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An irreducible submodule of a Noetherian module is primary

Statement

Let $R$ be a commutative ring, let $M$ be a Noetherian left $R$ -module, and let $Q ⊊ M$ be irreducible, meaning that whenever $Q = A \cap B$ with submodules $A, B \subseteq M$ , then $Q = A$ or $Q = B$ . Then $Q$ is primary.

Facts & Assumptions

Given: A commutative ring $R$ , a Noetherian left $R$ -module $M$ , and an irreducible proper submodule $Q ⊊ M$ .

[L1]

In a short exact sequence, a quotient of a Noetherian module is again Noetherian (Noetherian and Artinian conditions are each exact in short exact sequences).

[L2]

The quotient module $M / Q$ is formed from the cosets of $Q$ (Quotient module $M / N$ with scalar multiplication on additive cosets).

[L3]

A proper submodule is primary exactly when the classical power condition of the previous lemma holds (Primary submodules are exactly quotients with nilpotent zero divisors).

[L4]

If every submodule is finitely generated, every ascending sequence of submodules stabilizes. This implication uses no choice principle (Finite generation, ACC, and maximal-condition characterizations of Noetherian modules, condition 1 implies condition 2).

Proof

technique · direct

1.1L1L2givenalgebra

Let $N = M / Q$ . By [L1], the quotient module $N$ is Noetherian. The submodule $0 \subset N$ is irreducible: if $0 = A^{'} \cap B^{'}$ in $N$ , then taking inverse images in $M$ gives $Q = A \cap B$ with $Q \subseteq A, B$ , so the irreducibility of $Q$ forces $A = Q$ or $B = Q$ , hence $A^{'} = 0$ or $B^{'} = 0$ .

2.1L4step 1.1choosealgebra

Let $x \in R$ be a zero divisor on $N$ , and write $K_{k} : = {y \in N : x^{k} y = 0}$ . Since $R$ is commutative, multiplication by $x^{k}$ is $R$ -linear, so $K_{k}$ and $x^{k} N$ are submodules. Also $K_{1} \neq 0$ and $K_{k} \subseteq K_{k + 1}$ . By [L4] this ascending sequence stabilizes; choose $n \geq 1$ beyond its stabilization index, so $K_{n} = K_{2 n}$ . If $z \in K_{n} \cap x^{n} N$ , write $z = x^{n} y$ . Then $x^{2 n} y = x^{n} z = 0$ , hence $y \in K_{2 n} = K_{n}$ and $z = 0$ . Thus $0 = K_{n} \cap x^{n} N$ . This uses only the choice-free implication of [L4].

3.1L2L3step 1.1step 2.1algebra

Since $0 \neq K_{1} \subseteq K_{n}$ , irreducibility of $0 \subset N$ and $0 = K_{n} \cap x^{n} N$ force $x^{n} N = 0$ . Consequently, if $x m \in Q$ and $m \notin Q$ , the nonzero coset $m + Q$ shows that $x$ is a zero divisor on $N$ , so some $n \geq 1$ satisfies $x^{n} M \subseteq Q$ . By [L3], $Q$ is primary.

4.1step 1.1step 3.1∎

Hence every irreducible submodule of a Noetherian module is primary, without an additional choice assumption.

An irreducible submodule of a Noetherian module is primary

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources