Let M be an R module. Note that S is also an R module. The action of c, relative to S, is multiplication by h(c). Thus we can tensor M and S, giving a new R module B.
Let xy be a pair generator of B, where x is in M and y is in S. Now the elements of S act on xy by acting on y. This makes B an S module. By tensoring with S we have performed a base change; the result is B, which is an R module and an S module.
If c is in R, let c act on xy by applying it to either x or y. In the latter case, c acts on y via h(c)*y. This is the same as mapping c into S and then performing the action of S. In other words, the action of R and the action of S are compatible.
(M×N)×W = M×(N×W)
In either case, the module consists of M cross N cross W, with relations that pass the action of R between M and N, and the action of S between N and W. Review the original proof, and tweak it, so that R passes between M and N, and S passes between N and W.
In an earlier section we proved that tensor product was commutative and associative. Thus B1B2 = (M1S)(M2S) = (M1M2)ss, as R modules. In each case the generators are 4-tuples drawn from M1 cross M2 cross S cross S, and the relations of multi-linearity define the same kernel. Of course B1 and B2 are suppose to B S modules. This adds relations that pass the action of S between the two S components of M1S and M2S. What about the right hand side? Treat M1M2S as an S module, using the usual base change convention. Hence S acts on the third component. Tensor this with the ring S, giving a module that is isomorphic to the original. Once again the action of S can be passed between the third and fourth components. The left and right sides are isomorphic, and we can apply a base change before or after a tensor product.
Let S be the quotient ring of R mod its maximal ideal J. Thus S is a field. Change base with respect to S. Thus 0 tensor S = AB tensor S = (AS)(BS). Since BS is an S module it is an S vector space of a certain dimension. Remember the vector spaces are really free modules, and the tensor product of free modules is free, with rank equal to the product of the individual ranks. If this is to be zero, then either AS or BS is zero.
Assume without loss of generality that BS = 0. If BS is 0 as an S module it is 0 as an R module.
apply the formula for tensoring with a quotient ring. Thus BS equals B mod JB, where J is the maximal ideal. Since this is zero, B = JB. The action of J maps B onto B. Since J is the jacobson radical, apply nakiama's lemma to assert B = 0.
If either module is infinitely generated, all bets are off. Let R be the localization of Z about p, i.e. the rationals without p in the denominator. Now the integers mod p and the rationals are both R modules, yet their tensor product is 0. Equate any pair generator xy with xp cross y/p, which becomes 0.
Our result, A or B = 0, becomes true again if the local ring has a nilpotent maximal ideal. Use the same proof and close with a variation of nakiama's lemma.
Here is a wonderful application of base change with respect to a field. If M is a free R module it has a well defined rank. Two free modules that are isomorphic must have the same rank. We derived a rather technical proof for this theorem. We didn't realize it at the time, but we were actually tensoring with the quotient field. Let's prove the same results here, using the theorems at hand. Start with a short exact sequence and tensor with S.
0 → A → B → C → 0
? → AS → BS → CS → 0
Set A = 0 and let B and C be isomorphic free R modules. This leads to isomorphic vector spaces downstairs. We know from linear algebra that AS and BS have the same dimension. Since tensor and direct sum commute, the rank of B is the dimension of BS, and similarly for C and CS. Thus B and C have the same rank.
A free R module has a well defined rank. This is called the invariant dimension property, a term borrowed from linear algebra.
Next let A be unconstrained, while B and C are free. Now BS maps onto CS, and the dimension of BS is at least the dimension of CS, hence the rank of B is at least the rank of C. If B maps onto C it cannot have a smaller rank.
Now if C is spanned by precisely k generators, and k is finite, these generators form a basis. Before we prove this, let's see what goes wrong when k is infinite.
Let C be a free module of infinite rank, and build a set of generators from the basis, but toss in the first basis element twice. This set certainly spans C, and its cardinality agrees with the rank of C, but it does not form a basis, because two of the generators are precisely the same element in C.
Now, let C be free with finite rank k, and consider any k generators that span C. Let g() take the standard basis for C onto the k generators of C. Thus g maps C onto C. We want to show that this is an isomorphism - that the new generators of C look just like the original basis for C.
We have a homomorphism g whose image is all of C; we only need show the kernel is 0.
Suppose there is a nonzero element w in the kernel. Remember that C is a free module, so w ∈ C is nonzero across finitely many components. Select one of these components and look at the nonzero projection x, which is an element of R. Since zero is a local property, there is a prime ideal P in R such that x/1 remains nonzero in the localization RP.
Let S be the localization RP, and tensor 0 → A → C → C → 0 with S. Here A is the kernel of the map from C onto C. Since tensor and direct sum commute, CS becomes a free S module, having the same rank as C. The image of w cross 1 has a nonzero component of x cross 1, and remains nonzero in CS. Since w maps to 0 in C, we have an element w,1 in CS that maps to 0 in CS. Also, the map from CS to CS carries k basis elements onto k generators. These generators span all of C cross 1, and when viewed AS an S module, they span all of cs. Therefore the problem has been reduced to a simpler case. If A is nontrivial, then there is an induced map from CS onto CS with a nontrivial kernel, where R is a local ring. We will prove this is impossible, hence our original hypothesis fails.
Let R be a local commutative ring and let C be a free module that is spanned by a finite set of generators equal to its rank. The map from basis to generators now carries C onto C. Suppose this map has a nontrivial kernel A.
