Let $P$ be a prime ideal in the polynomial ring $K\left[x\_1,...,x\_m\right]$ and $Q$ be a prime ideal in the polynomial ring $K\left[y\_1,...,y\_n\right]$. 


Is $P+Q$ a prime ideal in $K\left[x\_1,...,x\_m,y\_1,...,y\_n\right]$ ?


For example for $Q=\left(y\_1,...,y\_n\right)$, it is easy to prove that. But how about general cases?



Is the sum of two prime ideals in different polynomial rings, K[X_i] and K[Y_i] a prime ideal in K[X_i Y_i]?
We have $K\left[x\_1,...,x\_m,y\_1,...,y\_n\right] \cong K\left[x\_1,...,x\_m\right] \otimes K\left[y\_1,...,y\_n\right]$ (where all tensor products are over $K$), and under this isomorphism, the ideal of $K\left[x\_1,...,x\_m,y\_1,...,y\_n\right]$ generated by $P+Q$ corresponds to $P\otimes K\left[y\_1,...,y\_n\right] + K\left[x\_1,...,x\_m\right]\otimes Q$. So your question is a particular case of the following one:


Let $K$ be a field, and $A$ and $B$ be two $K$-algebras, and $P$ and $Q$ be two prime ideals of $A$ and $B$, respectively. When can we conclude that $P\otimes B+A\otimes Q$ is a prime ideal of $A\otimes B$ ?


Since $P\otimes B+A\otimes Q$ is the kernel of the canonical map $A\otimes B\to \left(A/P\right)\otimes \left(B/Q\right)$ (this is a fact of linear algebra, and I can give a proof if needed), it is clear that $P\otimes B+A\otimes Q$ is a prime ideal of $A\otimes B$ if and only if $\left(A/P\right)\otimes \left(B/Q\right)$ is an integral domain. On the other hand, we know that $A/P$ and $B/Q$ are integral domains (since $P$ and $Q$ are prime ideals). So our question becomes: When is the tensor product of two integral domains (which both are $K$-algebras) again an integral domain?


This is always true when $K$ is algebraically closed and one of the two integral domains is a finitely-generated $K$-algebra (see Proposition 4.15 (b) in [J. S. Milne, *Algebraic Geometry*, Version 5.21](http://jmilne.org/math/CourseNotes/ag.html)). For $K$ not algebraically closed, the answer is "no" (not generally), even in your case as we can easily see (in fact, every finitely generated $K$-algebra which is an integral domain can be considered as a quotient of a polynomial ring $K\left[z\_1,z\_2,...,z\_k\right]$ for some $k$ modulo some prime ideal, so the problems are equivalent). For a particularly simple counterexample to your claim, take $K=\mathbb R$, $n=1$, $m=1$, $P=\left(x\_1^2+1\right)$ and $Q=\left(y\_1^2+1\right)$ (in this case, $P\otimes Q\cong \mathbb C\otimes\_{\mathbb R} \mathbb C$, and this is a commutative $\mathbb R$-algebra of dimension $4$, so it is clearly not an integral domain).


