Units and nilpotents in quotient ring.

Question

$A$ is a commutative ring and $N(A)$ is the nilradical of $A$. If $A/N(A)$ is a field, show that every $a \in A$ is invertible or nilpotent.

Answer 1

Let $a\in A$.
If $a\in N(A)$ then $a$ is nilpotent.
If $a\notin N(A)$ then $\bar a$, the residue class of $a$ modulo $N(A)$, is invertible in $A/N(A)$, so there is $\bar b\in A/N(A)$ such that $\bar a\bar b=\bar 1$, that is, $1-ab\in N(A)$. Since $1-ab$ is nilpotent there is an positive integer $m$ such that $(1-ab)^m=0$. Now use the binomial theorem and get $1-mab+\cdots+(-1)^ma^mb^m=0$, so $1=a(?)$ and therefore $a$ is invertible..

Answer 2

The nilradical is the intersection of all prime ideals, so in particular is contained in all maximal ideals. If $A/N(A)$ is a field, then $N(A)$ is maximal, and this shows that $N(A)$ is the unique maximal ideal of $A$. Now pick $a\in A$ outside of $N(A)$. If $a$ is not invertible, then it is contained in a maximal ideal, but since there is only one, it must be $a\in N(A)$, contradiction. Therefore $a$ is invertible and the claim is proved.