I have the following problem:
Let $K$ be a field and $R=K[X,Y]/(XY)$. For $P\in K[X,Y]$ we denote $[P]$ its class in $R$. Show that the Ideal $(XY)$ is not a prime ideal.
My Idea was the following:
Let us assume $(XY)$ is a prime ideal, this means that $$PQ\in (XY)\Rightarrow Q\in (XY)~~~or~~~P\in (XY)$$ Now take $PQ=XY$, then $X\in (XY)$ but this means that $X=P(X,Y)\cdot XY$ where $P(X,Y)\in K[X,Y]$. Now in the solution they took $Y=0$ then clearly $X=0$. But then they said that this is a contradiction. But I don't see why $X=0$ is a contradiction.
Thanks for your help
Here is one possible proof:
Let $\varphi : K[X,Y] \to K$ be the unique $K$-algebra homomorphism such that $\varphi(X) = 1$ and $\varphi(Y) = 0$. Then $\varphi(XY) = \varphi(X)\varphi(Y) = 1 \cdot 0 = 0$, so this factors through the quotient $K[X,Y] \to K[X,Y]/(XY) = R$ to give a homomorphism $\widetilde{\varphi} : R \to K$ such that $\widetilde{\varphi}([P]) = \varphi(P)$ for all $P \in K[X,Y]$.
Since $K$ is a field, $\widetilde{\varphi}([X]) = \varphi(X) = 1 \neq 0$, so $[X] \neq 0$. In other words, $X \notin (XY)$. Symmetrically, we must have $Y \notin (XY)$. But $XY \in (XY)$, showing that $(XY)$ is not prime.