TITLE: $(x,y)\subset \mathbb Q[x,y]$ is not a principal ideal
QUESTION [8 upvotes]: Is my proof that $(x,y)\subset \mathbb Q[x,y]$ is not a principal ideal correct?
Suppose $(x,y)=(f(x,y))$. Since $x,y\in (x,y)$, the degree of $f$ in $x$ and $y$ (separately) cannot exceed 1. So $f(x,y)=ax+by+c$. Since $x\in (x,y)$, there must exist $g(x,y)$ such that $f(x,y)g(x,y)=x$. The only event in which this happens is when $f(x,y)=ax,\ g(x,y)=1/a$. So $b=c=0$. Since $y\in (x,y)$, $a=c=0$. Thus $f(x,y)=0$. But $(x,y)$ is not the zero ideal, a contradiction.

REPLY [3 votes]: Your proof is almost fine, it just needs a couple of details. But you're overcomplicating things.
Consider $f(x,y)g(x,y)=x$ as an equality between polynomials in $y$ over $\mathbb{Q}[x]$. As this has degree $0$, both $f$ and $g$ must have degree $0$. Thus $b=0$. Similarly, $a=0$.
Therefore $f(x,y)=c$, which either generates the whole ring or the zero ideal. On the other hand, $(x,y)$ is a nonzero proper ideal, as $\mathbb{Q}[x,y]/(x,y)\cong\mathbb{Q}$.