## Characterization of primary ideals

Let $R$ be a commutative ring. An ideal $Q \subseteq R$ is called primary if whenever $ab \in Q$ and $a \notin Q$, then $b^n \in Q$ for some $n \geq 1$. Prove the following about primary ideals.

1. The proper primary ideals of $\mathbb{Z}$ are $0$ and $(p^k)$ where $p$ is a prime and $k \geq 1$.
2. Every prime ideal or $R$ is primary.
3. An ideal $Q \subseteq R$ is primary if and only if every zero divisor in $R/Q$ is nilpotent.
4. If $Q \subseteq R$ is a primary ideal then $\mathsf{rad}(Q)$ is a prime ideal.

1. Suppose $(n) \subseteq \mathbb{Z}$ is a nonzero primary ideal. Suppose further that $n = ab$ is composite, where $a$ and $b$ are both less than $n$ with $\mathsf{gcd}(a,b) = 1$. Now $ab \in (n)$ and $a \notin (n)$, since $n$ does not divide $a$. However, $n$ does not divide any power of $b$ since, in particular, $a$ does not divide any power of $b$. Thus we have a contradiction, and if $I \subseteq \mathbb{Z}$ is primary then either $I = 0$ or $I = (p^k)$ for some prime $p$ and $k \geq 1$.

Now we will verify that these ideals are in fact primary. $0$ is primary because if $ab = 0$, then either $a = 0$ or $b = 0$. Now suppose $ab \in (p^k)$. Then $p^k$ divides $ab$, and in particular $p$ divides either $a$ or $b$ because $p$ is prime. If $a \notin (p^k)$, then some power of $p$ must divide $b$. Thus $b^k \in (p^k)$. Hence $(p^k)$ is primary.

2. Suppose $P$ is a prime ideal. If $ab \in P$ and $a \notin P$, then $b = b^1 \in P$. Thus $P$ is primary.
3. $(\Rightarrow)$ Suppose $Q \subseteq R$ is primary. Let $x+Q \in R/Q$ be a zero divisor. Then $(x+Q)(y+Q) = 0$ for some $y$, where we have $x,y \notin Q$. Since $y \notin Q$, we have $x^n \in Q$ for some $n \geq 1$; thus $(x+Q)^n = 0$, and so $x+Q$ is nilpotent in $R/Q$.

$(\Leftarrow)$ Suppose every zero divisor in $R/Q$ is nilpotent. Suppose $xy \in Q$ with $x \notin Q$. If $y \notin Q$, then $y+Q$ is a zero divisor in $R/Q$, so that $(y+Q)^n = 0$ for some $n$; in this case, $y^n \in Q$. Thus $Q$ is primary in $R$.

4. Let $Q \subseteq R$ be a primary ideal. Suppose $xy \in \mathsf{rad}(Q)$, with $x \notin \mathsf{rad}(Q)$. Now $(xy)^n = x^ny^n \in Q$ for some $n \geq 1$. Since $x \notin \mathsf{rad}(Q)$, $x^n \notin \mathsf{rad}(Q)$. Since $Q$ is primary, $y^n \in Q$, so that $y \in \mathsf{rad}(Q)$. Thus $\mathsf{rad}(Q)$ is a prime ideal in $R$.
I am not so sure this $P$ is primary. For instance, $xy \in P$ and $x \notin P$, but no power of $y$ is in $P$.