The spectrum of $\mathbb{Q}[x,y]/(x^2,xy)$ Is the spectrum of $\mathbb{Q}[x,y]/(x^2,xy)$ irreducible?
I think $(x^2,xy)$ is a prime ideal of $\mathbb{Q}[x,y]$. If this is right, then the spectrum is trivially irreducible using the fact that the spectrum of an integral domain is irreducible. But, if the ideal is not prime, then how can we determine the nilradical of the given ring? Any hints? Thanks beforehand.
 A: The ideal $I=(x^2,xy)\subset\Bbb{Q}[x,y]$ is clearly not prime because $xy\in I$ but $x,y\notin I$.
As for the nilradical of $Q:=\Bbb{Q}[x,y]/I$; it is the ideal consisting of all nilpotent elements. To get you started; what can you say about $f\in\Bbb{Q}[x,y]$ if $f^2\in I$? Or if $f^k\in I$?
Alternatively, the nilradical of $Q$ is the intersection of all prime ideals of $Q$. Via the quotient map this corresponds to the intersection of all prime ideals of $\Bbb{Q}[x,y]$ containing $I$. So which are the prime ideals containing $I$?

Edit: As OP seems to have solved the problem, here is a sketch of a solution:
The nilradical of $\Bbb{Q}[x,y]/(x^2,xy)$ is the intersection of all prime ideals of $\Bbb{Q}[x,y]/(x^2,xy)$. Through the quotient map this corresponds to the intersection of all prime ideals of $\Bbb{Q}[x,y]$ containing $(x^2,xy)$. For every prime ideal $P$ containing $(x^2,xy)$ we have $x^2\in P$ and hence $x\in P$, so $x$ is in the intersection of all prime ideals containing $(x^2,xy)$. Conversely $(x^2,xy)\subset(x)$, which shows that the intersection of all prime ideals containing $(x^2,xy)$ is precisely $(x)$. It follows that the nilradical of $\Bbb{Q}[x,y]/(x^2,xy)$ is precisely $(x)$, which is prime, so the spectrum is irreducible.
