# Sums of powers of prime ideals in a Dedekind domain.

Let $P$ and $Q$ be distinct nonzero prime ideals of a Dedekind domain, $R$. Show that $P^{m} + Q^{n} = R$ for integers $m$, $n$.

It's clear to me that $P+Q=R$ since both $P$ and $Q$ are maximal and $P+Q$ is the smallest ideal containing both $P$ and $Q$; hence $R$. My thought was then that there exists $x \in P$, $y \in Q$ such that $x-y=1 \in P+Q$. Letting $d=\operatorname{lcm}(m,n)$, then $x^{d}-y^{d} \in P^{m}+Q^{n}$, so $$x^{d}-y^{d}=(x-y)(x^{d-1} + x^{d-2}y + \cdots + xy^{d-2} + y^{d-1}).$$ But since I don't know that $P^m + Q^n$ is prime, I don't see that this implies that $1 \in P^{m}+Q^{n}$. Am I on the right track? Any help would be greatly appreciated.

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This has nothing to do with Dedekind domains. And the question is a dublicate of math.stackexchange.com/questions/10400 – Martin Brandenburg Feb 11 '13 at 1:55
@MartinBrandenburg If you think this is a duplicate why don't vote to close it? – user26857 Feb 11 '13 at 2:11
It doesn't really matter, since the (way too complicated) answer has already been accepted. – Martin Brandenburg Feb 11 '13 at 3:24
@MartinBrandenburg I can't imagine, even if there was a trivial way to do this, that my solution could be called "way too complicated". It is two lines and uses ideas learned in the first class of a ring theory course. – Alex Youcis Feb 11 '13 at 8:57

Suppose that $P^m+Q^n$ is contained in some maximal ideal $M$. Note then $P^m,Q^n$ are contained in $M$ and so $P,Q$ are contained in $M$ because it's prime. Thus, $P+Q=R$ is contained in $M$. This is a contradiction.
But isn't the containment in the opposite direction, namely $P^m \subseteq P$? I'm not seeing why it should be that $P^m + Q^n \subseteq M \Rightarrow P + Q \subseteq M$. – JoeDub Feb 11 '13 at 0:10
@JoeDub $P^m+Q^n\subseteq M$ implies that $P^m,Q^n\subseteq M$ by definition of the sum (i.e. the smallest ideal containing both $P^m$ and $Q^n$). But, since $M$ is maximal, and thus prime, it's trivial that $P^m\subseteq M$ implies that $P\subseteq M$ and similarly $Q^n\subseteq M$ implies that $Q\subseteq M$. Thus, $P,Q\subseteq M$ and so $R=P+Q\subseteq M$. – Alex Youcis Feb 11 '13 at 0:11