) to (0.4,-\B+0.4) to (0.4,-\B-0.4) to (-\B+0.6,-\B-0.4); \filldraw[black!67, fill=black!20] (\B*\B/2-\B/2+0.6,-\B-0.4) to (\B*\B/2-\B/2+0.6,-\B+0.4) to (\B*\B/2+\B/2+0.4,-\B+0.4) to (\B*\B/2+\B/2+0.4,-\B-0.4) to (\B*\B/2-\B/2+0.6,-\B-0.4); \filldraw[black!20, fill=black!67] (\B*\B/2+\B/2-0.4,-\B-0.4) to (\B*\B/2+\B/2-0.4,-\B+0.4) to (\B*\B/2+\B/2+0.4,-\B+0.4) to (\B*\B/2+\B/2+0.4,-\B-0.4) to (\B*\B/2+\B/2-0.4,-\B-0.4); \foreach \i in {-9,-8,-7,-6,-5,-4,-3,-2,-1} \node[empt] at (\i,-\B) {}; \node[gelt] at (0,-\B) {}; \foreach \i in {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22} \node[empt] at (\i,-\B) {}; \node[gelt] at (23,-\B) {}; \foreach \i in {24,25,26,27,28,29} \node[empt] at (\i,-\B) {}; } \end{tikzpicture}}

]]> http://math.fontein.de/2010/01/29/finding-lattice-points-finite-abelian-groups-and-explaining-algorithms/feed/ 0 Homomorphisms, Tensor Products and Certain Canonical Maps. http://math.fontein.de/2010/01/29/homomorphisms-tensor-products-and-certain-canonical-maps/ http://math.fontein.de/2010/01/29/homomorphisms-tensor-products-and-certain-canonical-maps/#comments Fri, 29 Jan 2010 07:20:57 +0000 Felix Fontein http://math.fontein.de/?p=560 Let V and W be vector spaces over a field K and V^* = \Hom_K(V, K), W^* = \Hom_K(W, K) their duals. In case V is finite dimensional, one obtains a non-canonical isomorphism V \cong V^*, a canonical isomorphism V \cong V^{**} and a canonical isomorphism W^* \tensor_K V \cong \Hom_K(W, V).

In case \dim_K V = \infty, V and V^* are not isomorphic: a basis of V^* has a cardinality strictly larger than the one of V. Moreover, the canonical map V \to V^{**} is still a monomorphism, but no longer surjective. In the case of V \tensor_K V^*, one has as well a canonical monomorphism V \tensor_K V^* \to \Hom_K(V, V), but it is no longer surjective as well. We want to study the images of the canonical maps V \to V^{**} and V \tensor_K V^* \to \Hom_K(V, V).

We begin with an auxiliary lemma.

