Finding the irreducible components of $V(x_1-x_2 x_3, x_1 x_3 - x_2^2)\subset \mathbb{A}_\mathbb{C}^3$

Question

I want to find the irreducible components of $V(x_1-x_2 x_3, x_1 x_3 - x_2^2)\subset \mathbb{A}_\mathbb{C}^3$.

Now it's not hard to find that $$V(x_1-x_2 x_3, x_1 x_3 - x_2^2)=V(x_1,x_2)\cup V(x_2-x_3^2,x_1-x_3^3).$$ I know the first part is irreducible, but I'm having trouble proving the second part is irreducible.

For I would need to prove that $I(V(x_2-x_3^2,x_1-x_3^3))=\sqrt{(x_2-x_3^2,x_1-x_3^3)}$ is prime. Now I can prove that $(x_2-x_3^2,x_1-x_3^3)$ is prime, and I'm pretty sure that $\sqrt{(x_2-x_3^2,x_1-x_3^3)}=(x_2-x_3^2,x_1-x_3^3)$, but I don't know how to prove this.

So my question is: How do I find $\sqrt{(x_2-x_3^2,x_1-x_3^3)}$? (without using a computer algebra system of course).

Thanks

Answer 1

$$k[x_1,x_2,x_3]/(x_2-x_3^2,x_1-x_3^3)\cong k[x_3]$$ so the ideal is in fact prime, and a prime ideal is radical.