Are non-coprime ideals always contained in some prime ideal?

Question

I'm reading Milne's notes on Algebraic Number Theory and am confused about the proof of the following lemma

Let $D$ be a ring and $\mathfrak{a}$ and $\mathfrak{b}$ be coprime ideals of $D$. Then the ideals $\mathfrak{a}^m$ and $\mathfrak{b}^n$ are coprime for all $m, n \in \Bbb N$.

The proof starts as follows:

Suppose $\mathfrak{a}^m$ and $\mathfrak{b}^n$ are not coprime. Then they are both contained in some prime (even maximal) ideal $\mathfrak{p}$.

Why is this the case? I've seen a proof that every ideal is contained in some maximal ideal but I can't seem to find anything about the same statement for prime ideals.

Answer 1

Maximal ideals are prime ideals, so, if you can find a maximal ideal with some property, you've got a prime ideal with that property. That's why the notes say "(even maximal)".

Answer 2

You can also see it like this: there exist $a \in \frak{a}$, $b \in \frak{b}$ so that $a+b=1$. Now notice that in the binomial expansion of $1 = (a+b)^{m+n-1}$ each term is in $\frak{a}^m$ or in $\frak{b}^n$.