Show that $\mathbf{I}(\mathbf{V}(x^n, y^m)) = \langle x, y \rangle$. Where $\mathbf{I}$ is the ideal, and $\mathbf{V}$ is the affine variety.
I'm not sure how to even begin on this one. I know that I need to show $\mathbf{I}(\mathbf{V}(x^n, y^m)) \subset \langle x, y \rangle$ and $\langle x, y \rangle \subset \mathbf{I}(\mathbf{V}(x^n, y^m))$.
If we start with $p \in \langle x, y \rangle$ then we know that
$$ p(x, y) = ax+by $$
vanishes at $(0, 0)$, for all $a, b \in \mathbb{R}\left[x, y\right]$.
Any hints would be appreciated ...
Here is an answer that doesn't use Hilbert's Nullstellensatz.
Let $P =(x_P, y_P) \in V(x^n, y^n)$. Since $x_P^n = y_P^n = 0$ and we're doing this computation over a field, it follows that $x_P = y_P = 0$. Thus, $V(x^n, y^n) = \{(0, 0)\}$ and $(x, y) \subset I(V(x^n, y^n))$.
Let $f \in I(V(x^n, y^n))$. By the polynomial division algorithm, we have $$ f(x, y) = xp(x, y) + q(y). $$
Since $f(0, 0) = 0$, we have $q(0) = 0$. It follows that $q \in (y)$. We conclude that $f \in (x, y)$ and $I(V(x^n, y^n)) \subset (x, y)$.