Proposition

Given a field $k$, show that $\sqrt{\ev{x^2, y^2}} = \ev{x, y}$, and, more generally, show that $\sqrt{\ev{x^m, y^n}} = \ev{x, y}$ for any positive integers $n$ and $m$.

Solution

We have $x, y \in \sqrt{\ev{x^m, y^n}}$. Since the radical of an ideal is an ideal by Lemma 5 (P. 182, Ideals, Varieties, and Algorithms), $\ev{x, y} \subset \sqrt{\ev{x^m, y^n}}$. Let $f \in \sqrt{\ev{x^m, y^n}}$. Then $f^k = h_1x^m + h^2y^n$ for some $k \in \mathbb{N}$, $h_1, h_2 \in k[x, y]$. Since this implies that $f$ has no constant term, $f \in \ev{x, y}$. Therefore, $\ev{x, y} = \sqrt{\ev{x^m, y^n}}$.