In case one has , one has that is a root of as well, and one can combine the terms and to a term , multiplied by a polynomial in of degree . This allows to give a partial fraction decomposition of purely in term of real polynomials. Note that is the minimal polynomial of over .

More generally, one can consider polynomials over an arbitrary field . Now will in general not split into linear (or at most quadratic) factors over . But nonetheless, one can still do a partial fraction decomposition. Moreover, one can continue this to another level of abstractness by working with arbitrary principal ideal domains.

The main ingredient which we need is the Chinese Remainder Theorem:

Theorem (Chinese Remainder Theorem).

Let be a principal ideal domain and let . Write , where are pairwise coprime elements. Then is an isomorphism. Moreover, there exist such that divides , that the inverse of the above isomorphism is given by In case is Euclidean, one can choose the such that .

Proof.

Consider , . We first show that . For that, note that Now the are pairwise coprime, whence this is equivalent to dividing . Hence, . Therefore, induces a monomorphism . We have to show that is surjective, or alternatively, that it admits an inverse.

Let . We have that is coprime to ; hence, we have a Bézout identity, i.e. we can write with ; in case is Euclidean, we can also assume that , (see my post on the Extended Euclidean Algorithm). But this means that is zero in for all , in . Set ; therefore, , whence , as we had to show.

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We need another simple lemma, which works for a certain class of Euclidean rings such as with for some fixed , and with . The two properties we need are the following:

Definition.

We will say that an Euclidean domain is admissible if is multiplicative and for all , one can write with such that and .

Lemma.

Let be an admissible Euclidean domain and . Any can be written in the form with , , where .

Proof.

We show this by induction on . For , , whence . Hence, is of the required form. Now, assume that . Write with , and . Hence, and , whence by induction, we can write with suitable , . If we set , we obtain with as required.

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Now partial fraction decomposition is a direct corollary of these results:

Corollary (Partial Fraction Decomposition).

Let be a principal ideal domain with field of fractions , and let be two elements. Assume that , where , are pairwise coprime polynomials and . Then there exist elements such that In case is Euclidean with a multiplicative degree function, we can assume that for , .

As an example, we want to consider the Partial Fraction Decomposition for , i.e. for , , . Now and , whence is the partial fraction decomposition of in .

Proof.

We can assume that by replacing by .

By the Chinese Remainder Theorem, the map , is an isomorphism. The inverse is given by for some ; in case is Euclidean, can be chosen such that . Therefore, there exist (where in case is Euclidean, ) and such that Dividing by , we get This shows the claim in case is not Euclidean.

In case is Euclidean, we can write with ; by adding to , we can therefore assume that satisfies . By the previous lemma, we can now write with and ; as , we get . Therefore, we obtain with , which gives and by reindexing the second sum we obtain the formula from the statement of the theorem.

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Now assume that we have a rational function written as where are pairwise coprime, with . Then In case is prime, this can be considered an element in the localization of at the prime ideal . Let be the completion of . Note that both and have the residue field .

In the special case that , we see that . Hence, since is a prime element in , we obtain an isomorphism with . In , the element corresponds to In particular, we can use the Hasse derivative of respectively (the definition can be translated to power series rings as well) to obtain In the case , this is Heaviside’s formula.

We are interested in itself. Even though we know that for some , we do not know how this can be chosen.

First note that , where and that , where is Euler’s function. We want to find an explicit lower bound for , which allows us to find an explicit lower bound for itself.

As in the proof of Theorem 330 in Hardy and Wright, we have where is the Möbius function. Hardy and Wright compare that expression to and show that the error is .

More precisely, it is First, Next, Since we finally obtain

With a simple computer program, I verified that for all , and the above bound shows for . The computer verification took less than one second. We therefore obtain:

Theorem.

We have for all . Equality holds if and only if .

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The program for verification of the bound is the following C++ program:

01 #include <iostream>
02 #include <iomanip>
03 #include <cmath>
04 #include <vector>
05  
06 // Simple implementation of Euclid's algorithm
07 unsigned long long gcd(unsigned long long a, unsigned long long b)
08 {
09     while (true)
10     {
11         if (a == 0)
12             return b;
13         b %= a;
14         if (b == 0)
15             return a;
16         a %= b;
17     }
18 }
19  
20 // Compute all primes up to a bound
21  
22 std::vector<unsigned> primes;
23  
24 void create_primes(unsigned max)
25 {
26     // TODO: replace with sieve of Eratosthenes
27     primes.push_back(2);
28     for (unsigned p = 3; p <= max; p += 2)
29     {
30         bool prime = true;
31         for (unsigned i = 0; i < primes.size(); ++i)
32             if ((p % primes[i]) == 0)
33             {
34                 prime = false;
35                 break;
36             }
37         if (prime)
38             primes.push_back(p);
39     }
40 }
41  
42 // Compute the Euler phi function (for not too large arguments)
43 unsigned phi(unsigned n)
44 {
45     if (n < 2)
46         return 1;
47     unsigned r = n;
48     bool found_divisor = false;
49     for (unsigned i = 0; i < primes.size(); ++i)
50         if ((n % primes[i]) == 0)
51         {
52             r = r / primes[i] * (primes[i] - 1);
53             found_divisor = true;
54         }
55     if (!found_divisor)
56         --r;
57     return r;
58 }
59  
60 int main()
61 {
62     // Compute probabilities up to:
63     unsigned N = 932453;
64     // unsigned N = 10000000;
65  
66     // Compute enough primes
67     create_primes(sqrt(N));
68     std::cout << "Prepared " << primes.size() << " primes\n";
69  
70     // Compute probabilities
71     unsigned long long C = 1;
72     for (unsigned n = 2; n < N; ++n)
73     {
74         C += phi(n); // number of pairs (x, y) with 1 <= x <= y <= n such that gcd(x, y) == 1
75         // Compute rational number representing probability
76         unsigned long long num = 2 * C - 1;
77         unsigned long long den = n;
78         den *= n;
79         unsigned long long g = gcd(num, den);
80         num /= g;
81         den /= g;
82         // Compute floating point approximation of probability
83         long double P = (long double)num / (long double)den;
84         // Output if probability is <= 0.6079
85         if (num <= (unsigned long long)((long double)0.6079 * den))
86             std::cout << n << ": " << C << ", " << num << "/" << den << " \\approx " << std::setprecision(20) << P << "\n";
87     }
88 }

On my laptop, it runs in less than one second. For , it requires less than 22 seconds.

Assume that you have an algebraic structure with two binary operations and , where is distributive over . Further assume that (a) there exists a neutral element with respect to , and that (b) the operation is associative, and that (c) the operation is cancelable, i.e. for we have and . Note that (b) and (c) are satisfied for example if forms a group.

These rather weak requirements already imply that is commutative: if , then Using the cancellation property, implies .

Using the aforementioned special case that forms a group, we obtain:

Corollary.

Assume that the algebraic structure satisfies that is a group and is distributive over and has a neutral element. Then is already an abelian group.

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Therefore, if we loosen up the definition of a unitary ring by dropping the requirement that addition is commutative, the other axioms already force the commutativity of addition. Therefore, to get something more general than unitary rings (even if the multiplication is not associative or commutative), one has to make sure that does not imply , for example by asking for an addition not having the cancelation property.

Theorem.

Let be a field and such that for all . Then

It is very easy to prove it by induction:

Proof.

For , the left-hand side equals which equals the right-hand side for . Hence, the statement is true for . Now assume that it holds for . Then
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 , , where and is the standard orthonormal base of . It turns out that one can explicitly describe the Gram-Schmidt orthogonalization, namely it is with and the squared norm of is given by This one can also easily show by induction, using the above identity; it appears two times with . Note that the system already appeared once in this blog, namely when I tried to find the closest vector in its span to ; 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.

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

Let be any prime not dividing . Then modulo is invertible, and modulo , the system has a unique solution. Moreover, for any integer , the system has a unique solution modulo : this is true since is also invertible modulo – for that, it suffices to check that the determinant of is a unit, which it is since it is coprime to . Moreover, if is a solution to over the integers, then modulo is the unique solution of modulo . Hence, if we choose such that bounds all coefficients of the solution , we can recover a solution to over the integers from a solution to modulo , by chosing the unique preimages in .

This opens the question on how to solve modulo . For that, a Hensel-like lifting technique can be used. (In fact, this follows from Bourbaki’s generalization since the Jacobian of the map , equals .) Assume that we have an which satisfies . We want to find with . Write with . As , and as is divisible by , we obtain the linear system . Hence, it suffices to solve linear systems over the prime field to solve over .

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 .
   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 such that all coefficients are in .
   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 with square over a finite field, which returns information on the number of solutions. (Note that is not divisible by if and only if there is a unique solution.) If more information is known on the matrix , 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 the smallest prime not dividing , and by the time the linear system solver over needs to solve a system of size . Let (resp. resp. ) denote the largest absolute value of an coefficient of (resp. resp. ).

Clearly, the number of iterations is in . In each iteration, one linear system over of size has to be solved, and has to be evaluated. The former takes operations, and the latter involves multiplications and additions of integers of size and , and substractions of integers of size and . For simplicity, assume that . Finally, to compute , we need multipliations of integers of size and , and additions which can be neglected. Clearly, the multiplications can also be neglected, since the evaluation of already is slower.

Let denote the time a multiplication of two numbers of size needs. Then inside the main loop, we need time units, and the main loop alltogether needs time units. Finding needs time units (using the Prime Number Theorem).

