# Primary Ideals, Prime Ideals, and Radical Ideals

## Problem 247

Let $R$ be a commutative ring with unity. A proper ideal $I$ of $R$ is called **primary** if whenever $ab \in I$ for $a, b\in R$, then either $a\in I$ or $b^n\in I$ for some positive integer $n$.

**(a)** Prove that a prime ideal $P$ of $R$ is primary.

**(b)** If $P$ is a prime ideal and $a^n\in P$ for some $a\in R$ and a positive integer $n$, then show that $a\in P$.

**(c)** If $P$ is a prime ideal, prove that $\sqrt{P}=P$.

**(d)** If $Q$ is a primary ideal, prove that the radical ideal $\sqrt{Q}$ is a prime ideal.

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## Definition.

For an ideal $I$ of $R$, the **radical ideal** $\sqrt{I}$ is defined to be

\[\sqrt{Q}=\{ a\in R \mid a^n \in Q \text{ for some positive integer } n\}.\]

## Proof.

### (a) A prime ideal is primary

To show that $P$ is primary, suppose that $ab \in P$ for $a, b \in R$. Since $P$ is a prime ideal, we have either $a\in P$ or $b\in P$.

This implies that $P$ is a primary ideal. (We can always take $n=1$ for a prime ideal.)

### (b) If $a^n$ is in the prime ideal $P$, then $a\in P$

Suppose that we have $a^n\in P$ for some $a\in R$ and a positive integer $n$. We prove that $a\in P$ by induction.

The base case $n=1$ is trivial.

So assume that $a^k\in P$ implies $a\in P$ for some $k>1$.

When $n=k+1$, we want to show that $a^{k+1}\in P$ implies $a\in P$.

Since the product $a^{k+1}=a\cdot a^k$ of $a$ and $a^k$ is in the prime ideal $P$, we have either $a\in P$ or $a^k\in P$.

If the former is the case, we are done. If the latter is the case, then by the induction hypothesis, we also have $a\in P$.

Hence the induction step is completed, and the statement (b) is true for any positive integer $n$.

### (c) If $P$ is prime, then $\sqrt{P}=P$

Suppose that $P$ is a prime ideal. By definition, it is clear that $P \subset \sqrt{P}$. We prove that the other inclusion $\sqrt{P} \subset P$. Take an arbitrary element $a\in \sqrt{P}$.

Then there exists a positive integer $n$ such that $a^n \in P$.

It follows from part (b) that this implies that $a\in P$ since $P$ is a prime ideal. Thus we have proved $\sqrt{P} \subset P$, and hence $\sqrt{P}=P$.

### (d) If $Q$ is primary, then $\sqrt{Q}$ is prime

Suppose that $Q$ is a primary ideal of $R$. To show that the radical ideal $\sqrt{Q}$ is prime, suppose that $ab \in \sqrt{Q}$. Our goal is to show that either $a\in \sqrt{Q}$ or $b\in \sqrt{Q}$.

Since $ab\in \sqrt{Q}$, there exists a positive integer $n$ such that

\[(ab)^n \in Q.\]
Since $R$ is commutative, this implies that we have

\[a^n\cdot b^n \in Q.\]

Since $Q$ is primary, this yields by definition that either

\[a^n\in Q \quad\text{ or }\quad (b^n)^m \in Q\]
for some positive integer $m$.

By definition of the radical ideal, it follows that

\[a\in \sqrt{Q} \quad\text{ or }\quad b\in \sqrt{Q},\]
and thus $\sqrt{Q}$ is a prime ideal.

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