# 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 $f(p) = 0$.

-
Have you heard of the universal properties of polynomial rings and quotient rings? –  Martin Brandenburg Jun 29 '12 at 9:23
@MartinBrandenburg No. –  Rudy the Reindeer Jun 29 '12 at 9:28
@MartinBrandenburg I wonder why you don't answer this: math.stackexchange.com/questions/159699/… –  Makoto Kato Jun 29 '12 at 10:38
Bad form to use $p$ to stand for a polynomial in 3 variables and a polynomial in 1 variable in the same sentence. –  Gerry Myerson Jun 29 '12 at 12:14

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.

-
Let me know if I should expand any of this. I don't want to obscure the basic ideas, but it's important to know what you're using. –  Dylan Moreland Jun 29 '12 at 12:48
Thank you Dylan. Yes, please: For the first part we use the universal property of quotient rings to get $\tilde{f}$. Where do we use the universal property of polynomial rings (also mentioned by Martin)? And how does this save me work as opposed to doing what I tried in the question? (Is what I wrote there wrong, btw?) –  Rudy the Reindeer Jun 29 '12 at 14:12
@Clark Hey, just saw this. We use the properties of polynomial rings to get $f$ in the first place: to give a $k$-algebra map $k[x, y, z] \to A$ for any $A$ is to pick arbitrary images for each of $x, y, z$. I guess the point of the universal property is to know how to define maps out of polynomial rings. Of course, you could define $\tilde f$ straight away but I think that one should argue why it is well defined and universal properties are generally the cleanest way of doing that. –  Dylan Moreland Jun 29 '12 at 15:45
And the map you wrote down is the same as mine, so I think (hope!) it's correct. But going through these checks ensures that you don't attempt to write down, say, a map $k[x, y, z]/(xz - y^2) \to k[y]$, $p(x, y, z) \mapsto p(0, y, 0)$. That's no good, obviously. –  Dylan Moreland Jun 29 '12 at 15:48

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 $(\overline{x}, \overline{z})$.

-
What about the kernel? –  Makoto Kato Jun 29 '12 at 12:25