Maximal Radical implies Primary Ideal

Theorem

Let $R$ be a commutative ring with unity.

Let $\mathfrak a$ be an ideal of $R$.

Let $\map \Rad {\mathfrak a}$ be the radical of $\mathfrak a$.

Suppose that $\map \Rad {\mathfrak a}$ is a maximal ideal.

Then $\mathfrak a$ is a primary ideal.

Proof

Consider the quotient ring $R / \mathfrak a$.

By Definition 1 of Nilradical of Ring and Definition 1 of Radical of Ideal of Ring:

$\Nil {R / \mathfrak a} = \map \Rad {\mathfrak a} / \mathfrak a$

On the other hand, by hypothesis, $\map \Rad {\mathfrak a} / \mathfrak a$ is a maximal ideal of $R / \mathfrak a$.

Thus, in view of Definition 1 of Nilradical of Ring:

$\Nil {R / \mathfrak a}$

is the only maximal ring in $R / \mathfrak a$.

In view of Proper Ideal of Ring is Contained in Maximal Ideal, Definition 2 of Primary Ideal follows.

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