Let $f,g$ be relatively prime polynomials in $k[x,y]$, where $k$ is a field. Then is it true that every prime ideal $P$ of $k[x,y]$ containing $f$ and $g$ is maximal ?
The only thing I am able to show is that $\exists 0\ne u(x)\in k[x]$ and $0\ne v(y)\in k[y]$ such that $u,v \in (f,g)$. Please help. Thanks in advance
Let us build on what you have done so far. Let $P$ be a prime ideal containing $f,g$. In particular, we have $u,v \in P$.
Clearly $k[x,y]/(u,v)$ is finite-dimensional of dimension $\deg u \cdot \deg v$. Furthermore the inclusion $(u,v) \subset P$ gives rise to a a surjection $k[x,y]/(u,v) \twoheadrightarrow k[x,y]/P$.
In particular $k[x,y]/P$ is a finite-dimensional $k$-algebra, which is also an integral domain. By the well known standard proof, it is a field. Hence $P$ is maximal.
This statement has the geometric interpretation, that two plane curves - that do not share a common component - have only finitely many common points.
You could also rephrase it as follows: Two general equations in two variables have only finitely many common solutions.