Isomorphism of coordinate ring of smooth quadric and another subring of a polynomial ring. 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.
 A: 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 $k[x,y,z,w]/(xy-zw)$. 
A: 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 $q=r=0$.
