How can I prove that $(x^3-y^2)$ is a radical ideal in $\mathbb{F}_2[x,y]$?

Question

In an algebraically closed field, it's easy to verify that $(x^3-y^2)$ is a prime ideal, hence a radical ideal. However, $\mathbb{F}_2$ is not algebraically closed. So how can I prove this?

Answer 1

I was reasoning along the following lines but now am not so sure. RTP that if $p^r \in (x^3-y^2)$ then so is $p$. Clearly $r$ cannot be even, as a square in char 2 only has even powers (cross terms go). So it must be odd. If $r=1$ we are done (vacuously). The cube of a linear polynomial can't look like $x^3-y^2$ (just test all possibilities) and neither can any higher power. So the ideal is radical.