I have the homomorphism $$F : C[x,y] \to C[x] \times C[y],\space F(p(x,y)) = (p(x,0), p(0,y))$$
The kernel is clearly $\langle xy \rangle$.
If I want to show that $\langle xy \rangle = \ker(F)$ I need to use double inclusion.
First : To show that $\langle xy \rangle$ is in $\ker(F)$. I take a $p(x,y)$ generated by $\langle xy \rangle$, so $p(x,y) = xyq(x,y)$ ; $p(x,0) = 0$ and $p(0,y) = 0$. Therefore $(p(x,0),(0,y)) = (0,0)$ and this is in $\ker(F)$.
My trouble begins with the second part, when I need to show that $\ker(F)$ is included in $\langle xy \rangle$
I would appreciate some advice on how to solve this problem. Thanks.
Best Answer
Let $\alpha\in \ker(F)$ not divisible by $xy$.
It implies that $\alpha = Qxy+ax+by+c$ such that not all of $a,b,c$ are 0.
Since $\alpha(x,0)=0$, $ax+c=0$, which at $0$ implies $c=0$, and at $1$ implies $a=0$
Since $\alpha(y,0)=0$, $by+c=0 \implies by=0$ which at $1$ implies $b=0$ which is a contradiction.