Kernel of homomorphism Is it possible to explain the sentence "the kernel of $\phi$, where $\phi: p(x,y) \mapsto a_0$  consists of all the polynomials with no constant term, is $<x,y>$  ".
How do we find the kernel? Any answer would be appreciated.
 A: Sure. The kernel of $\phi: p(x,y) \mapsto a_{0}$ are the polynomials whose image is $0$ under $\phi$. Since $\phi(p(x,y))$ is the constant term of $p(x,y)$, it follows that $\ker \phi$ are the polynomials with no constant term. Now, we must show that this kernel is the ideal generated by $x$ and $y$, or $\langle x, y \rangle$. To do this, we show mutual containment. Suppose $p(x,y) \in \ker(\phi)$. Then every term of $p(x,y)$ is of the form $\alpha x^{i}y^{j}$ where at least one of $i, j$ is $> 0$. Hence, $p(x,y)$ is a finite sum of terms in $\langle x, y \rangle$, hence is in $\langle x, y \rangle$. This show the containment $\ker \phi \subset \langle x, y \rangle$. Can you show $\langle x, y \rangle \subset \ker \phi$? Feel free to comment if you need more hints!
Edit: a brief aside. First, while it may be obvious to you, I think that the kernel of $\phi$ might be more obvious if it were written a little differently. Really, $\phi$ is the evaluation homomorphism at the point $(0,0)$. Inuitively, then, the polynomials which are killed by this homomorphism are those with no constant term, since any powers of $x$ and $y$ are killed by plugging in $0$, leaving only the constant term. 
