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 $.