Show that $(x^3-y^3)$ is a radical ideal of $\mathbb{F}_2[x,y]$.

Question

Since $(x^3-y^3)\subset\sqrt{(x^3-y^3)}$, what we need to show is that $\sqrt{(x^3-y^3)}\subset(x^3-y^3)$. For any $i\in\sqrt{(x^3-y^3)}$，there exists $k\geq 1$，s.t. $i^k\in(x^3-y^3)$. Suppose that $i\notin(x^3-y^3)$. Now what should I do next to show the contradiction? Or is there any other way to prove this statement?

Answer 1

I'll expand a little on Joshua and Jyrki's comments.

It's really useful to know that $\mathbb F_2[x, y]$ is a unique factorization domain. In fact, if $k$ is any field, then the polynomial ring $k[x_1, \dots, x_n]$ is a UFD. (See here, for example.)

Now suppose that $R$ is a unique factorization domain and $I = (r) \subset R$ is a principal ideal. We want to determine whether $I$ is a radical ideal. To do this, we can write $r$ as a product of irreducibles, $r = p_1^{n_1} \dots p_k^{n_k}$. It's easy to see that $I$ is a radical ideal iff $n_1 = \dots = n_k = 1$ (i.e. iff each irreducible in the factorization of $r$ appears with multiplicity one). Rather than writing out a formal proof, it's probably more instructive if I illustrate this with some examples.

• Example 1: $r = p_1p_2$, where $p_1$ and $p_2$ are distinct irreducibles. Suppose $x^m \in (r)$. Then $r | x^m$. So $p_1 | x^m$ and $p_2 | x^m$. Thus $p_1 | x$ and $p_2 | x$. So $p_1p_2 | x$, i.e. $x \in (r)$. This shows that $(r)$ is a radical ideal.
• Example 2: $r = p_1^2$, where $p_1$ is an irreducible. Taking $x = p_1$, we have $x \notin (r)$ but $x^2 \in (r)$. So $(r)$ is not radical.

So a sensible approach would be to factorize $x^3 - y^3$ as a product of irreducibles, and check that each irreducible in this factorization appears with multiplicity one.