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Algebra II

ContentsArticles

Structure theorem for finitely generated modules over a principal ideal domain 1Torsion (algebra) 5Zero divisor 8Smith normal form 10Finitely-generated module 14Free module 17Chinese remainder theorem 19Bézout's identity 27

ReferencesArticle Sources and Contributors 29Image Sources, Licenses and Contributors 30

Article LicensesLicense 31

Structure theorem for finitely generated modules over a principal ideal domain 1

Structure theorem for finitely generated modulesover a principal ideal domainIn mathematics, in the field of abstract algebra, the structure theorem for finitely generated modules over aprincipal ideal domain is a generalization of the fundamental theorem of finitely generated abelian groups androughly states that finitely generated modules can be uniquely decomposed in much the same way that integers havea prime factorization. The result provides a simple framework to understand various canonical form results forsquare matrices over fields.

StatementWhen a vector space over a field F has a finite generating set, then one may extract from it a basis consisting of afinite number n of vectors, and the space is therefore isomorphic to Fn. The corresponding statement with the Fgeneralized to a principal ideal domain R is no longer true, as a finitely generated module over R need not have anybasis. However such a module is still isomorphic to a quotient of some module Rn with n finite (to see this it sufficesto construct the morphism that sends the elements of the canonical basis Rn to the generators of the module, and takethe quotient by its kernel.) By changing the choice of generating set, one can in fact describe the module as thequotient of some Rn by a particularly simple submodule, and this is the structure theorem.The structure theorem for finitely generated modules over a principal ideal domain usually appears in the followingtwo forms.

Invariant factor decompositionEvery finitely generated module M over a principal ideal domain R is isomorphic to a unique one of the form

where and . The order of the nonzero ideals is invariant, and the number ofis invariant.

The nonzero elements, together with the number of which are zero, form a complete set of invariants for themodule. Explicitly, this means that any two modules sharing the same set of invariants are necessarily isomorphic.The themselves are called invariant factors of M.The ideals are unique. In terms of the elements, this means that the are unique up to multiplication by aunit.The free part is visible in the part of the decomposition corresponding to the factors. These occur at the endof the sequence of 's, as everything divides zero.Some prefer to write the free part of M separately:

where the visible are nonzero, and f is the number of 's which are 0.

Structure theorem for finitely generated modules over a principal ideal domain 2

Primary decompositionEvery finitely generated module M over a principal ideal domain R is isomorphic to one of the form

where and the are primary ideals. The are unique (up to multiplication by units).The elements are called the elementary divisors of M. In a PID, primary ideals are powers of primes, and so

.

The summands are indecomposable, so the primary decomposition is a decomposition into indecomposablemodules, and thus every finitely generated module over a PID is a completely decomposable module. Since PID's areNoetherian rings, this can be seen as a manifestation of the Lasker-Noether theorem.As before, it is possible to write the free part (where ) separately and express M as:

where the visible are nonzero.

ProofsOne proof proceeds as follows:• Every finitely generated module over a PID is also finitely presented because a PID is Noetherian, an even

stronger condition than coherence.• Take a presentation, which is a map (relations to generators), and put it in Smith normal form.This yields the invariant factor decomposition, and the diagonal entries of Smith normal form are the invariantfactors.Another outline of a proof:• Denote by tM the torsion submodule of M. Then M/tM is a finitely generated torsion free module, and such a

module over a commutative PID is a free module of finite rank, so it is isomorphic to for a positive integer n.This free module can be embedded as a submodule F of M, such that the embedding splits (is a right inverse of)the projection map; it suffices to lift each of the generators of F into M. As a consequence .

• For a prime p in R we can then speak of for each prime p. This is asubmodule of tM, and it turns out that each Np is a direct sum of cyclic modules, and that tM is a direct sum of Npfor a finite number of distinct primes p.

• Putting the previous two steps together, M is decomposed into cyclic modules of the indicated types.

CorollariesThis includes the classification of finite-dimensional vector spaces as a special case, where . Since fieldshave no non-trivial ideals, every finitely generated vector space is free.Taking yields the fundamental theorem of finitely generated abelian groups.

Let T be a linear operator on a finite-dimensional vector space V over K. Taking , the algebra ofpolynomials with coefficients in K evaluated at T, yields structure information about T. V can be viewed as a finitelygenerated module over . The last invariant factor is the minimal polynomial, and the product of invariantfactors is the characteristic polynomial. Combined with a standard matrix form for , this yields variouscanonical forms:• invariant factors + companion matrix yields Frobenius normal form (aka, rational canonical form)• primary decomposition + companion matrix yields primary rational canonical form

Structure theorem for finitely generated modules over a principal ideal domain 3

• primary decomposition + Jordan blocks yields Jordan canonical form (this latter only holds over an algebraicallyclosed field)

UniquenessWhile the invariants (rank, invariant factors, and elementary divisors) are unique, the isomorphism between M andits canonical form is not unique, and does not even preserve the direct sum decomposition. This follows becausethere are non-trivial automorphisms of these modules which do not preserve the summands.However, one has a canonical torsion submodule T, and similar canonical submodules corresponding to each(distinct) invariant factor, which yield a canonical sequence:

Compare composition series in Jordan–Hölder theorem.

For instance, if , and is one basis, then is another basis, and the

change of basis matrix does not preserve the summand . However, it does preserve the summand,

as this is the torsion submodule (equivalently here, the 2-torsion elements).

Generalizations

GroupsThe Jordan–Hölder theorem is a more general result for finite groups (or modules over an arbitrary ring). In thisgenerality, one obtains a composition series, rather than a direct sum.The Krull–Schmidt theorem and related results give conditions under which a module has something like a primarydecomposition, a decomposition as a direct sum of indecomposable modules in which the summands are unique upto order.

Primary decompositionThe primary decomposition generalizes to finitely generated modules over commutative Noetherian rings, and thisresult is called the Lasker–Noether theorem.

Indecomposable modulesBy contrast, unique decomposition into indecomposable submodules does not generalize as far, and the failure ismeasured by the ideal class group, which vanishes for PIDs.For rings that are not principal ideal domains, unique decomposition need not even hold for modules over a ringgenerated by two elements. For the ring R = Z[√−5], both the module R and its submodule M generated by 2 and1 + √−5 are indecomposable. While R is not isomorphic to M, R ⊕ R is isomorphic to M ⊕ M; thus the images ofthe M summands give indecomposable submodules L1, L2 < R ⊕ R which give a different decomposition of R ⊕ R.The failure of uniquely factorizing R ⊕ R into a direct sum of indecomposable modules is directly related (via theideal class group) to the failure of the unique factorization of elements of R into irreducible elements of R.

Structure theorem for finitely generated modules over a principal ideal domain 4

Non-finitely generated modulesSimilarly for modules that are not finitely generated, one cannot expect such a nice decomposition: even the numberof factors may vary. There are Z-submodules of Q4 which are simultaneously direct sums of two indecomposablemodules and direct sums of three indecomposable modules, showing the analogue of the primary decompositioncannot hold for infinitely generated modules, even over the integers, Z.Another issue that arises with non-finitely generated modules is that there are torsion-free modules which are notfree. For instance, consider the ring Z of integers. A classical example of a torsion-free module which is not free isthe Baer–Specker group, the group of all sequences of integers under termwise addition. In general, the question ofwhich infinitely generated torsion-free abelian groups are free depends on which large cardinals exist. Aconsequence is that any structure theorem for infinitely generated modules depends on a choice of set theory axiomsand may be invalid under a different choice.