Lemma.
Let v \in V with v \neq 0. Then there exists some \varphi \in V^* with \varphi(v) = 1. Hence, if v \in V satisfies \varphi(v) = 0 for all \varphi \in V^*, then v = 0.
Proof.
Choose a K-basis (v_i)_{i \in I} of V such that there exists some t \in I with v_t = v. Define \pi_t : V \to K by \sum_{i \in I} \lambda_i v_i \mapsto \lambda_t. Then \pi_t(v) = \pi_t(v_t) = 1 and \pi_t \in V^*.
□
Proposition.
The map \displaystyle \Psi : V \to V^{**}, \qquad v \mapsto \begin{cases} V^* \to K, \\ \alpha \mapsto \alpha(v) \end{cases} is a monomorphism and its image is \displaystyle \biggl\{ \varphi \in V^{**} \;\biggm|\; \bigcap_{\alpha \in \ker \varphi} \ker \alpha \neq 0 \biggr\} \cup \{ 0 \}. In particular, if \bigcap_{\alpha \in \ker \varphi} \ker \alpha \neq 0 for some \varphi \in V^{**}, then \dim_K \bigcap_{\alpha \in \ker \varphi} \ker \alpha = 1.
Proof.
Clearly, for v \in V, \Psi(v) : V^* \to K is K-linear. Moreover, one quickly sees that \Psi is K-linear itself. To see that \Psi is injective, let v \in V with v \neq 0. Now, by the lemma, there exists a \pi \in V^* with \pi(v) = 1; this shows that \Psi(v)(\pi_t) = 1, whence \Psi(v) \neq 0. Therefore, \ker \Psi = 0 an \Psi is injective.
Now, if \alpha \in \ker \Psi(v), \alpha(v) = \Psi(v)(\alpha) = 0, whence v \in \bigcap_{\alpha \in \ker \Psi(v)} \ker \alpha. This shows that the image of \Psi is contained in the given set. Now assume that \varphi \in V^{**} \setminus \{ 0 \} satisfies \bigcap_{\alpha \in \ker\varphi} \ker \alpha \neq 0; say, v \in \bigcap_{\alpha \in \ker\varphi} \ker \alpha \setminus \{ 0 \}. Then \alpha(v) = 0 for all \alpha \in \ker\varphi, whence \ker \varphi \subseteq \ker \Psi(v). By the Homomorphism Theorem, there exists a homomorphism \tilde{\varphi} : V^* / \ker \varphi \to K such that \displaystyle \xymatrix{ V \ar[rr]^{\Psi(v)} \ar[dr]_{\pi} & & K \\ & V / \ker \varphi \ar[ru]_{\tilde{\varphi}} & } commutes. Now V^* / \ker \varphi \cong \varphi(V) = K, whence \dim_K V^* / \ker \varphi = 1. As \tilde{\varphi} \neq 0 (as v \neq 0), \tilde{\varphi} is an isomorphism and we must have \Psi(v) = \lambda \varphi for some \lambda \in K \setminus \{ 0 \}. But then, \varphi = \Psi(\lambda^{-1} v) lies in the image of \Psi.
Finally, if \dim_K \bigcap_{\alpha \in \ker \varphi} \ker \alpha > 0, we saw that we have \varphi = \lambda_v \Phi(v) for any non-zero v \in \bigcap_{\alpha \in \ker \varphi} \ker \alpha, with \lambda_v \in K \setminus \{ 0 \} depending on v. Since \Phi : V \to V^{**} is injective, this shows that we must have \dim_K \bigcap_{\alpha \in \ker \varphi} \ker \alpha = 1.
□

This allows us to show that V \to V^{**} is surjective if, and only if, \dim V < \infty.

Corollary.
We have that \Psi : V \to V^{**} is surjective if, and only if, \dim V < \infty.
Proof.
First, if \dim V < \infty, we see that \dim V^{**} = \dim V^* = \dim V. Since \Psi is injective, it follows that \Psi is in fact an isomorphism.
Now assume that \dim V = \infty. It suffices to construct a hyperplane H in V^* with \bigcap_{\alpha \in H} \ker \alpha = 0; this defines an element of V^{**} which is not contained in the image of \Psi by the above proposition. For that, chose a basis (v_i)_{i\in I} of V (using Zorn’s lemma). This defines a family of elements of V^* by \pi_i : V \to K, \sum_{j\in I} \lambda_j v_j \mapsto \lambda_i. Let H' be the subspace of V^* generated by the \pi_i‘s. If we would have H' \subsetneqq V^*, we could emply Zorn’s lemma a second time to obtain a hyperplane H \subseteq V^* with H' \subseteq H; this would prove our claim.
Hence, we have to show that H' \neq V^*. Note that for v \in V, v = \sum_{i\in I} \pi_i(v) v_i; in particular, for every v \in V, only finitely many of the \pi_i(v)‘s are non-zero. Hence, it makes sense to define \pi : V \to K, v \mapsto \sum_{i\in I} \pi_i(v). We claim that \pi \not\in H' in case \abs{I} = \infty: for that, note that \{ \pi_i \}_{i\in I} is a linear independent set in V^*, since for every linear combination \sum \lambda_i \pi_i = 0 \in V^*, we get 0 = \bigl(\sum \lambda_i \pi_i \bigr)(v_j) = \lambda_j for every j \in I.
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Note that in fact, the proof shows that V^* is isomorphic to a \dim V-fold direct product of K, while V is isomorphic to a \dim V-fold direct sum of K. In case \dim V < \infty, these are of the same dimension, but in case \dim V = \infty, they are not.

