◀ SubmodulesSums & Products ▶
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Operations on Submodules

Let \(M\) be a module. Let \(X\) be a subset of \(M\). Then define \(\langle X \rangle\) to be the submodule of \(M\) generated by \(X\), that is, the intersection of all submodules of \(M\) containing \(X\). (Note that the intersection of modules is itself a module.)

Let \(\{ M_i | i \in I\}\) be a family of submodules of \(M\) (for some indexing set \(I\)). Define their sum as for ideals: \(\sum M_i\) consists of all finite sums \(\sum x_i\) where \(x_i \in M_i\) and almost all \(x_i\) are zero. Note \(\sum M_i = \langle union M_i \rangle\).

Thus the submodules of \(M\) form a complete lattice with respect to inclusion (the glb is interesction and lub is the sum).

Isomorphism Theorems

1. If \(L \supset M \supset N\) are \(R\)-modules then

\[ (L/N)/(L/M) \cong L/M \]

2. If \(M_1, M_2\) are submodules of an \(R\)-module \(M\) then

\[ (M_1 + M_2)/M_1 \cong M_2 / M_1 \cap M_2 \]

Proof:

  1. The map \(\theta:L/N \rightarrow L/M\) given by \(x+N \mapsto x+M\) is a surjective \(R\)-module homomoprhism with kernel \(M/N\), thus the result follows by the fundamental homomoprhism theorem.

  2. Apply the same reasoning to the map \(\theta:M_2 \rightarrow (M_1 + M_2)/M_1\) given by \(x \rightarrow x + M_1\).

Let \(I\) be an ideal of \(R\). Then define the product \(I M\) to be the set of all finite sums \(\sum a_i x_i\) where \(a_i \in I, x_i \in M\). It is a submodule of \(M\).

Let \(N, P\) be submodules of \(M\). Define

\[ (N:P) = \{a \in R | a P \subset N\} \]

(similar to ideal quotients). It is an ideal of \(R\). Define the annihilator of \(M\) by

\[ \mathrm{Ann}(M) = (0 : M) \]

If \(I\) is an ideal of \(R\) contained in \(\mathrm{Ann}(M)\) then \(M\) can be viewed as an \(R/I\)-module by defining \((x+I) m = x m\) for all \(x \in R, m \in M\). This map is well-defined since \(I M = 0\).

An \(R\)-module is faithful if \(\mathrm{Ann}(M) = 0\). Note \(M\) is faithful as an \(R/\mathrm{Ann}(M)\)-module.

The following can be easily verified:

  1. \( \mathrm{Ann}(M+N) = \mathrm{Ann}(M) \cap \mathrm{Ann}(N) \)

  2. \( (N:P) = \mathrm{Ann}((N+P)/N) \)

Let \(x \in M\). Then define \(R x\) to be \(\langle x \rangle\), that is the set of all \(a x\) where \(a \in R\). A set \(X \subset M\) is a set of generators of \(M\) if \(M = \langle X \rangle\) (so every element of \(M\) can be written as a linear combination of elements of \(X\)). If \(M\) has a finite set of generators then \(M\) is said to be finitely-generated.

Ben Lynn blynn@cs.stanford.edu 💡