Since C is free it is projective, and the sequence is split exact. Tensor 0 → A → C → C → 0 with S, where S is R mod its maximal ideal. The result remains split exact. Since CS and CS have the same finite dimension, AS has to be zero.
View AS as an R module. Hence A tensor S = 0. Clearly S is finitely generated, since it contains 1. The kernel is spanned, hence generated, by the basis of C, which is finite. My notes don't say so explicitly, but I think we have to assume R is noetherian, whence the kernel is finitely generated. The tensor product of two finitely generated modules over a local ring is 0. We showed above that one of the two modules has to be 0. Since S is nonzero, the kernel is 0. This completes the proof. Any k generators that span C act as a basis for C.
But what if we don't know ahead of time that C is a free R module. Assume R is noetherian, and C is trapped between two R modules of rank n, a lower module Cl and an upper module Cu. Map the basis of Cl onto the generators of C, and suppose there is a kernel A. Write this as 0 → A → Cl → C → 0. Tensor with the localization RP as we did before. Naturally Cl becomes a free RP module of rank n. At the same time, C becomes trapped between two RP modules of rank n. So we have pushed the problem down to a local ring. Mod out by the maximal ideal, and the last two modules become vector spaces of dimension n. The kernel has to be 0, and at this point the proof proceeds as it did before.
If A and B are free modules, tensor 0 → A → B → C → 0 with S, the fraction field of R. I'm getting ahead of myself here, but S is flat, which means tensoring with S keeps the exact sequence exact. (This is described in the next section.) The free R module A becomes a vector space of the same rank - and similarly for B. The vector space AS cannot have a higher dimension than BS, hence the rank of A cannot exceed the rank of B.
This theorem does not require R to contain 1. That's because the fraction field of R does contain 1, in the form of w/w for any nonzero w. Take the fractions of the even integers, for instance, to get the rationals. Everything is invertible, the fraction field is really a field, and the theorems from vector spaces apply.
If R does not contain 1, yet it contains at least one prime ideal P, we can get the invariant dimension property back again. Tensor 0 → 0 → B → C → 0 with R/P and find two isomorphic modules over A ring with no zero divisors. By the above, the two modules have the same rank, hence B and C have the same rank. More generally, if B maps onto C, its rank is at least as large as the rank of C. In our earlier pathological example, where R = R*R, R has no prime ideals, since 0 drags in all of R.
If R contains at least one prime ideal, (it may or may not contain 1), we can resurrect the free submodule theorem. I've never seen this theorem anywhere else, so this is my proof; I hope it's correct.
Suppose a free R module embeds in another free module of lower rank. Let P be a minimal prime ideal. Localize about P, i.e. tensor with RP, and a free RP module embeds in another free RP module of lower rank. We have reduced the problem to a ring that contains 1, and has but one prime ideal P.
The radical of 0 is the intersection of all the primes containing 0, and that is just P. The radical of 0 is also the elements x such that some power of x = 0. These are called nilpotent elements. Thus every x in P is nilpotent.
Since P is maximal, R/P is a field. Take y outside of P and write yz = 1 in R/P. Thus yz = 1+u for some u in P. Since 1+u is a unit, and u is nilpotent, y is a unit. Everything outside of P is a unit. Every x in R is either nilpotent or a unit.
At this point we need a lemma. Any finite set of elements in P is killed by a nonzero element of P. For a single element x, of order n, x is killed by xn-1. Let u kill a finite set S, and bring in the element z. If u kills z we're done. If this is not the case, consider uz. This kills all of S, and if it kills z we're done. If not, consider uz2. Continue as far as necessary, until uzn-1 kills S and z.
Suppose Rn+1 embeds in Rn. In other words, there are n+1 vectors in Rn that are linearly independent. For convenience, arrange them in an n+1 by n matrix. A linear combination of these rows never equals 0, unless the coefficients are all 0.
Imagine subtracting w times the second row from the first. Suppose a linear combination of the original rows, with coefficients ci, yields 0. Add wc1 to c2, and the new rows span 0. If any coefficient other than c2 was nonzero it is still nonzero. If they are all zero (other than c2), then c2 has not change, and is still nonzero. Thus the new rows are linearly dependent.
Of course we can add w times the second row to the first, reversing the process. One matrix represents an embedding iff the other one does.
Suppose a row contains no units. By the previous lemma, something kills that row, and we don't have linear independence.
Assume there is a unit in the upper left. Use gaussian elimination to clear the first column. Everything below this unit is 0.
If the second row has no units the matrix does not represent an embedding, so assume the unit is in the second position. Subtract multiples of the second row from the rows below, thus clearing the second column. Continue this process until there are units down the main diagonal, and zeros below. This includes the bottom row, which is all zeros. Multiply this by 1 and contradict linear independence.
As long as R has a prime ideal, a free module cannot embed in a free module of lesser, finite rank. This result extends to infinite ranks using the corresponding proof from linear algebra.
If B is finite, and A embeds in B, then A is also finite. Furthermore, the ring R (which may or may not contain a prime ideal) is finite, with finitely many copies of R forming A and B. With |B| ≥ |A|, the rank of B is at least the rank of A. (If A = B = 0 then R could be an infinite ring, but no worries; as the rank of both modules is 0.)
Tensor with S, the fraction field of R, and declare BS = 0, as we did above when R was a local ring. Now if B has a nonzero element x, x/1 remains nonzero in BS. This because S has no zero divisors and B is torsion free. However, BS can have no nontrivial elements, hence B = 0.