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Problem: Let $k$ be an algebraically closed field and $V=V(xw-yz)=\{(x,y,z,w)\in\mathbb{A}^4(k): xw-yz=0\}$. Let $\Gamma(V)$ be the ring of coordinates of $V$ and $k(V)$ its field of fractions. Let $f=\bar{x}/ \bar{y}\in k(V)$. (Note that $\bar{x}/\bar{y}=\bar{z}/\bar{w}$, so $f$ is defined as a rational function on $V$ on the points $P$ with nonzero $y$ or $w$ coordinate, possibly more). Prove that:

There do not exist polynomials $a,b$ such that $f=\bar{a}/\bar{b}$ and $b(P)\neq 0$ for every $P$ where $f$ is defined.

The bar indicates the image of a polynomial under the quotient map.

My attempt: we have $\bar{a}/\bar{b}=\bar{x}/\bar{y}\implies ay-bx=c(xw-yz)$ for some polynomial $c\implies y(a+cz)=x(b+cw)\implies y|b+cw\implies b=dy-cw$ for some $d$. Let $\Omega=\{(x,y,z,w)\in\mathbb{A}^4(k): y=w=0\}$ and let $\Omega'=\mathbb{A}^4(k)-\Omega$. It would suffice to prove that $V(b,xw-yz)\cap\Omega'\neq\varnothing$. From the discussion above, $\Omega\subseteq V(b)$ and clearly $\Omega\subseteq V(xw-yz)$, so $\Omega\subseteq V(b,xw-yz)$. Suppose $\Omega = V(b,xw-yz)$. Then $(y,w)=I(\Omega)=I(V(b,xw-yz))=rad(b,xw-yz)$, by the Nullstellensatz. Then I'm stuck, assuming obviously that this leads to a solution.

Cantlog
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random
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1 Answers1

3

1) The rational function $f$ is not defined at the origin $O=(0,0,0,0)$
Since the pole set of a rational function is closed, it sufffices to prove that $f$ is not defined at $Q=(q, 0,0,0)$ as soon as $q \neq 0$.
Indeed if we could write $f=\bar{x}/ \bar{y}=\bar c/\bar d$ with $d(Q)\neq 0$, we would deduce $xd-yc =g\cdot (xw-yc)$ for some polynomial $g$.
Evaluating at $Q$ we would get $q \cdot d(Q)-0\cdot c(Q)=g(Q) (q\cdot 0-0\cdot 0)=0$, a contradiction with the assumption $d(Q)\neq 0$.

2) You can't write $f=a/b$ with $b(P)\neq 0$ whenever $f$ is defined at $P$
Suppose you could, i.e. suppose that such a $b$ exists and consider the plane $x=z=0$.
The rational function $f$ is certainly defined at every point $(0,y,0,w)$ of that plane for which $y$ or $w$ is non-zero, so that the polynomial $b(0,y,0,w)$ should be non-zero outside $y=w=0$.
But a polynomial in two variables (here $y,w $) cannot vanish only at the origin, so that we have $b(0,y,0,w)\neq 0$ for all values of $y,w$.
So the (incorrect!) assumption that $b$ exists implies that $f$ is defined at any point $(0,y,0,w)$ of the plane $x=z=0$, since $b$ is non-zero at such a point .
In particular $f$ would be defined at the origin $O=(0,0,0,0)$, which contradicts the result established in 1) .

Edit
My proof in 1) that $f$ is not defined at the origin $O$ is a bit roundabout.
Here is a more direct one:

Suppose we could write $f=\bar x/\bar y=\bar a/\bar b$ for some polynomials $a, b$ with $b$ satisfying $b(O)\neq0$.
This means that there exists a polynomial $h$ such that $bx-ay=h(xw-yz)$.
Taking the linear term of the polynomials on both sides of the equality sign we would get the equality of linear forms (= homogeneous polynomials of degree one) $$b(O)x-a(O)y=0$$ a contradiction with the requirement $b(O)\neq 0$.

Georges Elencwajg
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