Commutative Algebra - Finitely Generated Modules and Free Modules

Finitely Generated Modules and Free Modules

Let $R$ be a ring and $M$ an $R$ -module. Note if $M ≅ R$ as $R$ -modules we may regard $M$ as a ring isomorphic to $R$ : if $θ : M \to R$ we may define multiplication by $x y = θ (x) y$ 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 $M_{i} ≅ R$ . 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 $\Leftrightarrow$ $M$ is isomorphic to a quotient of $R^{n}$ for some $n \geq 0$ .

Proof: ( $\Rightarrow$ ) Suppose $M = ⟨ x_{1}, . . ., x_{n} ⟩$ . Define $ϕ : R^{n} \to M$ by

(a_{1}, . . ., a_{n}) \mapsto a_{1} x_{1} + . . . + a_{n} x_{n}

Then $ϕ$ is a surjective $R$ -module homomorphism, thus $M ≅ R^{n} / \ker ϕ$ .

( $\Leftarrow$ ) Suppose $M$ is isomorphic to a quotient of $R^{n}$ for some $n$ . Then we have an onto module homomorphism $ϕ : R^{n} \to M$ . Set $e_{i} = (0,...,0,1,0,...,0)$ , (the $i$ +++th coordinate is 1). Then the $e_{i}$ generate $R^{n}$ thus the $ϕ (e_{i})$ generate $M$ . $▪$ +.

Proposition: Let $M$ be a finitely generated $R$ -module, $I ◃ R$ be an ideal, and $ϕ \in {Hom}_{R} (M, M)$ be a homomorphism with $ϕ (M) \subset I M$ . Then $ϕ$ is the root of a monic polynomial with nonleading coefficients from $I$ .

Proof: Suppose $M = ⟨ x_{1}, . . ., x_{n} ⟩$ . Now

I M = {\sum_{j = 1}^{n} b_{j} y_{j} ∣ m \geq 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 \geq 1, c_{i} \in I}

For each $i$ write $ϕ (x_{i}) = \sum_{j = 1}^{n} a_{i j} x_{j}$ 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 $Chi (ϕ) = 0$ .

Alternatively, we may write

\sum_{j = 1}^{n} (δ_{i j} ϕ - a_{i j}) x_{j} = 0

where $δ_{i j}$ is the Kronecker delta. By multiplying on the left by the adjoint of $δ_{i j} ϕ - a_{i j}$ we see that $\det (δ_{i j} ϕ - a_{i j})$ 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 $ϕ$ to be the identity in the previous theorem, thus we have $ϕ^{n} + a_{n - 1} ϕ^{n - 1} + . . . + a_{0} = 0$ 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 $I ◃ R$ be an ideal such that $I \subset J (R)$ where $J (R)$ is the Jacobson radical of $R$ . Then $I M = M \Rightarrow M = 0$ .

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 $u_{n} \in ⟨ u_{1}, . . ., u_{n - 1} ⟩$ 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 $M = I M + N \Rightarrow M = N$ .

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 $M = ⟨ x_{1}, . . ., x_{n} ⟩$ .

Proof: Let $N = ⟨ x_{1}, . . ., x_{n} ⟩$ . Then the map $N \to M \to M / I M$ maps $N$ onto $M / I M$ thus $N + I M = M$ so by the above corollary $N = M$ .