Proposition

Show that $\ev{xy, xz, yz}$ is a radical ideal.

Solution

Let $f \in k[x, y, z]$. Then $f$ is a finite linear combination of $x^ay^bz^c$ for some $a, b, c \geq 0$. If at least two of $a, b, c$ are positive, it can be expressed as a multiple of $xy, xz$, or $yz$. For instance, $xyz^2 = (xy)z^2$.

Using this idea, $f$ can be expressed as $f_1xy + f_2xz + f_3yz + c_1x^{n_1} + c_2y^{n_2} + c_3y^{n_3}$ for some $f_1, f_2, f_3 \in k[x_1, \cdots, x_n], c_1, c_2, c_3 \in k, n_1, n_2, N_3 \geq 0$.

Then $f^m = g_1xy + g_2xz + g_3yz + c_1^mx^{mn_1} + c_2^my^{mn_2} + c_3^mz^{mn_3}$ for some $g_1, g_2, g_3 \in k[x_1, \cdots, x_n]$. (This can be shown rigorously by induction on $m$.)

Therefore, $f_m \in \ev{xy, xz, yz}$ if and only if $c_1^m = c_2^m = c_3^m = 0$. And this happens if and only if $c_1 = c_2 = c_3 = 0$, and that happens if and only if $f \in \ev{xy, xz, yz}$. Therefore, $\ev{xy, xz, yz}$ is a radical ideal.