UTM Ideals Varieties and Algorithm - Chapter 1 Section 4 Exercise 7

Problem:

Solution:

Recall the ideal of a variety is the set of all polynomials vanishes on $V$, but what is $V$ as a point set?

We claim $V(x^n, y^m) = \{(0,0)\}$, this is actually pretty obvious, for this is the only solution for $x^n = y^m = 0$.

So the problem becomes, what is the set of all polynomials that vanish on $\{(0,0)\}$?

For one thing, it must not have a constant term, otherwise $\{(0,0)\}$ is not a solution. With that, we pick all the terms with $x$ and factor out $x$, and then all the term without $x$ must have $y$ and therefore factor out $y$, we get

$f(x, y) \in I(V(x^n, y^m)) \implies f(x, y) = a(x, y) x + b(x, y) y \in \langle x, y \rangle$, this show $I(V(x^n, y^m)) \subset \langle x, y \rangle$.

The other inclusion is just as easy, notice if $f(x, y) = c_1 x + c_2 y$, then $f(0, 0) = 0 \in I(V(x^n, y^m))$.

Therefore we proved $I(V(x^n, y^m)) = \langle x, y \rangle$.