UTM Ideals Varieties and Algorithm - Chapter 1 Section 4 Exercise 12

Problem:

Solution:

Part (a) is almost completely analogous to Exercise 11. Skipped here for brevity.
Suffice to say the answer is $(t^2, t^3, t^4) = V(y^2 - x^3, z - x^2)$.

To complete the rest, we claim $I(V) = \langle y^2 - x^3, z - x^2 \rangle$.

By substituting $(t^2, t^3, t^4)$ to a polynomial $p \in \langle y^2 - x^3, z - x^2 \rangle$, we show that $\langle y^2 - x^3, z - x^2 \rangle \subset  I(V)$

For the reverse inclusion, the problem already hinted on we cannot use the same approach as Exercise 11 for part (b), so we try division algorithm.

A polynomial in $k[x, y, z]$ can be thought of as $k[x, y][z]$, the ring of polynomial on the ring of $k[x, y]$, that is, we treat any polynomial of the form $f(x, y)$ as a coefficient to the power of $z$, so we can use division to show for any polynomial

$a(x, y, z) = (z - x^2)b(x, y) + c(x, y)$.

This is really no different from showing $a(z) = (z - m)b(z) + c$, just simple division.

Doing this again on the remainder give $c(x, y) = (y^2 - x^3)d(x) + e(x)y + f(x)$, so we get

$a(x, y, z) = (z - x^2)b(x, y) + (y^2 - x^3)d(x) + e(x)y + f(x)$.

Putting the parametrized curve into the equation, we get

$a(x, y, z) = e(t^2)t^3 + f(t^2) = 0$.

Here is the interesting part, we claim $e(x)$ and $f(x)$ are both 0 polynomial.

The reason is that if we look at $e(t^2)t^3 + f(t^2)$ as a polynomial in $t$, it is identically zero so all coefficients are 0, the polynomial has no constant term and $t$ terms, all even terms have coefficients comes from $f$ and all odd terms have coefficients comes from $e$, so all the coefficients of $e$ and $f$ are zero!

This prove the tricky part for this problem.

Writing a polynomial in $a(x, y, ) \in I(V)$ as $a(x, y, z) = (z - x^2)b(x, y) + (y^2 - x^3)d(x)$, we proved that $I(V) \subset  \langle y^2 - x^3, z - x^2 \rangle$.

So we conclude $I(V) = \langle y^2 - x^3, z - x^2 \rangle$.