Projective Modules, Invariant Factor Theorem

Invariant Factor Theorem

Assume the quotient module U/V is killed by t^j.

Let W_i = tⁱU intersect V. Note that W₀ = V, and each successive W_i can only get smaller.

Let Z_i = W_i mod tV.

Remember that V is free of rank n. Mod out V by tV and get the vector space Kⁿ. Now W_i is an R module, and so is Z_i. Since t carries Z_i to 0, Z_i is a K module. This means Z_i is a subspace of our vector space Kⁿ.

Remember that t^jU lies in V. This means t^j+1U lies in tV. With all of W_j+1 in tV, Z_j+1 = 0. The descending chain of vector spaces becomes 0 at j+1.

Start at the smallest nonzero subspace of Kⁿ and give it a basis commensurate with its dimension. Move up to the next Z_i and extend the basis to cover Z_i. repeat until you reach the top, Z₀, which is Kⁿ. Call the resulting basis y. Thus each subspace Z_i is spanned by a subset of our comprehensive basis y.

If y₅ is the fifth element in our basis, it is actually a coset of tV in V. Let's select a coset representative, but not just any representative. We selected y₅ to span a subspace Z₃, which comes from W₃. Select x₅ from W₃ so that x₅ represents the basis element y₅ in Z₃. Do this across the board, and x is a string of representatives in V, which becomes our basis y when reduced mod tV.

Recall the quotient formula for tensor product. Tensoring V with K is the same as V mod tV. Thus tensoring with K carries the elements of x onto our basis y, which spans Kⁿ.

In the previous section we used a technique to demonstrate an isomorphism between modules; let's apply it here. In the following sequence, X represents the submodule of V spanned by the elements of x, and C is the cokernel V/X.

0 → X → V → C → 0

Tensor with K, and the embedding of X into V becomes an embedding of y into Kⁿ, which is an isomorphism. This means C×K = 0. In other words, C = tC. Apply nakiama's lemma, and C = 0. Thus x spans all of V.

If n generators span a free module of rank n, the generators form a basis. Thus x is a basis for the free module V. The same reasoning shows the i^th subset of x forms a basis for W_i.

Each x_i is a power of t times something in U. Recall our example x₅, living in W₃. Such an element is t³ times something in U. Suppose two different elements in U lead to x₅. Their difference is killed by t³, and that contradicts the fact that U is a free module. Thus the preimage is unique. There is a specific b₅ such that t³b₅ = x₅.

If the exponent associated with x_i is e_i, the basis for V is t^e_ib_i.

We would like to step back, and show that b spans all of U. If we can do this, the quotient U/V is well understood. It is isomorphic to the direct product of R mod M^e_i, as i runs from 1 to n. (Some of these may be 0, so there are at most n components in the direct product.) We just need to show b is a basis for U.

Let q be an element of U. Let l be the least exponent such that t^lq lies in V. Thus t^lq lies in W_l. Now W_l is spanned by the first few elements of x, or if you prefer, the first few elements of b multiplied by various powers of t, not to exceed t^l. Some of the basis vectors are multiplied by t^l, since l is minimal. Say b₇ and b₈ are multiplied by t^l. Divide by t^l, and b₇ and b₈ span a point that we will call q′. Surely q′ is spanned by b, so if q-q′ is spanned, then so is q. Since the terms with t^l have been subtracted away, q-q′ can be multiplied by t raised to a lesser power, such as t^l-2, giving an element in W_l-2. By induction, the points of U that lead to earlier submodules W_i are spanned by b, hence q is spanned by b.

Of course we need to start the inductive process. If q is already in V, i.e. already in W₀, it is spanned by b, since b spans all of V.

Once again n generators b₁ through b_n span a free module of rank n, and this implies b is a basis for U. As mentioned earlier, U/V is the direct product of quotient rings R/M^e_i. This is an R module, but it also happens to be a ring.

Unique

When multiplied by a power of t, U/V is going to produce the same submodule, no matter the representation. So if one representation has 7 instances of R/M^j, where j is maximal, multiply by t^j-1 and get K⁷. The other representation must also produce K⁷, so it has exactly 7 instances of R/M^j, where j is maximal.

The two representations lead to isomorphic submodules, 7 instances of R/M^j. Mod out by this "slice" in common, and the remaining summands, the components R/Mⁱ for i < j, must be isomorphic. Repeat the above process, multiplying by lower powers of t, and at each step the resulting vector spaces have the same rank, hence the number of quotient rings, per exponent, is the same. The representation of U/V is unique.

When R is a PID

Since R is an integral domain, tensoring with a fraction ring, such as R_M, embeds U into U_M, and also embeds V into V_M. Use the above to find a basis b in U_M that, when multiplied by appropriate powers of t, leads to a basis for V_M.

If the element b₁ in such a basis consists of a numerator from U and a denominator from R-M, change the denominator to 1. This does not change the submodule spanned by b₁, hence the result remains a basis for U_M. Furthermore, the adjusted basis, when multiplied by the appropriate powers of t, remains a basis for V_M. Thus the entire basis b can be "normalized" to lie in U.

If the adjusted elements of b, lying in U, are not linearly independent, embed the result in U_M and b would not be a basis for U_M. So if b spans U, we have a basis.

Suppose b does not span U. In the following exact sequence, B is the free R module spanned by b, and C is the cokernel.

0 → B → U → C → 0

Tensor with the fraction ring R_M, and the map from B into U becomes an isomorphism. That means C tensor R_M is 0. Suppose C has a nonzero element x. Now x/1 is equivalent to 0 in the tensor product. There are no zero divisors in R, so some z in R kills x, and z divides 1. In other words, a unit kills x; but this means 1 kills x. Modules are assumed unitary, so x must be 0, and C is 0. The basis b is a basis for U.

Similar reasoning shows that b, multiplied by the appropriate powers of t, becomes a basis for V.

With the basis in hand, U/V becomes the direct product of quotient rings R/Mⁱ, as we saw before. The representation is unique, as described above.

When n is 0

Assume Q is finitely generated, with n generators, and let U = Rⁿ. A homomorphism carries U onto Q, and has kernel V. Let R be a pid, and V becomes a free submodule of U.

Since R is an integral domain, the rank of V is no larger than the rank of U. Suppose it is smaller. Tensor with R/M^j, and a free module of rank less than n maps into a free module of rank n, (it cannot map onto a module of rank n), so Q tensor R/M^j is nonzero. Something in Q is not killed by t^j, which is a contradiction. Therefore V has rank n, just like U.

The conditions of the invariant factor theorem are met, and Q is a finite direct product of quotient rings. The number of component rings does not exceed the number of generators of Q.