Nonzero Prime Ideals are Maximal in Euclidean Domains

Question

Prove that every nonzero prime ideal in a Euclidean domain is maximal.

This is what I have so far:

Let $R$ be a euclidean domain and let $P$ be a nonzero prime ideal in $R$ generated by $a$. So, $P=(a)$ and since $P$ is not zero then $a$ is not zero.
Then for some ideal $M$: ($M$ isn't prime, right?)

$P$ is properly contained in $M$ (i.e $P$ not equal to $M$) So we must show that $M=R$.

(Sorry, I don't know how to use math symbols in here)

This is where I get stuck. Have I set the question up properly? What would I attempt next? Any help is much appreciated.

Answer 1

Let $R$ be a Euclidean Domain and let $0 \neq P$ be a prime ideal.

Since $R$ is a Euclidean Domain, it is an Principal Ideal Domain, so $P = (p)$ for some $p \in R$, but we have the following result:

Lemma: $p$ is prime if and only if $(p)$ is prime.

This tells us that $p$ is prime, given that $P = (p)$ is prime.

So suppose $P \subseteq M \subseteq (1)$. Since $R$ is Principal Ideal Domain, $M = (y)$ for some $y \in R$, so we have $(p) \subseteq (y) \subseteq (1)$.

Since $p \in (y)$, $p = ry$ for some $r \in R$. Since $p$ is prime, either $p \mid r$ or $p \mid y$.

• If $p \mid r$, then $r = ps$ for some $s \in R$ in which case $p = ry = (ps)y$ which means $1 = sy$, since a Euclidean Domain is an Integral Domain, and we can cancel, hence $y$ is a unit, so $(y) = (1)$.
• If $p \mid y$, then $y = ps$ for some $s \in R$ and $y \in (p)$ which means $(p) = (y)$.

Conclude by definition that $P$ is a maximal ideal.