References• Atiyah, Michael Francis; Macdonald, I.G. (1969), Introduction to Commutative Algebra, Westview Press,

ISBN 978-0-201-40751-8• Dummit, David S.; Foote, Richard M. (2004), Abstract algebra (3rd ed.), New York: Wiley,

ISBN 978-0-471-43334-7, MR 2286236 (http:/ / www. ams. org/ mathscinet-getitem?mr=2286236)• Hungerford, Thomas W. (1980), Algebra, New York: Springer, pp. 218–226, Section IV.6: Modules over a

Principal Ideal Domain, ISBN 978-0-387-90518-1• Jacobson, Nathan (1985), Basic algebra. I (2 ed.), New York: W. H. Freeman and Company, pp. xviii+499,

ISBN 0-7167-1480-9, MR  780184 (http:/ / www. ams. org/ mathscinet-getitem?mr=780184)• Lam, T. Y. (1999), Lectures on modules and rings, Graduate Texts in Mathematics No. 189, Springer-Verlag,

ISBN 978-0-387-98428-5

Torsion (algebra) 5

Torsion (algebra)In abstract algebra, the term torsion refers to elements of finite order in groups and to elements of modulesannihilated by regular elements of a ring.

DefinitionAn element m of a module M over a ring R is called a torsion element of the module if there exists a regular elementr of the ring (a non-zero element of the ring that is neither a left nor a right zero divisor) that annihilates m, i.e., r m =0. In an integral domain (a commutative ring without zero divisors), every non-zero element is regular, so a torsionelement of a module over an integral domain is one annihilated by a non-zero element of the integral domain. Someauthors use this as the definition of a torsion element but this definition does not work well over more general rings.A module M over a ring R is called a torsion module if all its elements are torsion elements, and torsion-free if zerois the only torsion element. If the ring R is commutative then the set of all torsion elements forms a submodule of M,called the torsion submodule of M, sometimes denoted T(M). If R is not commutative, T(M) may or may not be asubmodule. It is shown in (Lam 2007) that R is a right Ore ring if and only if T(M) is a submodule of M for all rightR modules. Since right Noetherian domains are Ore, this covers the case when R is a right Noetherian domain (whichmight not be commutative).More generally, let M be a module over a ring R and S be a multiplicatively closed subset of R. An element m of M iscalled an S-torsion element if there exists an element s in S such that s annihilates m, i.e., s m = 0. In particular, onecan take for S the set of regular elements of the ring R and recover the definition above.An element g of a group G is called a torsion element of the group if it has finite order, i.e., if there is a positiveinteger m such that gm = e, where e denotes the identity element of the group, and gm denotes the product of m copiesof g. A group is called a torsion (or periodic) group if all its elements are torsion elements, and a torsion-freegroup if the only torsion element is the identity element. Any abelian group may be viewed as a module over thering Z of integers, and in this case the two notions of torsion coincide.

Examples1. Let M be a free module over any ring R. Then it follows immediately from the definitions that M is torsion-free

(if the ring R is not a domain then torsion is considered with respect to the set S of non-zero divisors of R). Inparticular, any free abelian group is torsion-free and any vector space over a field K is torsion-free when viewedas the module over K.

2. By contrast with Example 1, any finite group (abelian or not) is periodic and finitely generated. Burnside'sproblem asks whether, conversely, any finitely generated periodic group must be finite. (The answer is "no" ingeneral, even if the period is fixed.)

3. In the modular group, Γ obtained from the group SL(2,Z) of two by two integer matrices with unit determinant byfactoring out its center, any nontrivial torsion element either has order two and is conjugate to the element S orhas order three and is conjugate to the element ST. In this case, torsion elements do not form a subgroup, forexample, S · ST = T, which has infinite order.

4. The abelian group Q/Z, consisting of the rational numbers (mod 1), is periodic, i.e. every element has finiteorder. Analogously, the module K(t)/K[t] over the ring R = K[t] of polynomials in one variable is pure torsion.Both these examples can be generalized as follows: if R is a commutative domain and Q is its field of fractions,then Q/R is a torsion R-module.

5. The torsion subgroup of (R/Z,+) is (Q/Z,+) while the groups (R,+),(Z,+) are torsion-free. The quotient of atorsion-free abelian group by a subgroup is torsion-free exactly when the subgroup is a pure subgroup.

Torsion (algebra) 6

6. Consider a linear operator L acting on a finite-dimensional vector space V. If we view V as an F[L]-module inthe natural way, then (as a result of many things, either simply by finite-dimensionality or as a consequence of theCayley–Hamilton theorem), V is a torsion F[L]-module.

Case of a principal ideal domainSuppose that R is a (commutative) principal ideal domain and M is a finitely-generated R-module. Then the structuretheorem for finitely generated modules over a principal ideal domain gives a detailed description of the module M upto isomorphism. In particular, it claims that

where F is a free R-module of finite rank (depending only on M) and T(M) is the torsion submodule of M. As acorollary, any finitely-generated torsion-free module over R is free. This corollary does not hold for more generalcommutative domains, even for R = K[x,y], the ring of polynomials in two variables. For non-finitely generatedmodules, the above direct decomposition is not true. The torsion subgroup of an abelian group may not be a directsummand of it.

Torsion and localizationAssume that R is a commutative domain and M is an R-module. Let Q be the quotient field of the ring R. Then onecan consider the Q-module

obtained from M by extension of scalars. Since Q is a field, a module over Q is a vector space, possibly,infinite-dimensional. There is a canonical homomorphism of abelian groups from M to MQ, and the kernel of thishomomorphism is precisely the torsion submodule T(M). More generally, if S is a multiplicatively closed subset ofthe ring R, then we may consider localization of the R-module M,

which is a module over the localization RS. There is a canonical map from M to MS, whose kernel is precisely theS-torsion submodule of M. Thus the torsion submodule of M can be interpreted as the set of the elements that 'vanishin the localization'. The same interpretation continues to hold in the non-commutative setting for rings satisfying theOre condition, or more generally for any right denominator set S and right R-module M.

Torsion in homological algebraThe concept of torsion plays an important role in homological algebra. If M and N are two modules over acommutative ring R (for example, two abelian groups, when R = Z), Tor functors yield a family of R-modulesTori(M,N). The S-torsion of an R-module M is canonically isomorphic to Tor1(M, RS/R). The symbol Tor denotingthe functors reflects this relation with the algebraic torsion. This same result holds for non-commutative rings as wellas long as the set S is a right denominator set.

Torsion (algebra) 7

Abelian varieties

The 4-torsion subgroup of an elliptic curve over the complex numbers.

The torsion elements of an abelian variety aretorsion points or, in an older terminology,division points. On elliptic curves they maybe computed in terms of divisionpolynomials.

References

•• Ernst Kunz, "Introduction to Commutativealgebra and algebraic geometry",Birkhauser 1985, ISBN 0-8176-3065-1

•• Irving Kaplansky, "Infinite abeliangroups", University of Michigan, 1954.