We continue with the canonical map W^* \tensor_K V \to \Hom_K(W, V).
Proposition.
The map \displaystyle \Phi : W^* \tensor_K V \to \Hom_K(W, V), \qquad \alpha \tensor v \mapsto \begin{cases} W \to V, \\ w \mapsto \alpha(w) v \end{cases} is a monomorphism and its image is \displaystyle \Hom_K^{fin}(W, V) := \{ \varphi \in \Hom_K(W, V) \mid \dim_K \varphi(W) < \infty \}, the K-vector space of finite dimensional K-homomorphisms W \to V.
Proof.
One quickly sees that w \mapsto \varphi(w) v defines an element of \Hom_K^{fin}(W, V), whence \Phi is well-defined and its image is contained in \Hom_K^{fin}(W, V). Moreover, one quickly sees that \Phi is a homomorphism.
Let x = \sum_{i=1}^n \alpha_i \tensor v_i \in W^* \tensor V with \Phi(x) = 0, i.e. with \sum_{i=1}^n \alpha_i(w) v_i = 0 for all w \in W. Without loss of generality, we can assume that our representation of x satisfies that the v_i‘s are linearly independent. In that case, \sum_{i=1}^n \alpha_i(w) v_i = 0 implies \alpha_i(w) = 0 for all i. But since this is true for all w \in W, it follows that \alpha_i = 0 for all i. But then, x = 0. Therefore, \ker \Phi = 0, whence \Phi is injective.
Now let \varphi \in \Hom_K^{fin}(W, V), and let (v_1, \dots, v_n) be a basis of \varphi(W). Let \pi_i : \varphi(W) \to K be the projections with \pi_i(v_i) = 1 an \pi_i(v_j) = 0 for i \neq j. Set \alpha_i := \pi_i \circ \varphi. Then \varphi(w) = \sum_{i=1}^n \alpha_i(w) v_i for all w \in W since v = \sum_{i=1}^n \pi_i(v) v_i for all v \in \varphi(W); therefore, \varphi = \Phi(\sum_{i=1}^n \alpha_i \tensor v_i). This shows that \Hom_K^{fin}(W, V) \subseteq \Phi(W^* \tensor_K V), whence we have equality.
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Now \Hom_K^{fin}(V, V) is a K-algebra, whence for \varphi, \psi \in \Hom_K^{fin}(V, V), it makes sense to define \varphi \circ \psi : V \to V. We are interested on how \Psi^{-1}(\varphi \circ \psi) can be described in terms of \Psi^{-1}(\varphi) and \Psi^{-1}(\psi). This is resolved by the following result:

Proposition.
Let V, W, U be K-vector spaces. The map m :{} & (W^* \tensor_K V) \times (U^* \tensor_K W) \to U^* \tensor_K V, \\ & \biggl(\sum_{i=1}^n \beta_i \tensor v_i, \sum_{j=1}^m \alpha_j \tensor w_j\biggr) \mapsto \sum_{i=1}^n \sum_{j=1}^m \alpha_j \tensor \beta_i(w_j) v_i is K-linear and the following diagram commutes: \xymatrix{ (W^* \tensor_K V) \times (U^* \tensor_K W) \ar[r]^{\qquad\quad m} \ar[d]^{\cong} & U^* \tensor_K V \ar[d]^{\cong} \\ \Hom_K^{fin}(W, V) \times \Hom_K^{fin}(U, W) \ar[r]_{\qquad\quad \circ} & \Hom_K^{fin}(U, V) }
Proof.
Let \Psi_1 : W^* \tensor_K V \to \Hom_K^{fin}(W, V), \Psi_2 : U^* \tensor_K W \to \Hom_K^{fin}(U, W) and \Psi_3 : U^* \tensor_K V \to \Hom_K^{fin}(U, V) be the canonical maps. Since these are isomorphisms, we have to show that for x = \sum_{i=1}^n \beta_i \tensor v_i \in W^* \tensor_K V, y = \sum_{j=1}^m \alpha_j \tensor v_j U^* \tensor_K W and z = \sum_{i=1}^n \sum_{j=1}^m \alpha_j \tensor \beta_i(w_j) v_i \in U^* \tensor_K V, we have \Psi_1(x) \circ \Psi_2(y) = \Psi_3(z). For that, let u \in U. Then (\Psi_1(x) \circ \Psi_2(y))(u) ={} & \Psi_1(x)(\Psi_2(y)(u)) = \Psi_1(x)\biggl( \sum_{j=1}^m \alpha_j(u) v_j \biggr) \\ {}={} & \sum_{i=1}^n \beta_i\biggl( \sum_{j=1}^m \alpha_j(u) v_j \biggr) v_i \\ {}={} & \sum_{i=1}^n \sum_{j=1}^m \alpha_j(u) \beta_i(v_j) v_i = \Psi_3(z)(u), what we had to show.
□

