Radical is a prime ideal (prove)

Question

Let $k \in \mathbb{N}$. Let $J_k:=\langle 2, x^k \rangle$ be an ideal of the polynomial ring $R:=\mathbb{Z}[x]$.

How can it be proved that the radical $\sqrt{J_k}:=\lbrace a \in \mathbb{R}|a^n \in J_k$ for an $n \in \mathbb{N}\rbrace$ is a prime ideal?

I started to prove that $\sqrt{J_k}=J_1$ for all $k \in \mathbb{N}$. Then it should follow that $\sqrt{J_k}$ is a prime ideal.

If $k=1$:

$\sqrt{J_1}=J_1$

For $k=2$:

$\sqrt{J_2}=J_1$, since $2^k \in J_k$.

Is this argumentation right?

I don't know how to continue now to show it for all $k \in \mathbb{N}$.

Answer 1

I would say your proof isn't complete. Here's a nice little technique for making these computations; recall the following

Fact: If $I$ and $J$ are ideals of a ring $R$, then $\sqrt{I+J}=\sqrt{\sqrt{I}+\sqrt{J}}.$

Let's say you've already shown that $\sqrt{\langle x^k\rangle}=\langle x\rangle$ (this is straightforward to check if you haven't already); then sneakily noticing that $\langle a,b\rangle=\langle a\rangle+\langle b\rangle$ for elements $a,b$ in a ring $R$, we can use the above fact to calculate

$$\sqrt{\langle 2,x^k\rangle}=\sqrt{\langle 2\rangle+\langle x^k\rangle}=\sqrt{\sqrt{\langle 2\rangle}+\sqrt{\langle x^k\rangle}}=\sqrt{\langle 2\rangle+\langle x\rangle}=\sqrt{\langle 2,x\rangle}=\langle 2,x\rangle.$$