• Michiel Hazewinkel (2001), "Torsionsubmodule" [1], in Hazewinkel, Michiel,Encyclopedia of Mathematics, Springer,ISBN 978-1-55608-010-4

• Lam, T. Y. (2007), Exercises in modulesand rings, Problem Books in Mathematics,New York: Springer, pp. xviii+412,doi:10.1007/978-0-387-48899-8 [2],ISBN 0-387-98850-5, MR 2278849 [3]

References[1] http:/ / www. encyclopediaofmath. org/ index. php?title=T/ t093330[2] http:/ / dx. doi. org/ 10. 1007%2F978-0-387-48899-8[3] http:/ / www. ams. org/ mathscinet-getitem?mr=2278849

Zero divisor 8

Zero divisorIn abstract algebra, two nonzero elements a and b of a ring are respectively called a left zero divisor and a rightzero divisor if a b = 0;[1] this is a partial case of divisibility in rings. An element that is a left or a right zero divisoris simply called a zero divisor.[2] An element w that is both a left and a right zero divisor[3] is called a two-sidedzero divisor. If the ring is commutative, then the left and right zero divisors are the same. A non-zero element of aring that is not a zero divisor is called regular.

Examples• The ring of integers has no zero divisors, but in the ring the number is a zero divisor:

as a divisor of , which is a composite number.• A nonzero nilpotent element is always a two-sided zero-divisor.• Any idempotent element is always a two-sided zero divisor since .• An example of a zero divisor in the ring of matrices (over any unital ring except trivial) is the matrix

, because for instance

• Actually, the simplest example of a pair of zero divisor matrices is

.

• A direct product of two or more non-trivial rings always has zero divisors similarly to the -matrix examplejust above (the ring of diagonal matrices over a ring  is the same as the direct product ).

One-sided zero-divisor

• Consider the ring of (formal) matrices with and . Then

and . If ,

then is a left zero divisor iff is even, since ; and it is a right zero

divisor iff is even for similar reasons. If either of is , then it is a two-sided zero-divisor.• Here is another example of a ring with an element that is a zero divisor on one side only. Let be the set of all

sequences of integers . Take for the ring all additive maps from to , with pointwiseaddition and composition as the ring operations. (That is, our ring is , the endomorphism ring of theadditive group .) Three examples of elements of this ring are the right shift

, the left shift , and theprojection map onto the first factor . All three of these additive maps arenot zero, and the composites and are both zero, so is a left zero divisor and is a right zerodivisor in the ring of additive maps from to . However, is not a right zero divisor and is not a leftzero divisor: the composite is the identity. Note also that is a two-sided zero-divisor since

, while is not in any direction.

Zero divisor 9

Non-examplesThe ring of integers modulo a prime number does not have zero divisors and this ring is, in fact, a field, as everynon-zero element is a unit.More generally, there are no zero divisors in division rings.A commutative ring with 0 ≠ 1 and without zero divisors is called an integral domain.

PropertiesIn the ring of n-by-n matrices over some field, the left and right zero divisors coincide; they are precisely thenon-zero singular matrices. In the ring of n-by-n matrices over some integral domain, the zero divisors are preciselythe non-zero matrices with determinant zero.Left or right zero divisors can never be units, because if a is invertible and a b = 0, then 0 = a−10 = a−1a b = b.Every non-trivial idempotent element  a in a ring is a zero divisor, since a2 = a implies that a (a − 1) = (a − 1) a = 0,with nontriviality ensuring that neither factor is 0. Nonzero nilpotent ring elements are also trivially zero divisors.The set of zero divisors is the union of the associated prime ideals of the ring.

Notes[1][1] See Hazewinkel et al. (2004), p. 2.[2][2] See Lanski (2005).[3] " is both a left and a right zero divisor" means and , but such and are not necessarily equal.

References• Hazewinkel, Michiel, ed. (2001), "Zero divisor" (http:/ / www. encyclopediaofmath. org/ index. php?title=p/

z099230), Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4• Michiel Hazewinkel, Nadiya Gubareni, Nadezhda Mikhaĭlovna Gubareni, Vladimir V. Kirichenko. (2004),

Algebras, rings and modules, Vol. 1, Springer, ISBN 1-4020-2690-0• Charles Lanski (2005), Concepts in Abstract Algebra, American Mathematical Soc., p. 342• Weisstein, Eric W., " Zero Divisor (http:/ / mathworld. wolfram. com/ ZeroDivisor. html)", MathWorld.

Smith normal form 10

Smith normal formIn mathematics, the Smith normal form is a normal form that can be defined for any matrix (not necessarily square)with entries in a principal ideal domain (PID). The Smith normal form of a matrix is diagonal, and can be obtainedfrom the original matrix by multiplying on the left and right by invertible square matrices. In particular, the integersare a PID, so one can always calculate the Smith normal form of an integer matrix. The Smith normal form is veryuseful for working with finitely generated modules over a PID, and in particular for deducing the structure of aquotient of a free module.

DefinitionLet A be a nonzero m×n matrix over a principal ideal domain R. There exist invertible and -matrices S, T so that the product S A T is

and the diagonal elements satisfy . This is the Smith normal form of the matrix A. Theelements are unique up to multiplication by a unit and are called the elementary divisors, invariants, or invariantfactors. They can be computed (up to multiplication by a unit) as

where (called i-th determinant divisor) equals the greatest common divisor of all minors of the matrixA.

AlgorithmOur first goal will be to find invertible square matrices S and T such that the product S A T is diagonal. This is thehardest part of the algorithm and once we have achieved diagonality it becomes relatively easy to put the matrix inSmith normal form. Phrased more abstractly, the goal is to show that, thinking of A as a map from (the freeR-module of rank n) to (the free R-module of rank m), there are isomorphisms and

such that has the simple form of a diagonal matrix. The matrices S and T can be foundby starting out with identity matrices of the appropriate size, and modifying S each time a row operation isperformed on A in the algorithm by the same row operation, and similarly modifying T for each column operationperformed. Since row operations are left-multiplications and column operations are right-multiplications, thispreserves the invariant where denote current values and A denotes the originalmatrix; eventually the matrices in this invariant become diagonal. Only invertible row and column operations areperformed, which ensures that S and T remain invertible matrices.For a in R \ {0}, write δ(a) for the number of prime factors of a (these exist and are unique since any PID is also aunique factorization domain). In particular, R is also a Bézout domain, so it is a gcd domain and the gcd of any twoelements satisfies a Bézout's identity.To put a matrix into Smith normal form, one can repeatedly apply the following, where t loops from 1 to m.

Smith normal form 11

Step I: Choosing a pivotChoose jt to be the smallest column index of A with a non-zero entry, starting the search at column index jt-1+1 if t >1.

We wish to have ; if this is the case this step is complete, otherwise there is by assumption some k with, and we can exchange rows and k, thereby obtaining .

Our chosen pivot is now at position (t, jt).

