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I'm looking at exercise 1.8(c) in David Cox' algebraic geometry notes which goes:

Let $V=\mathbf{V}(xy-zw) \subset \mathbb{C}^4$. Prove that $\mathbb{C}[V]\cong\mathbb{C}[ab,cd,ac,bd]\subset \mathbb{C}[a,b,c,d]$.

I know that $\mathbb{C}[V]=\mathbb{C}[x,y,z,w]/\langle xy-zw \rangle$, and that this isomorphism of $\mathbb{C}$-algebras can be established using an isomorphism of varieties. The hint is to show V can be "parametrized surjectively" by $(a,b,c,d) \mapsto (ab,cd,ac,bd)$ but I still don't see where this is going exactly...

More directly can see that we have a surjective map $\mathbb{C}[x,y,z,w] \to \mathbb{C}[ab,cd,ac,bd]$ which just matches the respective monomials. Presumably I could show the isomorphism by showing the kernel is exactly $\langle xy-zw\rangle$. This ideal is clearly contained in the kernel, but how do I prove the reverse containment?

Suggestions for either approach are appreciated, thanks in advance.

Billy O.
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  • You don't need to prove the reverse containment. Checking surjectivity and describing the kernel is enough. – Sasha Oct 23 '17 at 07:04
  • @Sasha You need the reverse containment to describe the kernel. The OP noted that $\langle xy-zw \rangle$ is clearly (I agree) contained in the kernel of the map. You are right that "describing the kernel is enough", but "describing the kernel" means proving the reverse containment, that the kernel is contained in $\langle xy-zw \rangle$. – Zach Teitler Oct 23 '17 at 07:20

2 Answers2

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One can show easily that $V(xy - z w)$ is the image of the map $$\mathbb{C}^4 \ni (a,b,c,d) \overset{\phi}{\mapsto} (ab, cd, ac, bd)=(x,y,z,w) \in \mathbb{C}^4$$

This works for every field. Basically it says: every $2\times 2$ matrix of rank $\le 1$ is of the form $$\left( \begin{matrix} x & z\\ w & y \end{matrix} \right ) = \left( \begin{matrix} a \\ d\end{matrix}\right )\cdot (b,c) $$

Let $f$ a polynomial map on $V$. Then $f\circ \phi$ is a polynomial map on $\mathbb{C}^4$. Now, polynomial maps on $V$ are restrictions of polynomial maps $P$ on $\mathbb{C}^4$. We get a map $$P(x,y,z,t) \mapsto P(ab,cd,ac,bd)$$ This apriori is a map from polynomials $P$ to polynomial functions on $\mathbb{C}^4$, but this can be indentified with the polynomial $P(ab,cd,ac,bd)$, since $\mathbb{C}$ is infinite.

Now, we don't need to know what the ideal of $V$ is, although it's easy to show it is $(xy-zw)$ ( again, use that the field is infinite).

$\bf{Added:}$ So what is really going on? We could consider this over any field $k$. But then we should still work in an infinite extension of $k$.

Here is the thing: To test whether for a polynomial $P$ the polypomial $P(ab, cd, ad, bc)$ is the $0$ polynomial, it is equivalent to testing whether $P(ab, cd, ad, bc)$, is $0$ when we substitute any values for $a$, $b$, $c$, $d$. That works if we have an infinite field $k$, or, we take values from an infinite extension of $k$.

Also: the ideal of $V$ is $xy - z w$, provided the field $k$ is infinite. So, we see the need for working in an infinite field. But in the end, the following is true, no matter what the field $k$ is: The ideal of polynomials $P(x,y,z,w)$ so that the polynomial $P(ab, cd, ad, bd)$ is the zero polynomial is the principal ideal $(xy- zw)$. So $k[ab,cd,ad,bc]\subset k[a,b,c,d]$ is isomorphic to $k[x,y,z,w]/(xy-zw)$.

orangeskid
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  • First, I reckon that $I(V)=(xy−zw)$ because this polynomial is irreducible hence the ideal is prime hence radical. I could see $V$ is the image of the of the suggested map ($\phi$) from $\mathbb{C}^4$. This surjection makes the comorphism $\phi^:\mathbb{C}[V]\to \mathbb{C}[a,b,c,d]$ injective as a result. Since $\phi^$ is actually the surjective ring map I first described, this finishes it off? – Billy O. Nov 01 '17 at 07:56
  • @Billy O.: It's pretty much done. I want to mention that to check that you can prove that the ideal of $V$ is $(xy -zw)$ even if you only assume $k$ infinite. – orangeskid Nov 01 '17 at 08:23
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Consider the mape $\mathbb{C}[x,y,z,w] \to \mathbb{C}[a,b,c,d]$ sending $x \mapsto ab$, $y \mapsto cd$, $z \mapsto ac$, $w \mapsto bd$. Let $I$ be the kernel. As you noted, $xy-zw \in I$. To show that $I = \langle xy-zw \rangle$, let $p = p(x,y,z,w) \in I$. Our goal is to show $p \in \langle xy-zw \rangle$. We will modify $p$ by elements of $\langle xy-zw \rangle$ until we reduce $p$ to an element of $\langle xy-zw \rangle$; this will achieve our goal.

The modifications are simple: every $xy$ in $p$ is replaced by $zw$. Then, modulo $\langle xy-zw \rangle$, $p$ is congruent to some $q(x,z,w) + r(y,z,w)$. In particular $q+r \in I$. We have $q(ab,ac,bd)+r(cd,ac,bd)=0$ for all $a,b,c,d$. Since $\mathbb{C}$ is an infinite field, $q+r=0$ identically, i.e., all the coefficients vanish in the sum. Let $i$ be the highest $xz$-degree of any term in $q$, equivalently the highest $a$-degree of any term of $q(ab,ac,bd)$. In order to cancel, $r$ must have terms with $z$-degree $i$. Then the $yz$-degree of $r$ is at least $i$, so the $c$-degree of $r(cd,ac,bd)$ is at least $i$. But then in order to cancel this power of $c$, the $z$-degree of $q$ is also at least $i$. The terms of both $q$ and $r$ that have $z^i$ can't have any $x$ or $y$, only $w$. But then they must match term for term, and we can cancel them and continue reducing, to eventually reduce to $q=r=0$.

Zach Teitler
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