Computing the radical of an ideal using Nullstellensatz

Question

How do you compute the radical of the ideal $\langle\,x^3+1\,\rangle$ in $\mathbb{C}[x]$? I know that you can use Nullstellensatz but I don't know.

Answer 1

You don't need the full Nullstellensatz for this example. Over $\mathbb{C}$, $x^3 + 1$ factors as $$ x^3 + 1 = (x - \alpha)(x - \alpha \zeta_3)(x - \alpha\zeta_3^2), $$ where $\zeta_3$ is a primitive cubic root of unity, and $\alpha^3 = -1$. These linear factors generate distinct maximal ideals of $\mathbb{C}[x]$, so by the Chinese Remainder Theorem, $\mathbb{C}[x]/(x^3 + 1) \simeq \mathbb{C} \times \mathbb{C} \times \mathbb{C}$ is reduced. Hence, $(x^3 + 1)$ is radical.

This example is simpler because $\mathbb{C}[x]$ is a PID. In general, for $k = \bar{k}$ an algebraically closed field, the Nullstellensatz implies that for $I \subseteq k[x_1, \dots, x_n]$ an ideal, $$ \sqrt{I} = \bigcap_{(a_1, \dots, a_n) \in V(I)} (x_1 - a_1, \dots, x_n - a_n). $$ Here, the intersection runs over simultaneous zeros of the ideal $I$. If you apply the theorem to your example, you find that $$ \sqrt{(x^3 + 1)} = (x - \alpha) \cap (x - \alpha \zeta_3) \cap (x - \alpha \zeta_3^2). $$ These ideals are coprime as noted above, so the intersection is equal to the product $(x - \alpha)(x - \alpha \zeta_3)(x - \alpha \zeta_3^2) = (x^3 + 1)$.