Step II: Improving the pivot

If there is an entry at position (k,jt) such that , then, letting , we know by theBézout property that there exist σ, τ in R such that

By left-multiplication with an appropriate invertible matrix L, it can be achieved that row t of the matrix product isthe sum of σ times the original row t and τ times the original row k, that row k of the product is another linearcombination of those original rows, and that all other rows are unchanged. Explicitly, if σ and τ satisfy the aboveequation, then for and (which divisions are possible by the definition of β) one has

so that the matrix

is invertible, with inverse

Now L can be obtained by fitting into rows and columns t and k of the identity matrix. By construction thematrix obtained after left-multiplying by L has entry β at position (t,jt) (and due to our choice of α and γ it also hasan entry 0 at position (k,jt), which is useful though not essential for the algorithm). This new entry β divides the entry

that was there before, and so in particular ; therefore repeating these steps must eventuallyterminate. One ends up with a matrix having an entry at position (t,jt) that divides all entries in column jt.

Step III: Eliminating entriesFinally, adding appropriate multiples of row t, it can be achieved that all entries in column jt except for that atposition (t,jt) are zero. This can be achieved by left-multiplication with an appropriate matrix. However, to make thematrix fully diagonal we need to eliminate nonzero entries on the row of position (t,jt) as well. This can be achievedby repeating the steps in Step II for columns instead of rows, and using multiplication on the right. In general thiswill result in the zero entries from the prior application of Step III becoming nonzero again.However, notice that the ideals generated by the elements at position (t,jt) form an ascending chain, because entriesfrom a later step always divide entries from a previous step. Therefore, since R is a Noetherian ring (it is a PID), theideals eventually become stationary and do not change. This means that at some stage after Step II has been applied,the entry at (t,jt) will divide all nonzero row or column entries before applying any more steps in Step II. Then wecan eliminate entries in the row or column with nonzero entries while preserving the zeros in the already-zero row orcolumn. At this point, only the block of A to the lower right of (t,jt) needs to be diagonalized, and conceptually thealgorithm can be applied recursively, treating this block as a separate matrix. In other words, we can increment t byone and go back to Step I.

Smith normal form 12

Final stepApplying the steps described above to the remaining non-zero columns of the resulting matrix (if any), we get an

-matrix with column indices where . The matrix entries arenon-zero, and every other entry is zero.Now we can move the null columns of this matrix to the right, so that the nonzero entries are on positions for

. For short, set for the element at position .The condition of divisibility of diagonal entries might not be satisfied. For any index for which ,one can repair this shortcoming by operations on rows and columns and only: first add column tocolumn to get an entry in column i without disturbing the entry at position , and then apply a rowoperation to make the entry at position equal to as in Step II; finally proceed as inStep III to make the matrix diagonal again. Since the new entry at position is a linear combinationof the original , it is divisible by β.The value does not change by the above operation (it is δ of the determinant of the upper

submatrix), whence that operation does diminish (by moving prime factors to the right) the value of

So after finitely many applications of this operation no further application is possible, which means that we haveobtained as desired.Since all row and column manipulations involved in the process are invertible, this shows that there exist invertible

and -matrices S, T so that the product S A T satisfies the definition of a Smith normal form. Inparticular, this shows that the Smith normal form exists, which was assumed without proof in the definition.

ApplicationsThe Smith normal form is useful for computing the homology of a chain complex when the chain modules of thechain complex are finitely generated. For instance, in topology, it can be used to compute the homology of asimplicial complex or CW complex over the integers, because the boundary maps in such a complex are just integermatrices. It can also be used to prove the well known structure theorem for finitely generated modules over aprincipal ideal domain.

ExampleAs an example, we will find the Smith normal form of the following matrix over the integers.

The following matrices are the intermediate steps as the algorithm is applied to the above matrix.

Smith normal form 13

So the Smith normal form is

and the elementary divisors are 2, 6 and 12.

SimilarityThe Smith normal form can be used to determine whether or not matrices with entries over a common field aresimilar. Specifically two matrices A and B are similar if and only if the characteristic matrices and

have the same Smith normal form.For example, with

A and B are similar because the Smith normal form of their characteristic matrices match, but are not similar to Cbecause the Smith normal form of the characteristic matrices do not match.

References• Smith, Henry J. Stephen (1861). "On systems of linear indeterminate equations and congruences" [1]. Phil. Trans.

R. Soc. Lond. 151 (1): 293–326. doi:10.1098/rstl.1861.0016 [2]. Reprinted (pp. 367–409 [3]) in The CollectedMathematical Papers of Henry John Stephen Smith, Vol. I [4], edited by J. W. L. Glaisher. Oxford: ClarendonPress (1894), xcv+603 pp.

• Smith normal form [5] at PlanetMath• Example of Smith normal form [6] at PlanetMath• K. R. Matthews, Smith normal form [7]. MP274: Linear Algebra, Lecture Notes, University of Queensland, 1991.

References[1] http:/ / www. jstor. org/ stable/ 108738[2] http:/ / dx. doi. org/ 10. 1098%2Frstl. 1861. 0016[3] http:/ / archive. org/ stream/ collectedmathema01smituoft#page/ 366/ mode/ 2up[4] http:/ / archive. org/ details/ collectedmathema01smituoft[5] http:/ / planetmath. org/ encyclopedia/ GausssAlgorithmForPrincipalIdealDomains. html[6] http:/ / planetmath. org/ encyclopedia/ ExampleOfSmithNormalForm. html[7] http:/ / www. numbertheory. org/ courses/ MP274/ smith. pdf

Finitely-generated module 14

Finitely-generated moduleIn mathematics, a finitely generated module is a module that has a finite generating set. A finitely generatedR-module also may be called a finite R-module or finite over R.[1]

Related concepts include finitely cogenerated modules, finitely presented modules, finitely related modules andcoherent modules all of which are defined below. Over a Noetherian ring the concepts of finitely generated, finitelyrelated, finitely presented and coherent modules all coincide.A finitely generated module over a field is simply a finite-dimensional vector space, and a finitely generated moduleover the integers is simply a finitely generated abelian group.

Formal definitionThe left R-module M is finitely generated if and only if there exist a1, a2, ..., an in M such that for all x in M, thereexist r1, r2, ..., rn in R with x = r1a1 + r2a2 + ... + rnan.The set {a1, a2, ..., an} is referred to as a generating set for M in this case.In the case where the module M is a vector space over a field R, and the generating set is linearly independent, n iswell-defined and is referred to as the dimension of M (well-defined means that any linearly independent generatingset has n elements: this is the dimension theorem for vector spaces).

Examples• Let R be an integral domain with K its field of fractions. Then every R-submodule of K is a fractional ideal. If R is

Noetherian, every fractional ideal arises in this way.• Finitely generated modules over the ring of integers Z coincide with the finitely generated abelian groups. These

are completely classified by the structure theorem, taking Z as the principal ideal domain.• Finitely generated modules over division rings[citation needed] are precisely finite dimensional vector spaces.

Some factsEvery homomorphic image of a finitely generated module is finitely generated. In general, submodules of finitelygenerated modules need not be finitely generated. As an example, consider the ring R = Z[X1, X2, ...] of allpolynomials in countably many variables. R itself is a finitely generated R-module (with {1} as generating set).Consider the submodule K consisting of all those polynomials with zero constant term. Since every polynomialcontains only finitely many terms whose coefficients are non-zero, the R-module K is not finitely generated.In general, a module is said to be Noetherian if every submodule is finitely generated. A finitely generated moduleover a Noetherian ring is a Noetherian module (and indeed this property characterizes Noetherian rings): A moduleover a Noetherian ring is finitely generated if and only if it is a Noetherian module. This resembles, but is not exactlyHilbert's basis theorem, which states that the polynomial ring R[X] over a Noetherian ring R is Noetherian. Both factsimply that a finitely generated algebra over a Noetherian ring is again a Noetherian ring.More generally, an algebra (e.g., ring) that is a finitely-generated module is a finitely-generated algebra. Conversely,if a finitely generated algebra is integral (over the coefficient ring), then it is finitely generated module. (See integralelement for more.)Let 0 → M′ → M → M′′ → 0 be an exact sequence of modules. Then M is finitely generated if M′, M′′ are finitelygenerated. There are some partial converses to this. If M is finitely generated and M'' is finitely presented (which isstronger than finitely generated; see below), then M′ is finitely-generated. Also, M is Noetherian (resp. Artinian) ifand only if M′, M′′ are Noetherian (resp. Artinian).

