Radical of the powers of an ideal

Question

I am asked to prove the following:

$$\sqrt{\mathfrak{a}^n} = \sqrt{\mathfrak{a}}$$

Here is my attempt so far:

$\sqrt{\mathfrak{a}^n} \subseteq \sqrt{\mathfrak{a}}:$ (By Induction)

Clearly the base case $n=1$ holds, i.e., $\sqrt{\mathfrak{a}} \subseteq \sqrt{\mathfrak{a}}$.

Let $n=k$. Assume that for $n=k$, $\sqrt{\mathfrak{a}^k} \subseteq \sqrt{\mathfrak{a}}$ holds. Let $n=k+1$. Then

$\sqrt{\mathfrak{a}^{k+1}} = \sqrt{\mathfrak{a}^k \mathfrak{a}} = \sqrt{\mathfrak{a}^k} \cap \sqrt{\mathfrak{a}}$. Since $\sqrt{\mathfrak{a}^k} \subseteq \sqrt{\mathfrak{a}}$, by the inductive hypothesis, then $\sqrt{\mathfrak{a}^k} \cap \sqrt{\mathfrak{a}} \subseteq \sqrt{\mathfrak{a}}$. Hence $\sqrt{\mathfrak{a}^{k+1}} \subseteq \sqrt{\mathfrak{a}}$. Thus $\sqrt{\mathfrak{a}^n} \subseteq \sqrt{\mathfrak{a}}$.

$\sqrt{\mathfrak{a}} \subseteq \sqrt{\mathfrak{a}^n}:$

Let $x \in \sqrt{\mathfrak{a}}$. Then $\exists n> 0$ such that $x^n \in \mathfrak{a}$.

I can't see where to go after this. Any hints or clues would be greatly appreciated. Also is the first part, done by induction correct?

Thank you for your help.

Answer 1

It is easier to see it directly, maybe $$x \in \sqrt{\mathfrak a^n} \Rightarrow x^k = \sum a_i^n, a_i \in \mathfrak a$$

But of course $\sum_i a_i^n \in \mathfrak a$ so $x^k \in \mathfrak a$, which means that $x \in \sqrt{\mathfrak a}$.

In the other direction, $$x \in \sqrt{\mathfrak a} \Rightarrow x^k = a \in \mathfrak a \Rightarrow (x^k)^n = x^{kn} = a^n \in \mathfrak a^n$$

And if $x^{nk} \in \mathfrak a^n$, then it is in the radical as well.

Answer 2

If you know (as it seems) that $\sqrt{IJ}=\sqrt I\cap\sqrt J$, I don't understand why don't show by induction on $n$ that $\sqrt{\mathfrak a^n}=\sqrt{\mathfrak a}$?

For $n=1$ there is nothing to prove, and for the general case $$\sqrt{\mathfrak a^{n+1}}=\sqrt{\mathfrak a^n\mathfrak a}=\sqrt{\mathfrak a^n}\cap\sqrt{\mathfrak a}=\sqrt{\mathfrak a}\cap\sqrt{\mathfrak a}=\sqrt{\mathfrak a}.$$