Prove that a maximal ideal is radical.

Question

Let $m$ be a maximal ideal of commutative ring $R$. Prove that $m$ is radical.

I understand that $m$ is maximal if it is proper and there are no other ideals (except $R$) that properly contain it. In other words, $m$ is the "highest" ideal, as in if we have an ideal $n$ and $m \subseteq n$ then $m = n$ or $n = R$. I also understand that an ideal, $m$, is radical if $m = \sqrt m $ where $\sqrt m = \{ f \in k[x_1,x_2,...x_m] | f^m \in m\}$.

What do I have to do to show that the if $m$ is maximal then it is also radical?

Answer 1

$\sqrt{\mathscr{m}} = \{ r \in R | r^n \in \mathscr{m} \, \text{for some integer} \, n\}$.

To show that $\mathscr{m} = \sqrt{\mathscr{m}}$, we simply show that $\sqrt{\mathscr{m}} \subseteq \mathscr{m}$ because we already have that $\mathscr{m} \subseteq \sqrt{\mathscr{m}}$ since we can take $n=1$ for every $x \in \mathscr{m}$.

Take any $x \in \sqrt{\mathscr{m}}$. This implies $x^n \in \mathscr{m}$ for some $n$. Thus $x \in \mathscr{m}$ because $\mathscr{m}$ is also a prime ideal.

Answer 2

Hint: You have to show the following, for a maximal ideal $\mathfrak m$ of $R$:

1. $\sqrt{\mathfrak m}\supseteq\mathfrak m$
2. $\sqrt{\mathfrak m} \ne R$.

To show, e.g., assertion 2, it suffices to prove $1_R\not\in\sqrt{\mathfrak m}$. What would happen if $1_R\in\sqrt{\mathfrak m}$?