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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](http://math.stackexchange.com/questions/100558/understanding-a-proof-on-residue-class-degrees) that need merging?2012-01-22

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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 $\mathfrak b^m$.

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    Brilliant, thanks very much.2012-01-22