Finitely-generated module 15

Let B be a ring and A its subring such that B is a faithfully flat right A-module. Then a left A-module F is finitelygenerated (resp. finitely presented) if and only if the B-module B ⊗A F is finitely generated (resp. finitelypresented)[2].

Finitely generated modules over a commutative ringFor finitely generated modules over a commutative ring R, Nakayama's lemma is fundamental. Sometimes, thelemma allows one to prove finite dimensional vector spaces phenomena for finitely generated modules. For example,if f : M → M is a surjective R-endomorphism of a finitely generated module M, then f is also injective, and hence isan automorphism of M.[3] This says simply that M is a Hopfian module. Similarly, an Artinian module M iscoHopfian: any injective endomorphism f is also a surjective endomorphism[4].Any R-module is an inductive limit of finitely generated R-submodules. This is useful for weakening an assumptionto the finite case (e.g., the characterization of flatness with the Tor functor.)An example of a link between finite generation and integral elements can be found in commutative algebras. To saythat a commutative algebra A is a finitely generated ring over R means that there exists a set of elements G = {x1,..., xn} of A such that the smallest subring of A containing G and R is A itself. Because the ring product may be usedto combine elements, more than just R combinations of elements of G are generated. For example, a polynomial ringR[x] is finitely generated by {1,x} as a ring, but not as a module. If A is a commutative algebra (with unity) over R,then the following two statements are equivalent[5]:• A is a finitely generated R module.• A is both a finitely generated ring over R and an integral extension of R.

Equivalent definitions and finitely cogenerated modulesThe following conditions are equivalent to M being finitely generated (f.g.):

• For any family of submodules {Ni | i ∈ I} in M, if , then for some finite subset F of

I.• For any chain of submodules {Ni | i ∈ I} in M, if , then Ni = M for some i in I.

• If is an epimorphism, then the restriction is an epimorphism for some

finite subset F of I.From these conditions it is easy to see that being finitely generated is a property preserved by Morita equivalence.The conditions are also convenient to define a dual notion of a finitely cogenerated module M. The followingconditions are equivalent to a module being finitely cogenerated (f.cog.):

• For any family of submodules {Ni | i ∈ I} in M, if , then for some finite subset F

of I.• For any chain of submodules {Ni | i ∈ I} in M, if , then Ni = {0} for some i in I.

• If is a monomorphism, then is a monomorphism for some finite subset F

of I.Both f.g. modules and f.cog. modules have interesting relationships to Noetherian and Artinian modules, and theJacobson radical J(M) and socle soc(M) of a module. The following facts illustrate the duality between the twoconditions. For a module M:• M is Noetherian if and only if every submodule of N of M is f.g.

Finitely-generated module 16

• M is Artinian if and only if every quotient module M/N is f.cog.• M is f.g. if and only if J(M) is a superfluous submodule of M, and M/J(M) is f.g.• M is f.cog. if and only if soc(M) is an essential submodule of M, and soc(M) is f.g.• If M is a semisimple module (such as soc(N) for any module N), it is f.g. if and only if f.cog.• If M is f.g. and nonzero, then M has a maximal submodule and any quotient module M/N is f.g.• If M is f.cog. and nonzero, then M has a minimal submodule, and any submodule N of M is f.cog.• If N and M/N are f.g. then so is M. The same is true if "f.g." is replaced with "f.cog."Finitely cogenerated modules must have finite uniform dimension. This is easily seen by applying thecharacterization using the finitely generated essential socle. Somewhat asymmetrically, finitely generated modulesdo not necessarily have finite uniform dimension. For example, an infinite direct product of nonzero rings is afinitely generated (cyclic!) module over itself, however it clearly contains an infinite direct sum of nonzerosubmodules. Finitely generated modules do not necessarily have finite co-uniform dimension either: any ring R withunity such that R/J(R) is not a semisimple ring is a counterexample.

Finitely presented, finitely related, and coherent modulesAnother formulation is this: a finitely generated module M is one for which there is an epimorphism

f : Rk → M.Suppose now there is an epimorphism,

φ : F → M.for a module M and free module F.• If the kernel of φ is finitely generated, then M is called a finitely related module. Since M is isomorphic to

F/ker(φ), this basically expresses that M is obtained by taking a free module and introducing finitely manyrelations within F (the generators of ker(φ)).

• If the kernel of φ is finitely generated and F has finite rank (i.e. F=Rk), then M is said to be a finitely presentedmodule. Here, M is specified using finitely many generators (the images of the k generators of F=Rk) and finitelymany relations (the generators of ker(φ)).

• A coherent module M is a finitely generated module whose finitely generated submodules are finitely presented.Over any ring R, coherent modules are finitely presented, and finitely presented modules are both finitely generatedand finitely related. For a Noetherian ring R, all four conditions are actually equivalent.Some crossover occurs for projective or flat modules. A finitely generated projective module is finitely presented,and a finitely related flat module is projective.It is true also that the following conditions are equivalent for a ring R:1. R is a right coherent ring.2. The module RR is a coherent module.3. Every finitely presented right R module is coherent.Although coherence seems like a more cumbersome condition than finitely generated or finitely presented, it is nicerthan them since the category of coherent modules is an abelian category, while, in general, neither finitely generatednor finitely presented modules form an abelian category.

Finitely-generated module 17

References[1][1] For example, Matsumura uses this terminology.[2][2] Bourbaki 1998, Ch 1, §3, no. 6, Proposition 11.[3][3] Matsumura 1989, Theorem 2.4.[4] Atiyah & Macdonald 1969, Exercise 6.1.[5][5] Kaplansky 1970, p. 11, Theorem 17.

Textbooks• Atiyah, M. F.; Macdonald, I. G. (1969), Introduction to commutative algebra, Addison-Wesley Publishing Co.,

Reading, Mass.-London-Don Mills, Ont., pp. ix+128, MR  0242802 (39 #4129) (http:/ / www. ams. org/mathscinet-getitem?mr=0242802+ (39+ #4129))

• Bourbaki, Nicolas, Commutative algebra. Chapters 1--7. Translated from the French. Reprint of the 1989 Englishtranslation. Elements of Mathematics (Berlin). Springer-Verlag, Berlin, 1998. xxiv+625 pp. ISBN 3-540-64239-0

• Kaplansky, Irving (1970), Commutative rings, Boston, Mass.: Allyn and Bacon Inc., pp. x+180, MR  0254021(http:/ / www. ams. org/ mathscinet-getitem?mr=0254021)

• Lam, T. Y. (1999), Lectures on modules and rings, Graduate Texts in Mathematics No. 189, Springer-Verlag,ISBN 978-0-387-98428-5

• Lang, Serge (1997), Algebra (3rd ed.), Addison-Wesley, ISBN 978-0-201-55540-0• Matsumura, Hideyuki (1989), Commutative ring theory, Cambridge Studies in Advanced Mathematics 8 (2 ed.),

Cambridge: Cambridge University Press, pp. xiv+320, ISBN 0-521-36764-6, MR  1011461 (90i:13001) (http:/ /www. ams. org/ mathscinet-getitem?mr=1011461+ (90i:13001)) Unknown parameter |note= ignored (help)

Free moduleIn mathematics, a free module is a free object in a category of modules. Given a set , a free module on is afree module with basis .Every vector space is free,[1] and the free vector space on a set is a special case of a free module on a set.

