## 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

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

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

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

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

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 $N = M$.