Modules, Conductor Ideal

The Product of an Ideal and a Module

For any left ideal H, H*M is the submodule generated by the products xy, where x is in H and y is in M. The products themselves are not sufficient; we must consider all finite sums of these products, in order to generate the submodule. Let's consider an example.

Set R = F[x,y], the polynomials in x and y with coefficients taken from the field F, say the reals. Let x and y generate H. Thus H contains the polynomials with 0 constant term. Let the module M be free with basis e₁ and e₂. The product submodule H*M has generators xe₁, xe₂, ye₁, ye₂, and must include the element xe₁+ye₂. Suppose this is the product of some polynomial in H and something in M. The polynomial must divide both x and y, which is impossible. Thus H*M is more than just products of elements. We saw the same thing when multiplying ideals. The product H² includes x²+y², which cannot be the product of two polynomials taken from H, assuming the coefficients are real numbers.

To show associativity of A*B*M, where A and B are ideals and M is a module, take finite sums of products of elements from the two ideals, then apply that to the module, and compare that with finite sums of products from B*M, and apply A on the left. In either case the result is the same; all finite sums of triple products from A B and M. Note that A could be a left ideal, but B must be a two sided ideal.

Conductor, Annihilator

Given submodules H and J of a left module M, the conductor ideal [H:J] is the set of elements y in R such that y*J lies in H. Note that this is a left ideal.

The annihilator of J is the conductor ideal [0:J]. This is the left ideal that drives J into 0.

Ideals as R Modules

When the ring R is viewed as a left R module, submodules and left ideals are synonymous. Some of these results can be generalized.

Let R and S be rings, and let a nontrivial ring homomorphism f carry R into S. This makes S an R module. Given x in R, apply the homomorphism f, then multiply by f(x) on the left.

Equivalently, let S be a bimodule, with R on the left and S on the right. Since all modules are unitary, 1*1 = 1, and R maps into S through 1. Let S act on S (from the right) by right multiplication.

The map from R into S respects addition and multiplication, and is a ring homomorphism. Since S is a bimodule, the action of x on y, or x*y, equals x*(1*y) = (x*1)*y = f(x)*y. Hence the action of R is left multiplication by the image of R, as described above.

Often R embeds in S, i.e. R is a subring of S.

Let T be a left ideal in S, possibly all of S. Since R*T lives in T, T is an R submodule.

Now let R and S be commutative. Let H be an ideal in R, possibly all of R. Let U = H*T, the submodule spanned by elements of H times elements of T. We will show that U is an ideal in S.

Since U is a submodule it is closed under addition, as an ideal should be.

Let c be an element of S and let xd be an element of U, where x is in H and d is in T. Consider the product (xd)c. We want this to lie in U. This is the same as x(dc). Since dc lies in T, xdc lies in U, and U is an ideal.