Is the sum of two prime ideals a prime ideal? Is the sum of two prime ideals (of a ring) again a prime ideal?
Please can someone give me a hint? I know that the intersection of two prime ideals is again a prime ideal iff one is contained in the other, but I'm not sure about the sum.
 A: Th answer is no, consider the ring $\mathbb{Z}$ and the prime ideals $(2)$ and $(5)$. Since two and five are relative primes then every number can be written as a linear combination over $\mathbb{Z}$ of 2 and 5. To see this. note that if if $k$ is any interger, since we know that
$$5-(2)2=1$$
by multiplying by $k$ we obtain the expression
$$k5-(2k)2=k$$
which is a $\mathbb{Z}$-linear combination of $k$ in therms of $2$ and $5$. Note that $k5\in(5)$ and $-2k(2)=-4k\in(2)$. This means that $(2)+(5)=\mathbb{Z}$, which is not a prime ideal.
Note that the sum of ideals is an ideal when one is contained in the other. This is not possible in $\mathbb{Z}$ with the exemption when of the ideals is the prime ideal $(0)$, but other rings can give more interesting examples. Have a look at polynomials rings for example.
A: I think (correct me if I'm wrong) that if you are not so restrictive in your definition of prime ideals (allowing 1 to belong to a prime ideal) the answer is positive.
For instance, let $p_{1}$ and $p_{2}$ be two prime ideals of $R$, and take the natural projection map $\pi: R \longrightarrow R/p_{1}$, then $\pi(p_{2})$ is a prime ideal because $\pi$ is a surjective map and $p_{2}$ is a prime ideal. But then, $\pi^{-1}(\pi(p_{2}))$ is a prime ideal in R, and $\pi^{-1}(\pi(p_{2}))=p_{1}+p_{2}$.
