Annihilator of Noetherian module has Noetherian quotient

Statement

Verbal statement

Consider a Noetherian module over a commutative unital ring. The quotient of the ring by the annihilator of this module, is a Noetherian ring.

Symbolic statement

Let $M$ be a Noetherian module over a commutative unital ring $R$ . Let $I$ be the annihilator of $M$ . Then the quotient ring $R / I$ is a Noetherian ring.

Proof

Let $m_{1}, m_{2}, \dots, m_{n}$ be a finite generating set for $M$ . Consider a $R$ -module map from $R$ to $M^{n}$ given by:

$a \mapsto (a m_{1}, a m_{2}, \dots, a m_{n})$

The kernel of this map is precisely $I$ , so the quotient is a submodule of $M^{n}$ .

Since $M$ is Noetherian, $M^{n}$ is Noetherian, and hence $R / I$ is Noetherian (as it is a submodule of a Noetherian module). But $R / I$ being Noetherian as a $R$ -module is equivalent to $R / I$ being a Noetherian ring.