Assuming that we use a naive Gaussian algorithm as well as naive multiplication, i.e. and , we obtain a total running time of Using fast multiplication, i.e. (using FFT methods), and fast linear system solving, i.e. , where , we obtain a total running time of

Now let us try to eliminate from this expression. Clearly, the the second part, we can use that . To eliminate from the first part, we need to find an upper bound. For that, let us first stick to , the smallest prime not dividing the integer . (Letting be a -matrix yields ; in general, using this notation.) Now is divisible by , whence for we have . Note that for integral , , where denotes the Chebyshev function. Using known bounds on , we get Therefore, becomes true for for every . This shows that eventually holds for , yielding and, thus, . Using the Leibniz formula, .

Finally, we can use some linear algebra to bound in terms of and . First note that , where denotes the identity matrix and denotes the adjungate matrix of . As , we see that it suffices to bound . Now the coefficients of are determinants of matrices with coefficients coming from , whence . Therefore, when assuming that .

This can be combined into the following theorem:

Theorem.

Assuming that and , the above algorithm needs time units to compute the unique solution of using naive arithmetic in , , and naive Gaussian elimination to solve linear systems over . Using fast linear algebra and fast multiplication, we only need time units for any .

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

Theorem.

Let be a field and . Then the matrix has determinant

Before I present the proofs, I want to say a few words on how this matrix comes up. Consider the linear system , where Assuming that all , one sees that this system has no solution. Instead, one can try to find a vector which minimizes . A well-known fact in Linear Algebra says that this is the case for a unique , and that is the solution to the system . Now and , whence we can describe the unique solution using Cramer’s rule by Now the above theorem gives an easy to evaluate formula for the determinants, namely The solution vector can be evaluated in operations, and in case the are rational numbers, one can hence efficiently compute an exact (i.e. rational) solution.

Proof (First proof).

For this is easy to verify. Hence, assume that and that the statement is true for . Using the multilinearity of the determinant and Lagrange expansion, both for the first row, we see that The same argument applied to the second row of the second determinant, we obtain that the second determinant Since has two identical rows – namely the first two –, its determinant is 0. Moreover, by induction hypothesis, and Plugging this in yields which shows the claim.

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Proof (Second proof).

For this is clear. Hence, assume that and that the statement is true for . We do a Lagrange expansion by the last column. This yields a term , which by induction hypothesis equals The other terms are of the form , where is with the last column and the -th row removed. Note that by cyclically permuting rows to of , we obtain the matrix . The permutation has sign , whence we see that Combining everything so far, we see that what we wanted to show.

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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 is an integer, and we are given integers with . Define the quantity as One can show that this is always a non-negative integer. Now I claim that is either 0, or .

Where does this strange expression comes from? Consider the function field over defined by (In fact, one can do this over any field whose characteristic is coprime to , and which has at least elements. Moreover, over an algebraically closed field , any function field extension which is cyclic of order , where is coprime to the characteristic of , can be written in this form, since this is a Kummer extension.) One can show that this equation is irreducibe if and only if , and that the genus of this function field is given by . This also explains why we must have that , 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 , or in case and , at least five of the terms must be since is not divisible by . Therefore, .

Moreover, if , we have , and therefore and . Hence, Therefore, it suffices to consider the case and .

For , note that implies . Hence, there is nothing to show.

For , let me consider three cases.

1. for some prime ;
2. for some prime ;
3. for two distinct prims .

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 is of the form , , . Therefore, for all . Since not all ‘s cannot be the same – this would contradict – we must have

Let me demonstrate how to do . (In fact, this case suffices if one does not wants the constant , but one is happy with the constant , since .) Assume that for some prime . Since we must have , with in case . In both cases, we have .

In case , implies . Therefore, . In case , implies . Therefore, we also have This yields The latter inequality is true if and only if If we write , this translates to if , , and if , . Hence, the only cases were the above argument does not work are () and (, ).

In case , we must have and . In that case, , whence In case , since we do by assumption , we obtain , whence since .

The cases and are proven analogously, with a few more case distinctions.

So we are left with the case . Here, one can proceed in a similar, painful way. Or one increases the constant to , since we know that not all ‘s can be , whence one is at least . Hence, which yields the claim.

To sum everything up, we showed the following theorem:

Theorem.

Let be an integer, , and satisfy . Then , defined as satisfies , and if it is strictly larger than zero,

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As mentioned, if one invests more work, one can actually show . For my preprint though, this is not worth the trouble.

Assuming that it is known that there is, up to scale, only one alternating -linear form (i.e. that ), one can proceed as follows. Given , consider the map , . This is clearly -linear and alternating, whence there exists some such that . Evaluating at the identity matrix gives . Evaluating at gives .

Of course, using the trick similar to the first lemma here, it suffices to show this for 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).