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.