# 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.

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