# Polynomial in two variables with zero constant coefficient form principal ideal?

Let $F$ be a field, and $F[x,y]$ the ring of polynomials in $x,y$. Let $J$ be the subset of all polynomials $P(x,y)$ in $F[x,y]$ such that $P(0,0)=0$. Then $J$ is an ideal. Is $J$ a principal ideal?

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The ideal you describe is called the irrelevant ideal and is given by $J=(x,y)$. It is evident that $J$ is not a principal ideal, since it is generated by two coprime irreducible polynomials $x,y$. Note though that $J$ is a maximal ideal, since $F[x,y]/(x,y) \cong F$.
@MikaH.: If it were a principal ideal, then it would be of the form $J=(p)$ for some polynomial $p$. Then $p$ would divide both $x$ and $y$. But $x$ is divisible only by $x$ and $1$. Similarly $y$ is divisible only by $y$ and $1$. This shows that $p$ must be $1$, contradiction. – Manos Jan 16 '14 at 16:33
Hint $\$ Notice $\,x\,$ is prime since $\,F[x,y]/(x) = F[y]\,$ is a domain. Similarly $\,y\,$ is prime. Being prime, $\,x,y\,$ are irreducible. So if $\,J = (x,y) = (f)\,$ then either $\,f\,$ is a unit, or $\,f\,$ is associate to $\,x\,$ and $\,y,\,$ so $\,x\,$ and $\,y\,$ are associate: $\,x=uy,\,$ so evaluating at $\,x,y = 1,0\,\Rightarrow\, 1=0,\,$ contra $\,F\,$ is a field. Hence $\,f\,$ is a unit, $\,(f)= (1) = (x,y)\,\Rightarrow 1 = x f(x,y) + y\, g(x,y)\,\Rightarrow\, 1=0\,$ by evaluating at $\,x,y = 0,0.$