Show that a variety is irreducible How do I show that the variety $V = \{(x,y)\in k^2 \mid x-y=0\}$ is irreducible for an algebraically closed field $k$?
One approach, I think, is to view $f(x) = x-y$ as an element in $R[x]$, where $R=k[y]$ and since $k$ is algebraically closed, for $f$ to be reducible we must have that $\deg (f)\ge2$, which is not the case here.
Is there a better approach to prove this claim?
Thanks.
 A: The vanishing ideal of $V$ is the radical of $\langle x-y\rangle$ in $k[x,y]$. Since $k[x,y]$ is a factorial ring or even for reasons of polynomial total degree, you should have no problem showing that this ideal is already radical. 
We are left to show that it is prime. In other words, we have to show that $f\cdot g = h\cdot (x-y)$ implies $f\in\langle x-y\rangle$ or $g\in\langle x-y\rangle$. If we can show that $x-y$ is irreducible, then this follows because $k[x,y]$ is a factorial ring and therefore, the factor $(x-y)$ must appear in either $f$ or $g$ in the above equation. 
Hence, assume $x-y = f\cdot g$. You may proceed as suggested and view this as an equality over $k[y][x]$, where $x-y$ has degree $1$ so without loss of generality, $g\in k[y]$. You can also view this as an equality over $k[x][y]$, so we either have $g\in k[x]$ or $f\in k[x]$ by the same argument. If $g\in k[x]$, then $g\in k$ and we are done. Otherwise, $g\in k[y]$ and $f\in k[x]$ can not both have nonzero degree, because $x-y$ has total degree $1$. Therefore, either $f$ or $g$ must be constant.
You could also just argue that $k[x,y]/\langle x-y\rangle\cong k[x]$ is an integral domain, but the above is the most elementary argument I could think of.
