Let $R$ be a commutative ring and $I $ be a maximal ideal of $R$. Then $I$ is a radical ideal of $R$

Question

Let $R$ be a commutative ring and $I $ be a maximal ideal of $R$. Then $I$ is a radical ideal of $R$.

Attempt at a proof:

Let $\sqrt{I}$ denote the radical of the ideal $I$. By its definition we have that $I\subseteq\sqrt{I}$. So it remains to show that $\sqrt{I} \subseteq I$ and we are done. Take an element $a\in\sqrt{I}$. There are now two possibilities for $a$: either $a\in I$ or $a \notin I$. If $a \in I$ then there is nothing to show, so suppose instead that $a\notin I$. Then the ideal generated by $a$ and $I$ properly contains $I$ and, because $I$ is maximal, this must be the whole of $R$, that is $aR+I=R$. Thus $ar + i = 1$ for some $r\in R$ and $i\in I$. If $r\in I$ then the LHS is an element of $I$, which contradicts the choice of $I$ (since $I$ is maximal so $I\subset R$ by definition). On the other hand, if $r\notin I$ then $i = 1 - ar\notin I$ which again gives a contradiction, since we chose $i$ to be an element of $I$. Thus $a\in I$ and $\sqrt{I} \subseteq I$.

Is this a correct proof?

Answer 1

You always have $I\subseteq \sqrt{I}$. If $\sqrt{I} = R$ then $1 \in \sqrt{I}$ so $1 = 1^k \in I$ and $I=R$, a contradiction. Thus $I=\sqrt{I}$.

Edit: A concern about your proof is that you never use what $\sqrt{I}$ is, you just use that $I$ is maximal. I'm not sure that if two elements don't belong to an ideal, their difference doesn't belong either: in $\mathbb{Z}$ clearly 7 and 5 are not even but $7-5 = 2\in 2\mathbb{Z}$. I don't think that step is correct.

Answer 2

Another way:

If you already know maximal ideals are prime, and that prime ideals are radical, then you are done.

Answer 3

Alternatively, an ideal $I$ is radical iff $R/I$ has no nilpotent elements. Then if $M$ is maximal, $R/M$ is a field, and therefore certainly has no nilpotent elements. Hence $M$ is radical.

Similarly, the same argument shows prime ideals are radical, because domains can't have nilpotent elements either.

Also as to whether or not your proof is correct, I'm not seeing where you used $a\in\sqrt{I}$. More specifically, I'm not following why $r\not\in I$ implies $1-ar\not\in I$.