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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-adic integers and numbers

The -adic numbers.

Let us begin with the -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 respectively the archimedean absolute value Recall that a sequence is said to be convergent with limit if for every , there exists some with for all . Note that every convergent series is a Cauchy sequence, i.e. it satisfies But not every Cauchy sequence in converges. One reason to use the real numbers is to add limits of Cauchy sequences, so that every Cauchy sequence (with coefficients in ) converges. More precisely, consider the set of Cauchy sequences, ; this is an -subspace of , the space of all functions (i.e. all sequences). Consider the subspace of sequences converging to ; note that . Therefore, we can consider the quotient ; embeds via the diagonal embedding, i.e. maps to . One quickly checks that 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 on extends to ; this allows to define and for in the same manner as for rational numbers. Moreover, one sees that all Cauchy sequences in actually converge.

The -adic numbers can be constructed in a very similar way. Fix a prime number , and condider the -adic valuation on , defined by if are not divisible by . Moreover, set . Then, we can define the -adic absolute value , ; then if, and only if, ; we have that the strict triangle inequality is satisfied; and we have that ; here, are arbitrary. Such absolute values which satisfy the strict triangle inequality are called non-archimedean absolute values.

Remark.

In fact, one can show that these -adic absolute values together with the above Archimedean absolute value are all absolute values on up to equivalence; here, we say that two absolute values are equivalent if there exists some number with for all .

Now let us continue with the -adic numbers. We can define Cauchy sequences and the notion of convergence as above, by replacing by . As above, we obtain that has a completion with respect to which forms a field, denoted by . This is the field of -adic numbers. Also, and extend onto . As opposed to the case of the real numbers, the image of resp. do not change; the reason is that is a discrete valuation, i.e. attains only integers. Actually, this field has several interesting properties, which we want to collect.

Theorem.

1. For every , .
2. The set is a subring of ; denote this subring by .
3. For any , we have that and are both open and closed in . In particular, is both open and closed.
4. We have that
5. The ring is local with maximal ideal , and . In fact, is a discrete valuation ring.
6. The series with converges if, and only if, .
7. Every non-zero element can be written uniquely in the form , where for all , and .
8. For every , .
9. If is equipped with the topology induced by the metric , then is the maximal compact subring of .

Proof.

1. Let , sequences in with , . Then, for every , . Now is a topological ring, and is continuous, whence the result follows from applying .
2. From (a), we see that this set is closed under addition. That it is closed under multiplication is clear, and are contained in it as well.
3. Write with and . Now , and
4. Let be a sequence of elements in which converges with respect to . Then for all as ; as is continuous, .
   Conversely, let and let be a sequence of elements in with . Now is open; therefore, there exists some with for all . Without loss of generality, we can assume that , i.e. all lie in . Moreover, we can assume that for all . We want to construct a sequence in with . For that, write with , where is not divisible by . Let be such that and set ; moreover, write with . Then whence . Therefore, one obtains that .
5. One quickly checks using 1. that is closed under addition. It is clearly also closed under multiplication by elements of , and contains 0; therefore, it is an ideal. Now consider the map ; clearly, , whence is contained in the kernel of this map. But is not contained in the kernel, as ; therefore, contains a copy of . To see that this is everything, let ; let be a sequence in with . Let be such that for all , ; then , whence . But is contained in the copy of .
6. Clearly, if the series converges, we must have that .
   Now, conversely, assume that . We show that with is a Cauchy series. For that, let be given. Choose such that for all , . Now, if with , we have that by 1., what we had to show.
7. First, for any choice of the ‘s and ‘s, we obtain an element in . Now assume that with . Assume that there exists some with , and further assume that is chosen to be minimal under this condition, i.e. for all . Then . Multiplying with gives with . Moreover, , and . But now , whence . But by construction , a contradiction. Therefore, the representations are unique.
   We have to show that every element in can be written in this way. For that, let ; now satisfies ; in particular, . We have to show that we can write with . For that, let be a sequence in with ; without loss of generality, we can assume that for every . Then we also have for every , whence . Therefore, we can choose inductively such that , and we obtain that . Finally, since (which follows from the strict triangle inequality), it follows that .
8. Clearly, ; therefore, using 7., we see that every residue class in is uniquely described by , where . Hence, . Now, as in the proof of 5., we see that injects into , whence this injection is in fact a bijection. Thus .
9. Let be any compact subring of and . If , then for . Hence, is unbounded, a contradiction.
   Hence, it is left to show that is compact. For that, it suffices to show that any sequence in has at least one accumulation point. Let be any sequence in . We claim that for any , there exists some such that contains infinitely many elements of the sequence. For this is clear for any choice of ; hence, assume that this is the case for some . Now ; now there must be some such that infinitely many of the ‘s lie in (otherwise, only finitely many can lie in ); set . We see that is a Cauchy sequence, whence exists. Now by construction, is an accumulation point of . Therefore, is compact.