DefinitionA free module is a module with a basis:[2] a linearly independent generating set.For an -module , the set is a basis for if:1. is a generating set for ; that is to say, every element of is a finite sum of elements of multiplied

by coefficients in ;2. is linearly independent, that is, if for distinct elements

of , then (where is the zero element of and is the zero element of).

If has invariant basis number, then by definition any two bases have the same cardinality. The cardinality of any(and therefore every) basis is called the rank of the free module , and is said to be free of rank n, or simplyfree of finite rank if the cardinality is finite.Note that an immediate corollary of (2) is that the coefficients in (1) are unique for each .The definition of an infinite free basis is similar, except that will have infinitely many elements. However thesum must still be finite, and thus for any particular only finitely many of the elements of are involved.In the case of an infinite basis, the rank of is the cardinality of .

Free module 18

ConstructionGiven a set , we can construct a free -module over . The module is simply the direct sum of copiesof , often denoted . We give a concrete realization of this direct sum, denoted by , as follows:• Carrier: contains the functions such that for cofinitely many (all but finitely

many) .• Addition: for two elements , we define by

.• Inverse: for , we define by .• Scalar multiplication: for , we define by

.A basis for is given by the set where

(a variant of the Kronecker delta and a particular case of the indicator function, for the set ).Define the mapping by . This mapping gives a bijection between and the basisvectors . We can thus identify these sets. Thus may be considered as a linearly independent basis for

.

Universal propertyThe mapping defined above is universal in the following sense. If there is an arbitrary -module

and an arbitrary mapping , then there exists a unique module homomorphismsuch that .

GeneralisationsMany statements about free modules, which are wrong for general modules over rings, are still true for certaingeneralisations of free modules. Projective modules are direct summands of free modules, so one can choose aninjection in a free module and use the basis of this one to prove something for the projective module. Even weakergeneralisations are flat modules, which still have the property that tensoring with them preserves exact sequences,and torsion-free modules. If the ring has special properties, this hierarchy may collapse, i.e. for any perfect localDedekind ring, every torsion-free module is flat, projective and free as well.

See local ring, perfect ring and Dedekind ring.

Free module 19

Notes[1][1] Keown (1975),[2][2] Hazewinkel (1989),

References• Adamson, Iain T. (1972). Elementary Rings and Modules. University Mathematical Texts. Oliver and Boyd.

pp. 65–66. ISBN 0-05-002192-3. MR  0345993 (http:/ / www. ams. org/ mathscinet-getitem?mr=0345993).• Keown, R. (1975). An Introduction to Group Representation Theory. Mathematics in science and engineering

116. Academic Press. ISBN 978-0-12-404250-6. MR  0387387 (http:/ / www. ams. org/mathscinet-getitem?mr=0387387).

• Govorov, V. E. (2001), "Free module" (http:/ / www. encyclopediaofmath. org/ index. php?title=Free_module&oldid=13029), in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4.

External linksThis article incorporates material from free vector space over a set on PlanetMath, which is licensed under theCreative Commons Attribution/Share-Alike License.

Chinese remainder theoremThe Chinese Remainder Theorem is a result about congruences in number theory and its generalizations in abstractalgebra. It was first published in the 3rd to 5th centuries by Chinese mathematician Sun Tzu.In its basic form, the Chinese remainder theorem will determine a number n that when divided by some givendivisors leaves given remainders.For example, what is the lowest number n that when divided by 3 leaves a remainder of 2, when divided by 5 leavesa remainder of 3, and when divided by 7 leaves a remainder of 2? A common introductory example is a woman whotells a policeman that she lost her basket of eggs, and that if she makes three portions at a time out of it, she was leftwith 2, if she makes five portions at a time out of it, she was left with 3, and if she makes seven portions at a time outof it, she was left with 2. She then asks the policeman what is the minimum number of eggs she must have had. Theanswer to both problems is 23.

Theorem statementThe original form of the theorem, contained in the 5th-century book Sunzi's Mathematical Classic (孫 子 算 經) bythe Chinese mathematician Sun Tzu and later generalized with a complete solution called Dayanshu (大 衍 術) inQin Jiushao's 1247 Mathematical Treatise in Nine Sections (數 書 九 章, Shushu Jiuzhang), is a statement aboutsimultaneous congruences.Suppose n1, n2, …, nk are positive integers that are pairwise coprime. Then, for any given sequence of integers a1,a2,…, ak, there exists an integer x solving the following system of simultaneous congruences.

Furthermore, all solutions x of this system are congruent modulo the product, N = n1n2…nk.Hence for all , if and only if .

Chinese remainder theorem 20

Sometimes, the simultaneous congruences can be solved even if the ni's are not pairwise coprime. A solution x existsif and only if:

All solutions x are then congruent modulo the least common multiple of the ni.Sun Tzu's work contains neither a proof nor a full algorithm. What amounts to an algorithm for solving this problemwas described by Aryabhata (6th century; see Kak 1986). Special cases of the Chinese remainder theorem were alsoknown to Brahmagupta (7th century), and appear in Fibonacci's Liber Abaci (1202).A modern restatement of the theorem in algebraic language is that for a positive integer with prime factorization

we have the isomorphism between a ring and the direct product of its prime power parts:

ExistenceExistence can be seen by an explicit construction of . We will use the notation to denote the multiplicativeinverse of as calculated by the Extended Euclidean algorithm. It is defined exactly when and arecoprime; the following construction explains why the coprimality condition is needed.

Case of two equationsGiven the system (corresponding to )

Since , we have from Bézout's identity

This is true because we agreed to use the inverses that came out of the Extended Euclidian algorithm; for any otherinverses, it would not necessarily hold true, but only hold true .Multiplying both sides by , we get

If we take the congruence modulo for the right-hand-side expression, it is readily seen that

But we know that

thus this suggests that the coefficient of the first term on the right-hand-side expression can be replaced by .Similarly, we can show that the coefficient of the second term can be substituted by .We can now define the value

and it is seen to satisfy both congruences by reducing. For example

Chinese remainder theorem 21

General caseThe same type of construction works in the general case of congruence equations. Let be theproduct of every modulus then define

and this is seen to satisfy the system of congruences by a similar calculation as before.

