# Finitely Generated Modules and Free Modules

Let $$R$$ be a ring and $$M$$ an $$R$$-module. Note if $$M \cong R$$ as $$R$$-modules we may regard $$M$$ as a ring isomorphic to $$R$$: if $$\theta:M\rightarrow R$$ we may define multiplication by $$x y = \theta(x) y$$ for all $$x,y \in M$$.

$$M$$ is free if $$M$$ is isomorphic to an $$R$$-module of the form $$\oplus_{i\in I}M_i$$ where each $$M_i \cong R$$. Sometimes this is denoted $$R^{(I)}$$. A finitely generated free module is isomorphic to $$R \oplus ... \oplus R$$ where there are $$n$$ summands, and is written $$R^n$$. By convention $$R^0$$ is the zero module.

Proposition: $$M$$ is a finitely generated $$R$$-module $$\iff$$ $$M$$ is isomorphic to a quotient of $$R^n$$ for some $$n \ge 0$$.

Proof: ($$\Rightarrow$$) Suppose $$M = \langle x_1,...,x_n \rangle$$. Define $$\phi:R^n \rightarrow M$$ by

$(a_1,...,a_n) \mapsto a_1 x_1 +...+a_n x_n$

Then $$\phi$$ is a surjective $$R$$-module homomorphism, thus $$M \cong R^n / ker \phi$$.

($$\Leftarrow$$) Suppose $$M$$ is isomorphic to a quotient of $$R^n$$ for some $$n$$. Then we have an onto module homomorphism $$\phi : R^n \rightarrow M$$. Set $$e_i = (0,...,0,1,0,...,0)$$, (the $$i$$th coordinate is 1). Then the $$e_i$$ generate $$R^n$$ thus the $$\phi(e_i)$$ generate $$M$$. $$∎$$.

Proposition: Let $$M$$ be a finitely generated $$R$$-module, $$I \triangleleft R$$ be an ideal, and $$\phi \in Hom_R(M,M)$$ be a homomorphism with $$\phi(M) \subset I M$$. Then $$\phi$$ is the root of a monic polynomial with nonleading coefficients from $$I$$.

Proof: Suppose $$M = \langle x_1,...,x_n \rangle$$. Now

$I M = \{ \sum_{j=1}^n b_j y_j | m\ge 1, b_j \in I, y_j \in M \}$

Since every element of $$M$$ can be written as a linear combination of the $$x_i$$, and since $$I$$ is an ideal, we have

$I M = \{ \sum_{i=1}^n a_i x_i | n\ge 1, c_i \in I\}$

For each $$i$$ write $$\phi(x_i) = \sum_{j=1}^n a_{i j} x_j$$ for some $$a_{i j} \in I$$.

Now we can use the Cayley-Hamilton Theorem: let $$A$$ is the matrix $$(a_{i j})$$ and let $$\Chi(x)$$ be its characteristic polynomial, that is $$\Chi(x) = |x I - A |$$. Then by the Cayley-Hamilon Theorem $$\Chi(A) = 0$$ hence $$\Chi(\phi) = 0$$.

Alternatively, we may write

$\sum_{j=1}^n (\delta_{i j} \phi - a_{i j}) x_j = 0$

where $$\delta_{i j}$$ is the Kronecker delta. By multiplying on the left by the adjoint of $$\delta_{i j}\phi - a_{i j}$$ we see that $$det (\delta_{i j} \phi - a_{i j})$$ must map each $$x_i$$ to zero, hence is the zero endomorphism of $$M$$. The determinant is an equation of the required form. $$∎$$

Corollary: Let $$M$$ be a finitely generated $$R$$-module and $$I$$ be and ideal of $$R$$ with $$I M = M$$. Then for some $$x\in 1+I$$ we have $$x M = 0$$.

Proof: Take $$\phi$$ to be the identity in the previous theorem, thus we have $$\phi^n + a_{n-1} \phi^{n-1} + ... + a_0 = 0$$ for some $$a_i \in I$$. Then set $$x = 1+a_{n-1} +...+a_0$$.

Nakayama’s Lemma: Let $$M$$ be a finitely generated $$R$$-module and $$I\triangleleft R$$ be an ideal such that $$I\subset J(R)$$ where $$J(R)$$ is the Jacobson radical of $$R$$. Then $$I M = M \implies M = 0$$.

Proof: By the previous corollary $$x M = 0$$ for some $$x \in 1 + J(R)$$. Then $$1-x \in J(R)$$, thus $$x = 1 - (1-x)$$ is a unit in $$R$$, hence $$M = x^{-1} x M = 0$$.

Alternatively, suppose $$M$$ is nonzero. Then let $$u_1,...,u_m$$ be a minimal set of generators of $$M$$. Then since $$u_n \in I M$$ we have $$u_n = a_1 u_1 +...+ a_n u_n$$ for some $$a_i \in I$$. Then

$(1-a_n) u_n = a_1 u_1 +...+ a_{n-1} u_{n-1}$

and since $$a_n \in I \subset J(R)$$ we have that $$1-a_n$$ is a unit, implying that $$u_n \in \langle u_1,...,u_{n-1} \rangle$$ contradicting the minimality of the set of generators. $$∎$$

Corollary: Let $$M$$ be a finitely generated $$R$$-module. Let $$N$$ be a submodule of $$M$$ and $$I$$ an ideal contained in $$J(R)$$. Then $$M = I M + N \implies M = N$$.

Proof: Note $$I (M/N) = (I M + N)/N = M/N$$ so by the lemma $$M/N = 0$$.

Let $$R$$ be a local ring with maximal ideal $$I$$. Let $$K = R/I$$ be the residue field of $$R$$. Let $$M$$ be a finitely generated $$R$$-module. Note $$I \subset Ann(M / I M)$$ so $$M/I$$ is can be viewed as an $$R/I$$-module, that is $$M/I$$ is a finite-dimensional $$K$$-vector space.

Proposition: Let $$x_1,...,x_n\in M$$ be elements such that $$\{x_1+ I M,...,x_n + I M\}$$ is a basis for the vector space $$M/I M$$. Then $$M = \langle x_1,...,x_n \rangle$$.

Proof: Let $$N = \langle x_1,...,x_n \rangle$$. Then the map $$N \rightarrow M \rightarrow M/ I M$$ maps $$N$$ onto $$M/ I M$$ thus $$N + I M = M$$ so by the above corollary $$N = M$$.

Ben Lynn blynn@cs.stanford.edu 💡