Surjective homomorphism from $k[x,y,z]/(xy-z^2) $ onto $ k[y]$ This is a question related to this question here.
Can you tell me if this is right:
I'd like to define a surjective homomorphism from $k[x,y,z]/(xy-z^2)$ onto $k[y]$ with kernel $(\bar{x}, \bar{z})$. To this end I define
$$ f: p(x,y,z) + (xy-z^2) \mapsto p(0,y, 0)$$
Then $f$ is surjective since if $q(y) \in k[y]$ we have $f(p(0,y, 0) + (xy- z^2)) = q(y)$. Also, $f$ has kernel $(\bar{x}, \bar{z})$ since if $p \in (\bar{x}, \bar{z})$ we have $f(p) = 0$. 
 A: By the universal property of polynomial rings there is a (surjective) ring homomorphism $f:k[x,y,z]\longrightarrow k[y]$ that sends $x,z$ to $0$, $y$ to $y$, and which is identity on $k$. On the other side, there is a canonical (surjective) ring homomorphism $\pi:k[x,y,z]\longrightarrow k[x,y,z]/(xy-z^2)$. Now use the universal property of the quotient rings in order to find the required surjective ring homomorphism. 
The kernel of this homomorphism is given by the classes of polynomials $p\in k[x,y,z]$ with the property that $p(0,y,0)=0$. Obviously $p(0,y,0)=0$ iff $p\in (x,z)$, and this shows that the kernel of our homomorphism coincides with the ideal $(\overline{x}, \overline{z})$.
A: You should learn the universal properties that Martin mentions, since they will save you some work in the future. See this answer of Bill Dubuque's and Wikipedia.
To begin, define a $k$-algebra homomorphism $f\colon k[x, y, z] \to k[y]$ by $x \mapsto 0$, $y \mapsto y$, $z \mapsto 0$. In other words, $f$ sends $p(x, y, z)$ to $p(0, y, 0)$. Since $f(xy - z^2) = 0 \cdot y - 0^2 = 0$, we get an induced map $\tilde f\colon k[x, y, z]/(xy - z^2) \to k[y]$ such that the composition
\[
k[x, y, z] \to k[x, y, z]/(xy - z^2) \to k[y]
\]
is $f$. It's easy to check — and I think you've done this — that $\bar x$ and $\bar z$ are contained in the kernel of $\tilde f$. You want to show that these two elements generate the kernel as an ideal. At this point there are a few ways to proceed. You could note that the above sequence induces
\[
k[x, y, z]/(x, z) \stackrel{\sim} \to (k[x, y, z]/(xy - z^2))/(\bar x, \bar z) \to k[y]
\]
and that this composition is an isomorphism.