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

Definition.

Let be a directed set and for every , let be a ring. Assume that for every with there exists a homomorphism such that for all with , we have , and we have . Such a tuple is called a projective system. A projective limit of is a ring together with homomorphisms such that if , which satisfies the following universal property:
If is any other ring and any other family of ring homomorphisms with whenever , there exists a unique homomorphism with for all .

We have the following, classical result:

Theorem.

For every projective system of rings, there exists a projective limit which is unique up to unique isomorphism; i.e., for any two projective limits and there exists a unique isomorphism such that for all .
Moreover, a projective limit can be constructed as where is the restriction of the canonical projection to the subset .

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Definition.

Let be a projective system. Define the projective limit of it as any projective limit, and denote it by .

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 be a ring and an ideal of . We say that is nilpotent if for some . We say that is a nilideal if every is nilpotent, i.e. if for every there is some with .

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 be a ring and a nilideal in . Let and such that and . Then there exists a unique with and .

Proof (Part I).

First, assume that both and satisfy and . Using , we see that for some using Taylor expansion. Since is a unit in , there exist some , with . How is nilpotent, whence , i.e. . But this means that , whence . This shows uniqueness.
Moreover, we want to show that it suffices to require that is nilpotent. Cosider the subring of which is the smallest (unitary) subring containing the coefficients of and , and some fixed with . This ring is clearly finitely generated, whence Noetherian, and is a nilideal in as well. Since is Noetherian, is in fact nilpotent. Moreover, since . Hence, it suffices to prove the lemma for instead of .

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We will complete the proof later. First, let us discuss some implications of this result. Consider the ring and , where . Then clearly is nilpotent in , whence we can apply Hensel’s lemma to this situation. Assume that is a polynomial and with ; if then , there exists a unique with and . Doing this construction for and all , we obtain a sequence of elements with and , i.e. we obtain an element of !

In particular, let be any element with . Consider . Now there exists some with ; therefore, . But this implies that there exists a unique with and . (Since is unique modulo , it follows that is uniquely defined by , i.e. by .) Hence, we have shown that any integer is invertible in if . But how to compute ? We will discuss this later; for now, note that we can use the Extended Euclidean Algorithm to find with ; then , i.e. we can approximate up to arbitrary precision.

What about ? In that case, : if would have an inverse in , say , then we would have , a contradiction. Therefore, we get:

Proposition.

We have that

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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 be a polynomial, a nilpotent ideal, and such that and . Fix some with , and consider the sequence We claim that for all ; since is nilpotent, this means that the sequence becomes stationary eventually and gives a root of . Moreover, we claim that , whence for all .

Assume that this is true for some , i.e. we have . Then . Now , whence . To show that , we again use the Taylor expansion; by it, for some . Now , whence . Moreover, . Combining this gives .

Therefore, the proof of Hensel’s lemma is completed. Moreover, we obtained an algorithm to refine an approximation of without using the Extended Euclidean Algorithm: as soon as with is known, we can compute modulo by applying the Newton iteration only times. Moreover, we can start with some intermediate result, say modulo , to compute modulo (with ) in at most iterations (which each need one multiplication and one addition modulo ). This is considerably faster than applying the Extended Euclidean Algorithm for and .

So, What About in ?

Assume that ; then we know that . Consider the polynomial ; clearly, in , and is a unit in . Therefore, we can use the methods from above to compute .

First, we need an approximation of modulo . For that, use the Extended Euclidean Algorithm to compute with ; then satisfies , i.e. .

Set and . Then, by the above, . Hence, this allows to approximate up to an error of in 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 in , i.e. , and . Clearly, , whence we get . Hence, . Now and , whence , . We rapidly get Hence, it seems that i.e. we have And indeed: since Note that follows from the fact that , whence converges in (see the theorem); this is a geometric series, whence its value is . Hence, if we multiply by , we obtain .

Finally, let us consider another example, namely in , i.e. , , ; then , i.e. . Hence, we obtain and . We then obtain 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 , where are coprime and not divisible by , and . Now let be the order of in the multiplicative group ; hence, it is the smallest non-negative rational number with . Then . Now, by the geometric series (as above), whence Moreover, write with such that . Then , and . Set ; then, if we write , we see that We are left to consider the part. In case , the -adic expansion of has finite length, whence adding it does not change the periodicity of . But what if ?

Lemma.

Let , where . Then

Proof.

Clearly, whence adding results in the statement.

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

Conversely, assume that with and ; we claim that . Clearly, without loss of generality, we can assume that . But then, if we set , Hence, we proved:

Proposition.

An element with lies in if, and only if, there exists some such that for all , .

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Finally, note that the order of 17 in is ; hence, the period length of is probably 616. We would have had to compute a high amount of -adic digits of to see this.