In particular, V^* \tensor_K V is a K-algebra isomorphic to \Hom_K^{fin}(V, V); it posseses a 1 if, and only if, \dim_K V < \infty.

Now consider transposition \displaystyle T : \Hom_K(V, W) \to \Hom_K(W^*, V^*), \quad \varphi \mapsto \begin{cases} W^* \to V^*, \\ \psi \mapsto \psi \circ \varphi. \end{cases} Clearly, transposition is injective:

Lemma.
The map T is K-linear and injective.
Proof.
It is clear that T is K-linear. To see that it is injective, let \varphi \in \Hom_K(V, W) with T(\varphi) = 0. Let v \in V; then \psi(\varphi(v)) = 0 for all \psi \in W^*, whence \varphi(v) = 0 by the first lemma. But that means \varphi = 0.
□

We show that transposition restricts to the subspaces of the homomorphism spaces of homomorphisms with finite-dimensional image.

Lemma.
Let \varphi \in \Hom_K(V, W). The map \displaystyle \varphi(V)^* \to T(\varphi)(W^*), \qquad \alpha \mapsto \alpha \circ \varphi is an isomorphism. In particular, T^{-1}(\Hom_K^{fin}(W^*, V^*)) = \Hom_K^{fin}(V, W).
Proof.
Let \varphi \in \Hom_K(V, W). The map \psi : \varphi(V)^* \to T(\varphi)(W^*), \alpha \mapsto \alpha \circ \varphi is well-defined and a homomorphism as T(\varphi)(W^*) \subseteq V^* and \varphi(V) \subseteq W. Now let \alpha \in \varphi(V)^* with \psi(\alpha) = 0, i.e. with \alpha \circ \varphi = 0. But since \alpha is defined on \varphi(V), this means that \alpha = 0. Hence, \psi is injective.
Now let \beta \in T(\varphi)(W^*), i.e. there exists some \hat{\psi} \in W^* with \beta = \hat{\psi} \circ \varphi. Set \alpha := \hat{\psi}|_{\varphi(V)}; then \alpha \in \varphi(V)^* and \psi(\alpha) = \hat{\psi}|_{\varphi(V)} \circ \varphi = \hat{\psi} \circ \varphi = \beta. Therefore, \psi is injective.
Finally, in case \dim_K \varphi(V) < \infty, we have \dim_K \varphi(V)^* < \infty and \dim_K T(\varphi)(W^*) < \infty, whence T(\varphi) \in \Hom_K^{fin}(W^*, V^*). On the contrary, if \dim_K \varphi(V) = \infty, we have \infty = \dim_K \varphi(V)^* = \dim_K T(\varphi)(W^*), whence T(\varphi) \not\in \Hom_K^{fin}(W^*, V^*).
□

Now we have seen that \Hom_K^{fin}(V, W) \cong V^* \tensor_K W and \Hom_K^{fin}(W^*, V^*) \cong W^{**} \tensor_K V^* in a canonical way, and we have the canonical monomorphism \Psi : W \to W^{**}. We show that these maps behave nicely with transposition:

