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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