I know that if $f(x) \in \mathbb{C}[x]$ is such that $f(0) = 0$, then $f(x) \in \langle x \rangle$.
But while solving a problem, I encountered a polynomial $f(x,y) \in \mathbb{C}[x,y]$ such that $f(0,0) = 0$. Is there a nicer way to represent such polynomials ?
Relying on symmetry, intuition says that $f(x,y) \in \langle x,y \rangle$ but I can't think of a rigorous way to show this.
I think this is right because $f(0,0)=0$ $\forall$ $f(x,y) \in \langle x,y \rangle$. But how do I prove the converse ?
Write the polynomial explicitly:
$f(x,y)=\sum\limits_{0\leq i,j} a_{ij}x^iy^j$
Where the sum is finite. Since $f(0,0)=0$ we have $a_{00}=0$. Thus every monomial of $f$ must be divisible either by $x$ or by $y$, and so every such monomial belongs to $(x,y)$. And so $f\in (x,y)$ as well, as a sum of elements in $(x,y)$.