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

2 Answers

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

-