Proposition.
The map \displaystyle T : V^* \tensor_K W \to W^{**} \tensor_K V^*, \qquad \sum_{i=1}^n v_i^* \tensor w_i \mapsto \sum_{i=1}^n \Psi(w_i) \tensor v_i^* is the unique homomorphism which makes the diagram \displaystyle \xymatrix{ \Hom_K^{fin}(V, W) \ar[r]^T \ar[d]_{\cong} & \Hom_K^{fin}(W^*, V^*) \ar[d]^{\cong} \\ V^* \tensor_K W \ar[r]_T & W^{**} \tensor_K V^* } commuting.
Proof.
Let x = \sum_{i=1}^n v_i^* \tensor w_i \in V^* \tensor_K W and y = \sum_{i=1}^n \Psi(w_i) \tensor v_i^* \in W^{**} \tensor_K V^*. Then \Phi(x)(v) = \sum_{i=1}^n v_i^*(v) w_i \in W for v \in V, and & T(\Phi(x))(w^*)(v) = (w^* \circ \Phi(x))(v) = w^*(\Phi(x)(v)) \\ {}={} & w^*\biggl(\sum_{i=1}^n v_i^*(v) w_i\biggr) = \sum_{i=1}^n v_i^*(v) w^*(w_i) \\ {}={} & \sum_{i=1}^n v_i^*(v) \Psi(w_i)(w^*) = \biggl( \sum_{i=1}^n w^*(w_i) v_i^* \biggr)(v) \\ {}={} & \biggl( \sum_{i=1}^n \Psi(w_i)(w^*) v_i^* \biggr)(v) = \biggl( \sum_{i=1}^n \Phi(\Psi(w_i) \tensor v_i^*)(w^*) \biggr)(v) \\ {}={} & \Phi(y)(w^*)(v) for all w^* \in W^* and v \in V. Hence, T(\Phi(x)) = \Phi(y), what we had to show.
□

Now consider double transposition, i.e. \displaystyle T \circ T : \Hom_K(V, W) \to \Hom_K(V^{**}, W^{**}), and its finite-dimensional image restriction \displaystyle T \circ T : \Hom_K^{fin}(V, W) \to \Hom_K^{fin}(V^{**}, W^{**}). The above shows that using the canonical isomorphisms \Hom_K^{fin}(V, W) \cong V^* \tensor_K W and \Hom_K^{fin}(V^{**}, W^{**}) \cong V^{***} \tensor_K W^{**}, we can describe double transpotition by the following commuting diagram: \displaystyle \xymatrix@R-0.85cm{ \Hom_K^{fin}(V, W) \ar[r]^{T \circ T \;\;} \ar[dddd]_{\cong} & \Hom_K^{fin}(V^{**}, W^{**}) \ar[dddd]^{\cong} \\ {\vphantom{x}} \\ {\vphantom{y}} \\ {\vphantom{z}} \\ V^* \tensor_K W \ar[r]^{T \circ T \;\;} & V^{***} \tensor_K W^{**} \\ \sum_{i=1}^n v_i^* \tensor w_i \ar@{|->}[r] & \sum_{i=1}^n \Psi(v_i^*) \tensor \Psi(w_i) }

If \psi \in \Hom_K(W^*, V^*), we obtain a map \displaystyle H(\psi) : V \to W^{**}, \qquad v \mapsto \begin{cases} W^* \to K \\ \alpha \mapsto \psi(\alpha)(v). \end{cases}

Lemma.
The map H : \Hom_K(W^*, V^*) \to \Hom_K(V, W^{**}) is K-linear and injective.
Proof.
First, if \psi \in \Hom_K(W^*, V^*) is fixed, H(\psi)(v + \lambda v') = H(\psi)(v) + \lambda H(\psi)(v') for all v, v' \in V, \lambda \in K; hence, H(V) \subseteq W^{**}. Now, if \psi, \psi' \in \Hom_K(W^*, V^*) and \lambda \in K, v \in V, we have H(\psi + \lambda \psi')(v) ={} & (\psi + \lambda \psi')(\alpha)(v) = \alpha((\psi + \lambda \psi')(v)) \\ {}={} & \alpha(\psi(v) + \lambda \psi'(v)) = H(\psi)(v) + \lambda H(\psi)(v'), whence H is K-linear.
To see that H is injective, let \psi \in \Hom_K(W^*, V^*) be such that H(\psi) = 0. Let \alpha \in W^* and v \in V; since \psi(\alpha)(v) = 0 for all v, we see that \psi(\alpha) = 0, but since this is the case for all \alpha we get \psi = 0.
□

