Ring morphism, find the kernel

Question

I am trying to solve the following problem

Let $k$ be a field and consider the ring morphism $f:k[x,y] \to k[t]$ defined by $f(x)=t$ and $f(y)=q(t)$ with $q(t) \in k[t]$. Find the kernel of $f$.

Well, I could show one easy inclusion which is $\langle q(x)-y \rangle \subseteq \ker(f)$. I suppose the other inclusion also holds but I really don't know how to prove the equality of these two ideals. I would appreciate some help.

Answer 1

Indeed, $q(x)-y\in \ker f$, because $f(q(x)-y)=q(t)-q(t)=0$.

Suppose $p(x,y)\in k[x,y]$. You can carry out the division with remainder of $p$ by $q(x)-y$ in $k[x][y]$ because the coefficient of $y$ is invertible; in other words you can write $$ p(x,y)=p_1(x,y)(q(x)-y)+r(x) $$ If $p(x,y)\in\ker(f)$, then also $r(x)\in\ker(f)$, that is, $r(t)=0$.

Answer 2

Consider $I$ the ideal generated by $q(x)-y$, let $g:k[x]\rightarrow k[x,y]/I$ defined by $g(x)$ is the class of $x$ in $k[x,y]/I$. Show that $g$ is an isomorphism and $\bar f\circ g$ is an isomorphism where $\bar f:k[x,y]/I\rightarrow k[t]$ is induced by $f$. This implies that the kernel of $f$ is $I$.