UTM Ideals Varieties and Algorithm - Chapter 1 Section 4 Exercise 10

Problem:

Solution:

First, let's look at what is $V(x - y)$, it is the point set such that $x - y = 0$, so it is simply the straight line $(t, t)$.

Next, we look at what $\langle x - y \rangle$ is, it is the set of all polynomial with a factor $(x - y)$, so it must vanish on the straight line, so we showed $\langle x - y \rangle \subset I(V(x - y))$

For the reverse inclusion, we consider the polynomials in $I(V(x - y))$, they must vanishes on $(t, t)$. The challenge is to show one such polynomial must have $(x - y)$ as a factor.

In order to solve the challenge, we need this result:

We claim that any polynomial $f(x, y)$ can be written as $f(x, y) = h(x, y)(x - y) + r(x)$.

The prove the claim, it suffice to show that the trick works when $f(x, y)$ are monomials. For general polynomials we just add the terms up.

So for a general monomial, we see that we can indeed express it in the claim form as follow:

$x^a y^b = x^a(x - (x - y))^b = x^a(x^b + c(x, y)(x-y)) = x^{a+b} + x^ac(x, y)(x - y)$.

Armed with the claim, we write any polynomial $f(x, y) \in I(V)$, $f(x, y) = h(x, y)(x - y) + r(x)$.

It must vanishes on $(t, t)$, so $h(t, t)(t - t) + r(t) = 0 \implies r(t) = 0$.

Therefore we proved any polynomial $f(x, y) \in I(V) \implies f(x, y) \in \langle x - y \rangle$.

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Here we have the previous wrong proof  - just for reference for a mistake I had.

An easy fact is that such polynomial must vanish on $(0, 0)$, so the polynomial must not have a constant term, any such polynomial can be written as

$f(x, y) = x a(x, y) + y b(x, y)$.

Next, we substitute $(t, t)$ and get $a(t, t) = -b(t, t)$ whenever $t \ne 0$.

So we have two polynomials that agree on infinite number of points (as the field $k$ is infinite). The two polynomials must be equal, so we have $a(x, y) = -b(x, y)$.

Putting back in, we have $f(x, y) = x a(x, y) - y a(x, y) = (x - y) a(x, y)$, so we have proved that any such polynomial must have $(x - y)$ as a factor, or $I(V(x, y)) \subset \langle x - y \rangle$.

The key mistake is in the red line. In general, two polynomials of two or more variables, even when they agree on infinite number of points, can be different.

See:
http://math.stackexchange.com/questions/623981/check-that-two-function-fx-y-and-gx-y-are-identical

This is a good story learnt.