Note that we have the following diagram: \displaystyle \xymatrix{ & & \Hom_K(V, W) \ar[dl]_T \\ & \Hom_K(W^*, V^*) \ar[dl]_T \ar[dr]^H & \\ \Hom_K(V^{**}, W^{**}) & & \Hom_K(V, W^{**}) } Moreover, using the canonical embeddings \Psi : V \to V^{**} and \Psi : W \to W^{**}, we can define a map \Hom_K(V^{**}, W^{**}) \to \Hom_K(V, W^{**}) by \varphi \mapsto \varphi \circ \Phi, and a map \Hom_K(V, W) \to \Hom_K(V, W^{**}) by \varphi \mapsto \Phi \circ \varphi. It turns out that these map make the diagram commute:

Lemma.
The maps \hat{H} : \Hom_K(V^{**}, W^{**}) \to \Hom_K(V, W^{**}), \varphi \mapsto \varphi \circ \Phi and \tilde{H} : \Hom_K(V, W) \to \Hom_K(V, W^{**}), \varphi \mapsto \Phi \circ \varphi are K-linear and make the diagram \displaystyle \xymatrix{ & & \Hom_K(V, W) \ar[dl]_T \ar[dd]^{\tilde{H}} \\ & \Hom_K(W^*, V^*) \ar[dl]_T \ar[dr]^H & \\ \Hom_K(V^{**}, W^{**}) \ar[rr]_{\hat{H}} & & \Hom_K(V, W^{**}) } commute. In particular, \tilde{H} is injective.
Proof.
That \hat{H} and \tilde{H} are K-linear is clear. For the lower triangle, let \varphi \in \Hom_K(W^*, V^*); we have to show that \hat{H}(T(\varphi)) = H(\varphi). For that, let v \in V and \alpha \in W^*; then H(\varphi)(v)(\alpha) ={} & \varphi(\alpha)(v) = \Phi(v)(\varphi(\alpha)) = (\Phi(v) \circ \varphi)(\alpha) \\ {}={} & T(\varphi)(\Phi(v))(\alpha) = \hat{H}(T(\varphi))(v)(\alpha). For the right triangle, let \varphi \in \Hom_K(V, W); we have to show that H(T(\varphi)) = \tilde{H}(\varphi). For that, let v \in V and \alpha \in W^*; then H(T(\varphi))(v)(\alpha) ={} & T(\varphi)(\alpha)(v) = (\alpha \circ \varphi)(v) = \alpha(\varphi(v)) \\ {}={} & \Phi(\varphi(v))(\alpha) = (\Phi \circ \varphi)(v)(\alpha) = \tilde{H}(\varphi)(v)(\alpha).
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Now note that H is injective. We can use this to determine the image of T. For example, for \psi \in \Hom_K(V^*, W^*), & \exists \varphi \in \Hom_K(V, W) : T(\varphi) = \psi \\ {}\Leftrightarrow{} & \forall v \in V : H(\psi)(v) \in \Phi(W) \\ {}\Leftrightarrow{} & \forall v \in V : (\alpha \mapsto \psi(\alpha)(v)) \in \Phi(W) \\ {}\Leftrightarrow{} & \forall v \in V : \bigcap_{\alpha \in V^* : \psi(\alpha)(v) = 0} \ker \alpha = 0 \text{ implies } \psi(\bullet)(v) = 0; the last equivalence follows from the first proposition. Unfortunately, this criterion does not really helps in practice.

In case anyone knows a better description of the image of T or \Psi, I’d be happy to know.

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