Finding the solution with basic algebra and modular arithmeticFor example, consider the problem of finding an integer x such that

A brute-force approach converts these congruences into sets and writes the elements out to the product of 3×4×5 =60 (the solutions modulo 60 for each congruence):

x ∈ {2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, …}x ∈ {3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, …}x ∈ {1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, …}

To find an x that satisfies all three congruences, intersect the three sets to get:x ∈ {11, …}

Which can be expressed as

Another way to find a solution is with basic algebra, modular arithmetic, and stepwise substitution.We start by translating these congruences into equations for some t, s, and u:• Equation 1: • Equation 2: • Equation 3: Start by substituting the x from equation 1 into congruence 2:

meaning that for some integer s.Plug t into equation 1:

Plug this x into congruence 3:

Casting out fives, we get

meaning that

Chinese remainder theorem 22

for some integer u.Finally,

So, we have solutions 11, 71, 131, 191, …Notice that 60 = lcm(3,4,5). If the moduli are pairwise coprime (as they are in this example), the solutions will becongruent modulo their product.

A constructive algorithm to find the solutionThe following algorithm only applies if the 's are pairwise coprime. (For simultaneous congruences when themoduli are not pairwise coprime, the method of successive substitution can often yield solutions.)Suppose, as above, that a solution is required for the system of congruences:

Again, to begin, the product is defined. Then a solution x can be found as follows.For each i the integers and are coprime. Using the extended Euclidean algorithm we can find integers and such that . Then, choosing the label , the above expression becomes:

Consider . The above equation guarantees that its remainder, when divided by , must be 1. On the other hand,since it is formed as , the presence of N guarantees a remainder of zero when divided by any when .

Because of this, and the multiplication rules allowed in congruences, one solution to the system of simultaneouscongruences is:

For example, consider the problem of finding an integer x such that

Using the extended Euclidean algorithm, for x modulo 3 and 20 [4×5], we find (−13) × 3 + 2 × 20 = 1; i.e., e1 = 40.For x modulo 4 and 15 [3×5], we get (−11) × 4 + 3 × 15 = 1, i.e. e2 = 45. Finally, for x modulo 5 and 12 [3×4], weget 5 × 5 + (−2) × 12 = 1, i.e. e3 = −24. A solution x is therefore 2 × 40 + 3 × 45 + 1 × (−24) = 191. All othersolutions are congruent to 191 modulo 60, [3 × 4 × 5 = 60], which means they are all congruent to 11 modulo 60.Note: There are multiple implementations of the extended Euclidean algorithm which will yield different sets of

, , and . These sets however will produce the same solution; i.e., (−20)2 + (−15)3 +(−24)1 = −109 = 11 modulo 60.

Chinese remainder theorem 23

Statement for principal ideal domainsFor a principal ideal domain R the Chinese remainder theorem takes the following form: If u1, …, uk are elements ofR which are pairwise coprime, and u denotes the product u1…uk, then the quotient ring R/uR and the product ringR/u1R× … × R/ukR are isomorphic via the isomorphism

such that

This map is well-defined and an isomorphism of rings; the inverse isomorphism can be constructed as follows. Foreach i, the elements ui and u/ui are coprime, and therefore there exist elements r and s in R with

Set ei = s u/ui. Then the inverse of f is the map

such that

This statement is a straightforward generalization of the above theorem about integer congruences: the ring Z ofintegers is a principal ideal domain, the surjectivity of the map f shows that every system of congruences of the form

can be solved for x, and the injectivity of the map f shows that all the solutions x are congruent modulo u.

Statement for general ringsThe general form of the Chinese remainder theorem, which implies all the statements given above, can be formulatedfor commutative rings and ideals. If R is a commutative ring and I1, …, Ik are ideals of R that are pairwise coprime(meaning that for all ), then the product I of these ideals is equal to their intersection, and thequotient ring R/I is isomorphic to the product ring R/I1 × R/I2 × … × R/Ik via the isomorphism

such that

Here is a version of the theorem where R is not required to be commutative:Let R be any ring with 1 (not necessarily commutative) and be pairwise coprime 2-sided ideals. Then thecanonical R-module homomorphism is onto, with kernel . Hence,

(as R-modules).

Applications• In the RSA algorithm calculations are made modulo n, where n is a product of two large prime numbers p and q.

1,024-, 2,048- or 4,096-bit integers n are commonly used, making calculations in very time-consuming. Bythe Chinese remainder theorem, however, these calculations can be done in the isomorphic ring instead. Since p and q are normally of about the same size, that is about , calculations in the latterrepresentation are much faster. Note that RSA algorithm implementations using this isomorphism are moresusceptible to fault injection attacks.

• The Chinese remainder theorem may also be used to construct an elegant Gödel numbering for sequences, whichis needed to prove Gödel's incompleteness theorems.

Chinese remainder theorem 24

• The following example shows a connection with the classic polynomial interpolation theory. Let r complex points("interpolation nodes") be given, together with the complex data , for all and .The general Hermite interpolation problem asks for a polynomial taking the prescribed derivatives ineach node :

Introducing the polynomials

the problem may be equivalently reformulated as a system of simultaneous congruences:

By the Chinese remainder theorem in the principal ideal domain , there is a unique such polynomial with degree . A direct construction, in analogy with the above proof for the integernumber case, can be performed as follows. Define the polynomials and .

The partial fraction decomposition of gives r polynomials with degrees such that

so that . Then a solution of the simultaneous congruence system is given by the polynomial

and the minimal degree solution is this one reduced modulo , that is the unique with degree less than n.• The Chinese remainder theorem can also be used in secret sharing, which consists of distributing a set of shares

among a group of people who, all together (but no one alone), can recover a certain secret from the given set ofshares. Each of the shares is represented in a congruence, and the solution of the system of congruences using theChinese remainder theorem is the secret to be recovered. Secret Sharing using the Chinese Remainder Theoremuses, along with the Chinese remainder theorem, special sequences of integers that guarantee the impossibility ofrecovering the secret from a set of shares with less than a certain cardinality.

• The Good-Thomas fast Fourier transform algorithm exploits a re-indexing of the data based on the Chineseremainder theorem. The Prime-factor FFT algorithm contains an implementation.

• Dedekind's theorem on the linear independence of characters states (in one of its most general forms) that if M is amonoid and k is an integral domain, then any finite family of distinct monoid homomorphisms (where the monoid structure on k is given by multiplication) is linearly independent; i.e., every family ofelements satisfying must be equal to the family .

Proof using the Chinese Remainder Theorem: First, assume that k is a field (otherwise, replace the integraldomain k by its quotient field, and nothing will change). We can linearly extend the monoid homomorphisms

to k-algebra homomorphisms , where is the monoid ring of M over k. Then, thecondition yields by linearity. Now, we notice that if are two elements ofthe index set I, then the two k-linear maps and are not proportional to each other(because if they were, then and would also be proportional to each other, and thus equal to each othersince (since and are monoid homomorphisms), contradicting the assumption that theybe distinct). Hence, their kernels and are distinct. Now, is a maximal ideal of forevery (since is a field), and the ideals and are coprime whenever

(since they are distinct and maximal). The Chinese Remainder Theorem (for general rings) thus yieldsthat the map

Chinese remainder theorem 25

given by

for all is an isomorphism, where . Consequently, the map

given by

for all is surjective. Under the isomorphisms , this map corresponds to the map

given by

for every Now, yields for every vector in the image of the map . Since issurjective, this means that for every vector . Consequently, ,QED.

