- Overview
- The Submodule of a Free Module is Free
- Torsion, Order
- Finitely Generated and Torsion Free is Free
- The Structure of a Finitely Generated Module

This chapter explores modules over a ring R, where R is a pid. Since R is commutative, there is no distinction between left and right modules; they are just modules.

The focus of this chapter is the characterization of a finitely generated module over a pid.
If R is the integers, for instance, which happens to be a pid,
then R modules are abelian groups.
Every finite abelian group has
already been characterized;
G is the direct product of cyclic groups, each cycle a prime power.
Change prime number to prime element,
and change finite to finitely generated,
and the same is true for modules over a pid.
Since R could be any pid, not just **Z**, this is a substantial generalization.

If M is finitely generated, M is noetherian. This means all submodules and quotient modules are noetherian, and finitely generated. So I will assert, from time to time, that various modules derived from M are finitely generated.

Let R be a pid and let F be a free module over R with basis b. Let M be a submodule of F.

Well order the basis b, so that we can consider the basis elements b0 b1 b2 b3 etc in order. Yes, this requires the axiom of choice, unless b is countable, or can otherwise be well ordered.

For any ordinal j, let Fj be the free submodule of F spanned by b0 through bj.

Let Mj be the intersection of Fj with M. These are the members of M that do not use any basis elements beyond bj.

Let Pj be the projection of Mj onto the jth copy of R. Note that Pj is an ideal of R. Since R is a pid, let gj generate Pj. If gj = 0 there is nothing to do. If gj is nonzero, let uj be anything in Mj whose jth component is equal to gj. Since uj lives in Mj, it has nonzero coefficients on b, up to gjbj, but not beyond bj. As we shall see, uj is a basis for M, whence M is free.

Demonstrate linear independence first. Suppose a finite linear combination ∑ cjuj = 0. Look at the last term, the largest j. Only uj contributes gj to the jth component. Thus cj has to be zero, which is a contradiction. The vectors uj are independent by construction. Each one brings in a component that was not there before, like vectors building a lower triangular matrix.

Any linear combination of vectors ending in uj produces something in Mj, and no smaller. This is because uj includes something times bj, and everything else lies below.

Try to span some x in M, where x is in Mj. Thus x employs basis elements up to bj, and no farther. The span must include uj, and cannot include anything higher, else x lives in some Mk for k > j. Either x is spanned by u0 through uj, or it is not.

Write the jth component of x as sbj, for some s in R. If s is 0 then x belongs to a smaller Mj, hence s is nonzero. Write s as tgj. Let y = x-tuj. By induction, y is spanned, and therefore x is spanned as well.

The arbitrary submodule M of F has a basis, and is free. Furthermore, the rank of M is less than or equal to the rank of F.

Set R to the integers, which is a pid, and the subgroup of a free abelian group is free abelian.

Let M be a module over an integral domain R, and let c be a member of M. Let H be the annihilator of c. In other words, H is the ideal in R that drives c to 0.

If H is nonzero, c is a torsion element. Thus a torsion element is a bit like a zero divisor. There are no zero divisors in R, it's an integral domain, but within the module M, c is killed by something in R.

The torsion submodule MT is the submodule consisting of all torsion elements in M. Verify this is a submodule. If c is killed by x then yc is also killed by x, (xyc = yxc = 0), so scaling is no problem. If x kills c and y kills d, then xy kills c+d. Since R is an integral domain, xy is nonzero, and c+d is torsion. This makes MT a submodule.

A module is torsion free if it has no torsion elements, and torsion if it equals its torsion submodule.

Let h be a module homomorphism on M. If xc = 0 then xh(c) = h(xc) = 0, and the image of a torsion element c is torsion. Turn this around and the preimage of a torsion free element is torsion free.

If h(c) is torsion then h(xc) = 0, and xc lies in the kernel of h.

As a special case, consider M/MT. If h(c) is torsion then c lies in the kernel, which is torsion, hence c is torsion. Therefore the image is torsion free.

Let R be a pid and let p be a prime element, such that pi kills c. Let x generate the ideal that annihilates c. Thus x divides pi, and since R is a ufd, x = pj for some j ≤ i.

The generator of the annihilator of c is called the order of c. It is zero iff c is torsion free. The order of c cannot be a unit, unless c = 0, which isn't very interesting.

Of course an ideal kills c, but in a pid, this ideal is principal,
generated by a prime element in R that is unique up to associates.
You can use a canonical associate, like +5 for the ideal 5**Z** in the integers,
or you can just say the order is the ideal, and carry this along.
This is superior in some ways, because it allows R to be dedekind, which is more general than a pid.

