Is the following correct? I couldn't find a statement of this, much less a proof, when searching Google, so I am not sure how else to double-check this.
Also, does anyone know of a direct proof, or does the result depend inextricably on the Law of the Excluded Middle?
Let $I$ be a maximal ideal of a (commutative) ring $R$. Assume by means of contradiction that $Rad(I) = R$. Then $1 \in Rad(I) \implies 1^m = 1 \in I$ for some $m>0 \implies I = R$. This is a contradiction because $I$ is a maximal ideal, hence proper.
Therefore $I \subset Rad(I) \subsetneq R$ which implies that $I = Rad(I)$ because $I$ is maximal. $\square$