# Are distinct prime ideals in a ring always coprime? If not, then when are they?

Essentially as the title suggests - in some commutative ring $K$ (with 0,1), if we have 2 distinct proper prime ideals $\mathfrak{p}_1 \neq \mathfrak{p}_2$, is it necessarily the case (or if not, when is it the case?) that $\mathfrak{p}_1 + \mathfrak{p}_2 = K$ ?

I ask because I'm working through some lecture notes which frequently take some $K$ complete with respect to a discrete valuation, a prime ideal in a valuation ring $\mathfrak{p} \subset \mathcal{O}_K$, a finite extension $L/K$, and then use the fact $\mathfrak{p}\mathcal{O}_L$ factors uniquely as some $\mathcal{P}_1^{a_1} \ldots \mathcal{P}_r^{a_r}$ (where $\mathcal{O}_L$ is the integral closure of $\mathcal{O}_K$ in $L$). Then we say $\mathcal{P}_i + \mathcal{P}_j = \mathcal{O}_L$ $(\forall$ $i \neq j)$.

However it isn't clear to me, because we've made so many different assumptions along the way, what the reason is that $\mathcal{P}_i + \mathcal{P}_j = \mathcal{O}_L$. Is it because these primes occur in the factorisation of $\mathfrak{p}$, is it because of some property of Dedekind domains or fields, or is it not really possible to say out of context? Many thanks, and if anything needs clarification please ask.

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By the way, do you have some accounts that need merging? – Dylan Moreland Jan 22 '12 at 18:38

This won't happen in general: in $\mathbf Z$, which happens to be a Dedekind domain, I could take $0$ and $p\mathbf Z$. Even if we avoid containment there are still examples: in $k[x, y]$ the prime ideals $(x)$ and $(y)$ generate a maximal ideal.
However, in a Dedekind domain the non-zero primes are maximal, and distinct maximal ideals are always coprime. Another nice fact to know is that in general if two ideals $\mathfrak{a}$ and $\mathfrak{b}$ are coprime then so are $\mathfrak a^n$ and $\mathfrak b^m$.
I take it that's because the sum $\mathfrak{p}_1 + \mathfrak{p}_2$ of the 2 ideals is another ideal, which contains both maximal ideals and therefore must be the entire ring? – Warner Jan 22 '12 at 18:43
@Warner Right, we know that $\mathfrak p_2 \not\subset \mathfrak p_1$, so the sum contains $\mathfrak p_1$ and some element not in $\mathfrak p_1$. – Dylan Moreland Jan 22 '12 at 18:48