# In a PID, every nonzero prime ideal is maximal

## Proof:

Let $R$ be a PID. Let $(x)$ be a nonzero prime ideal. Suppose $(x)\subsetneq (y)$. Then $x=yr$ for some $r\in R$.

Note that $yr\in(x)$ implies $y\in(x)$ or $r\in(x)$. Since $(x)\neq(y)$, we have $r\in(x)$, so $r=xz$ for some $z\in R$. Then, $\displaystyle x=yr=yxz$ which implies $1=yz$, thus $y$ is a unit. Hence $(y)=R$.

## Author: mathtuition88

http://mathtuition88.com

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