Non-commutative case: a caveatSometimes in the commutative case, the conclusion of the Chinese Remainder Theorem is stated as

. This version does not hold in the non-commutative case, since, as can be seen from the following example

Consider the ring R of non-commutative real polynomials in x and y. Let I be the principal two-sided ideal generatedby x and J the principal two-sided ideal generated by . Then but .

ProofObserve that I is formed by all polynomials with an x in every term and that every polynomial in J vanishes underthe substitution . Consider the polynomial . Clearly . Define a term in R as anelement of the multiplicative monoid of R generated by x and y. Define the degree of a term as the usual degree ofthe term after the substitution . On the other hand, suppose . Observe that a term in q of maximumdegree depends on y otherwise q under the substitution can not vanish. The same happens then for anelement . Observe that the last y, from left to right, in a term of maximum degree in an element of ispreceded by more than one x. (We are counting here all the preceding xs. E.g., in the last y is preceded by xs.) This proves that since that last y in a term of maximum degree ( ) is preceded by only one x.Hence .On the other hand, it is true in general that implies . To see this, note that

, while the opposite inclusion is obvious. Also, we have in general that, providedare pairwise coprime two-sided ideals in R, the natural map

is an isomorphism. Note that can be replaced by a sum over all orderings of of their product(or just a sum over enough orderings, using inductively that for coprime ideals ).

Chinese remainder theorem 26

References• Donald Knuth. The Art of Computer Programming, Volume 2: Seminumerical Algorithms, Third Edition.

Addison-Wesley, 1997. ISBN 0-201-89684-2. Section 4.3.2 (pp. 286–291), exercise 4.6.2–3 (page 456).• Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms,

Second Edition. MIT Press and McGraw-Hill, 2001. ISBN 0-262-03293-7. Section 31.5: The Chinese remaindertheorem, pp. 873–876.

• Laurence E. Sigler (trans.) (2002). Fibonacci's Liber Abaci. Springer-Verlag. pp. 402–403. ISBN 0-387-95419-8.• Kak, Subhash (1986), "Computational aspects of the Aryabhata algorithm" [1], Indian Journal of History of

Science 21 (1): 62–71.• Thomas W. Hungerford (1974). Algebra. Springer-Verlag. pp. 131–132. ISBN 0-387-90518-9.• Cunsheng Ding, Dingyi Pei, and Arto Salomaa (1996). Chinese Remainder Theorem: Applications in Computing,

Coding, Cryptography. World Scientific Publishing. pp. 1–213. ISBN 981-02-2827-9.

External links• Hazewinkel, Michiel, ed. (2001), "Chinese remainder theorem" [2], Encyclopedia of Mathematics, Springer,

ISBN 978-1-55608-010-4• "Chinese Remainder Theorem" [3] by Ed Pegg, Jr., Wolfram Demonstrations Project, 2007.• Weisstein, Eric W., "Chinese Remainder Theorem [4]", MathWorld.• C# program and discussion [5] at codeproject• University of Hawaii System [6] CRT by Lee Lady• Full text of the Sunzi Suanjing [7] (Chinese) — Chinese Text Project

References[1] http:/ / www. ece. lsu. edu/ kak/ AryabhataAlgorithm. pdf[2] http:/ / www. encyclopediaofmath. org/ index. php?title=p/ c022120[3] http:/ / demonstrations. wolfram. com/ ChineseRemainderTheorem/[4] http:/ / mathworld. wolfram. com/ ChineseRemainderTheorem. html[5] http:/ / www. codeproject. com/ KB/ recipes/ CRP. aspx[6] http:/ / www. math. hawaii. edu/ ~lee/ courses/ Chinese. pdf[7] http:/ / ctext. org/ sunzi-suan-jing

Bézout's identity 27

Bézout's identityBézout's identity (also called Bezout's lemma) is a theorem in the elementary theory of numbers: let a and b beintegers, not both zero, and let d be their greatest common divisor. Then there exist integers x and y such that

In addition, i) d is the smallest positive integer that can be written as ax + by, and ii) every integer of the form ax +by is a multiple of d. x and y are called Bézout coefficients for (a, b); they are not unique. A pair of Bézoutcoefficients (in fact the ones that are minimal in absolute value) can be computed by the extended Euclideanalgorithm.Bézout's lemma is true in any principal ideal domain, but there are integral domains in which it is not true.

HistoryFrench mathematician Étienne Bézout (1730–1783) proved this identity for polynomials.[1] However, this statementfor integers can be found already in the work of another French mathematician, Claude Gaspard Bachet de Méziriac(1581–1638).[2][3][4]

Non-uniqueness of solutionsAfter one pair of Bézout coefficients (x, y) has been computed (using extended Euclid or some other algorithm), allpairs may be found using the formula

ExampleLet a = 12 and b = 42, gcd(12, 42) = 6. Then

GeneralizationsBézout's identity can be extended to more than two integers: if

then there are integers such that

and 1) d is smallest positive integer of this form, and 2) every number of this form is a multiple of d.As noted in the introduction, Bézout's identity works not only in the ring of integers, but also in any other principal ideal domain (PID). That is, if R is a PID, and a and b are elements of R, and d is a greatest common divisor of a and b, then there are elements x and y in R such that ax + by = d. The reason: the ideal Ra+Rb is principal and indeed is

Bézout's identity 28

equal to Rd. An integral domain in which Bézout's identity holds is called a Bézout domain.

ProofBézout's lemma is a consequence of the Euclidean division defining property, namely that the division by a nonzerointeger b has a remainder strictly less than |b|. The proof that follows may be adapted for any Euclidean domain. Forgiven nonzero integers a and b there is a nonzero integer d = as + bt of minimal absolute value among all those ofthe form ax + by with x and y integers; one can assume d > 0 by changing the signs of both s and t if necessary. Nowthe remainder of dividing either a or b by d is also of the form ax + by since it is obtained by subtracting a multipleof d = as + bt from a or b, and on the other hand it has to be strictly smaller in absolute value than d. This leaves 0 asonly possibility for such a remainder, so d divides a and b exactly. If c is another common divisor of a and b, then calso divides as + bt = d. Since c divides d but is not equal to it, it must be less than d. This means that d is thegreatest common divisor of a and b; this completes the proof.

Notes[1] Bézout, Théorie générale des équations algébriques (http:/ / books. google. fr/ books?id=FoxbAAAAQAAJ& hl=en& pg=PP5#v=onepage&

q& f=false) (Paris, France: Ph.-D. Pierres, 1779).[3] On these pages, Bachet proves (without equations) “Proposition XVIII. Deux nombres premiers entre eux estant donnez, treuver le moindre

multiple de chascun d’iceux, surpassant de l’unité un multiple de l’autre.” (Given two numbers [which are] relatively prime, find the lowestmultiple of each of them [such that] one multiple exceeds the other by unity (1).) This problem (namely, ax - by = 1) is a special case ofBézout’s equation and was used by Bachet to solve the problems appearing on pages 199 ff.

[4][4] See also:

External links• Online calculator (http:/ / wims. unice. fr/ wims/ wims. cgi?module=tool/ arithmetic/ bezout. en) of Bézout's

identity.• Weisstein, Eric W., " Bezouts Identity (http:/ / mathworld. wolfram. com/ BezoutsIdentity. html)", MathWorld.

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Image Sources, Licenses and Contributors 30

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