# 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 \(N = M\).

*blynn@cs.stanford.edu*💡