If x is an element of R, let M be the quotient R mod {x}. This is an R module that is killed by x. The module, as a whole, has order x, or if you prefer, 1 in M has order x. Again, you might change x to a canonical associate, or think of x as the principal ideal generated by x.

Let R be a pid and let M be a torsion free R module. Furthermore, let M be finitely generated, with generators g1 g2 g3 etc.

Start with g1. Since M is torsion free, nothing kills g1, hence g1 spans a free submodule of M.

Bring in the subsequent generators, g2 g3 etc, one by one, as long as the resulting submodule is free. Let S be the largest free submodule in this ascending chain. The next generator produces a larger submodule that is not free. Call this troublesome generator g, and let J be the submodule spanned by S and g.

By assumption, g and S are linearly dependent. Find a nontrivial linear combination that equals 0. For some x in R, xg lies in S. Since g is outside of S, x is not 1.

Multiplication by x implements a module homomorphism on M. Since S is free, x drives S into S. Bring in g, and x drives J into S. The image is a submodule of a free module, and is free. The kernel is those elements of J that are killed by x. Since all of M is torsion free, nothing is killed by x. There is no kernel, and J is isomorphic to its image. J is free after all.

All the generators can be folded in, and M is free. Furthermore, the rank of M is bounded by the number of generators of M.

In contrast, **Q** is an infinitely generated torsion free **Z** module that is
not free.

If R is a pid, and M is a finitely generated R module, then M is a finite direct product of cyclic submodules, and this decomposition is unique up to isomorphism.

Remember, a cyclic submodule is generated by a single element, e.g. R*c for some c in M, and M is isomorphic to R/H, where H is the annihilator of c. In this case H is generated by some principal element x, so we may write the cyclic module R*c as R/x.

When R = **Z**, we are talking about **Z** modules, or abelian groups.
In this case a cyclic module is the integers, or the integers mod n for some positive integer n.
A finitely generated abelian group is the direct product of these cyclic groups.

Let Q be the quotient module M/MT. Remember that Q is torsion free. And Q is finitely generated via the generators of M. That makes Q a free module.

If M has n generators, then M is the homomorphic image of a free module of rank n. apply a second homomorphism from M onto Q, and Q, a free module, is the image of a free module of rank n. Therefore the rank of Q is at most n. Whatever its rank, it is well defined, and determined only by the structure of M.

Select a basis for Q, then let b1 b2 b3 etc be the preimage of these basis elements in M. It doesn't matter how you choose the preimages, as long as they induce a basis for Q. The preimage of torsion free is torsion free, and that means b spans a finitely generated torsion free submodule of M, which is free. If a linear combination of elements of b is 0, the same is true in Q, and that is a contradiction. Thus b is a basis for a free module in M. This free module is isomorphic to Q; let's just call it Q. It is after all a copy of Q in M.

Q maps forward, via the homomorphism M/MT, to a perfect copy of Q. Does this remind you of a semidirect product? Sure, but when groups are abelian, a semidirect product is a direct product. M is the direct product of Q and MT. Seen another way, every x in Q maps to a unique element in the quotient module, and thus represents a unique coset x+MT. The additive group is abelian, thus giving a direct product. The task remaining, then, is to analyze MT.

For notational convenience, let M stand for MT, a torsion module. I'm setting Q to the side for now.

For each prime element p in R, let Wp be the set of elements in M whose order is some power of p. Remember that the order of c in M is pk if the ideal pk*R is the annihilator of c. Think of p as a prime element generating a prime ideal, or as the prime ideal itself. Verify that Wp is a submodule.

If pk kills c, then it kills yc. The order of yc is a factor of pk, which is still a power of p. That takes care of scaling.

If pk kills c and pj kills d, then pj+k kills c+d, and the order of c+d is a factor of pj+k, which is a power of p.

The order of c in M depends only on the structure of M as an R module, hence the various submodules Wp are well defined. They are uniquely determined by M. When R is the integers and M is a finite abelian group, these are the p sylow subgroups.

The generators of M have specific orders, each carrying a finite number of primes, and since there are finitely many generators, there are finitely many submodules Wp to consider. We want to show that M is the direct product of these submodules.

If the generators are spanned then M is spanned. Consider a generator g, having order x, where x is the product of pi to the ki, over the relevant primes pi. If there is just one prime involved then g is part of Wp, and is spanned. Otherwise let xi be x divided by pi to the ki, that is, the product of the other primes. These values have no gcd, and some linear combination sixi equals 1. (This is bezout's identity.) (If you are thinking about ideals instead of elements, perhaps in a dedekind world, the coprime ideals xi span an ideal that contains them, which is R, and once again a linear combination of elements drawn from these ideals is equal to 1.) Write 1*g as the sum of sixig. If the ith term in the sum is multiplied by pi to the ki, the result is sixg, which is 0 (since xg is 0). The ith term is killed by a power of pi, and belongs to the ith submodule. Therefore g is spanned by these submodules. This holds for each generator, hence the finite collection of submodules Wp spans M.

