Let $K$ be a field. I want to show that the ideal $(XY) \subset K[X,Y]$ is not a prime ideal and deduce that $K[X,Y]/(XY)$ is not an integral domain.
My approach:
To show that $(XY)$ is not a prime ideal, I assume that it is one and try to get a contradiction.
We note that $(XY) = \{XY \cdot P |P \in K[X,Y]\}$.Since $(XY)$ is a prime ideal I get that $X \in (XY)$ or $Y \in (XY)$. Let us assume without loss of generality that $X \in (XY)$. This means that there exists $P' \in K[X,Y]$ such that $X = XY \cdot P'$.
How can I derive a contradiction from here?
Suppose, I have successfully shown that $(XY)$ is not a prime ideal. Then to show that $K[X,Y]/(XY)$ is not an integral domain I would have to assume that $\bar{X}\bar{Y} = 0$ and show that $\bar{X} \neq 0$ and $\bar{Y} \neq 0$. So let $Q + (XY) \in K[X,Y]/(XY)$ and $P + (XY) \in K[X,Y]/(XY)$ where $P,Q \in K[X,Y]$. Then $(Q + (XY))(P + (XY)) = PQ +(XY)$. So, suppose that $PQ +(XY) = 0 \iff PQ + XY\cdot R = 0$ and that $\forall R \in K[X,Y]$. Here, I do not know as well how to proceed to get the contradiction.
Any help/hint is appreciated.
Best Answer
Since $K[X,Y]$ is an integral domain you can cancel $X$ from both sides in the equation. You then get: $Y \cdot P' = 1$, which is a contradiction as $Y$ is not a unit in $K[X,Y]$.