Next, demonstrate linear independence. Suppose b is in Wp and in V, where V is the span of Wq for each q ≠ p. Since b is in Wp its order is a power of p. Yet b also belongs to V. Let z be the product of qi to the ki for all q ≠ p. Note that z kills each Wq, hence z kills all of V. In particular, z kills b. The ideal that kills b contains z and some power of p. These to elements are relatively prime, hence they span 1. The order of b is 1, and b = 0. All the submodules Wp are linearly independent, and M is the direct product of these submodules.

The task remaining, then, is to understand a module, M once again for notational convenience, having order pk.

The rest of the proof is a consequence of the invariant factor theorem, but that is many chapters ahead, so let's press on, using a technique similar to that used to deconstruct the p sylow subgroup.

Since M is part of a finitely generated noetherian module, it too is finitely generated. Yet M is more than an R module. Since pk kills M, M is a well defined S module, where S = R/pk. The action of R on M is exactly the same as the action of S on M, and all the submodules of M are the same.

The ideals of S are the ideals of R containing pk, which have generators dividing pk, which (by unique factorization) are the powers of p with exponent ≤ k. This is a finite chain of ideals, hence S is noetherian and artinian. The finitely generated module M is also noetherian and artinian, and has a unique composition series.

A simple module corresponds to a maximal ideal in S, and there is only one such ideal in S, generated by p. The quotient S/p is a field that I will call F. Thus each simple module in the composition series is F.

By induction, the penultimate submodule, which I will call U, is a direct product of cyclic modules R/pj for various values of j. Let y be a nontrivial cosrep of U in M. Any coset will do, since the quotient M/U is a simple module F.

Since y is an element of F, it looks like an element of R/p, and py is 0 in R/p, hence py lies in U. py attains a certain value in each quotient ring, in each component of U.

Suppose py lies inside the maximal ideal of the third component. Thus py produces px. Subtract x from y. Since x is in U, this does not change the coset of y; but py now produces 0 in the third component. Do this across the board, and py is 0 in each component, or it is a nontrivial cosrep for the maximal ideal pj.

If py is 0 in all components then y generates an independent submodule isomorphic to F. This is another cycle in the direct product. Otherwise there is more work to do.

To illustrate, assume py is nonzero in the first three components, and 0 in the rest, and assume the first component has the longest maximal ideal. Let py = x in the first component. Relabel the elements of the first component, so that x is 1. Why is this ok; why is this an isomorphism? Multiplication by x is a perfectly good module homomorphism that takes us from 1 to x. Since x is not divisible by p, x and pj are coprime. Apply bezout's identity, and wx + some multiple of pj = 1. Within the quotient ring R/pj, w is the inverse of x. The relabeling is an isomorphism; so assume yp is 1 in the first component. Do the same in the second and third components, and yp = [1,1,1,0,0,0,0,…].

The order of y is pj+1, and since pk kills all of M, j < k.

Let T be the quotient ring R/pj+1. Build an isomorphism between T and the submodule generated by y. Map 1 in T to y, and everything follows from there. Thus p maps to py, which was previously denoted [1,1,1]. The multiples of p in T correspond 1 for 1 with the elements in the first component, generated by 1, and the other two components ride along in parallel. Therefore T, generated by y, is part of M.

Replace the first component with T, a cycle of order pj+1. The second and third components are tied to the first by T, yet within U, they can run independently. So retain the generators [0,1,0] and [0,0,1]. These span the second and third cycles, just as they did before, so that these cycles can, once again, run independently of the first cycle.

That completes the inductive step. M is now a direct product of quotient rings, where each component is R/pj for some j.

In summary, a finitely generated module over a pid is the direct product of a free module whose rank is bounded by the number of generators, times a finite number of cyclic modules, which are quotients of various prime powers in R.

The separation of M into its cyclic modules is determined by the structure of M alone. MT consists of elements killed by things in R, and the quotient follows from there. Two isomorphic modules would produce isomorphic torsion submodules, and isomorphic free quotient modules having the same rank.

The primes of R that kill various submodules in MT are also determined by M. Focusing on one of these primes p, the number of cyclic components of length j or less is determined by those elements that are killed by pj. An isomorphic module would produce the same direct product of cyclic components at every level, from p to pk.

M has been separated into a direct product of cyclic submodules, and this decomposition corresponds uniquely to M, up to isomorphism.