## Idempotent elements outside the nilradical

Here is a cute exercise in Introduction to Commutative Algebra by Atiyah & Macdonald (A&M), that I recently solved, and felt like writing it up. The problem reads as follows:

Exercise 6 (Chapter 1). A ring $A$ is such that every ideal not contained in the nilradical contains non-zero idempotent (that is, an element $e$ such that $e^2=e\neq 0$). Prove that the nilradical and Jacobson radical of $A$ are equal.

Before proceeding to solution, let’s define what each of these terms mean. Even before that, let me note that throughout the word “ring” will mean commutative ring with 1 (multiplicative identity). This is the assumption made in the beginning of A&M. The nilradical $\mathfrak{N}$ of a ring $A$ is the set of all nilpotent elements of $A$ (an element $x\in A$ is called nilpotent if $x^{k}=0$ for some positive integer $k$.) Proposition 1.8 in A&M gives the following characterization:

Proposition 1.8. The nilradical of $A$ is the intersection of all the prime ideals of $A$.

On the other hand, Jacobson radical $\mathfrak{R}$ of a ring $A$ is defined to be the intersection of all maximal ideals of $A$. Proposition 1.9 in A&M gives the following characterization:

Proposition 1.9. $x\in\mathfrak{R} \Longleftrightarrow 1-xy$ is a unit in $A$ for all $y\in A$.

Using these two results, we can proceed to the solution of the exercise above.

Solution of Exercise 6 (Chapter 1). Since every maximal ideal is a prime ideal, and nilradical is the intersection of all prime ideals, it is clear that nilradical is contained in the intersection of all maximal ideals, that is, nilradical is contained in Jacobson radical. This proves one of the inclusions (note that this inclusion, namely $\mathfrak{N}\subset\mathfrak{R}$ holds generally in any ring). Now we need to show that Jacobson radical is subset of the nilradical. Let $x\in\mathfrak{R}$. We want to show that $x\in\mathfrak{N}$. Assume, to the contrary, $x\notin\mathfrak{N}$. By assumption, the ideal generated by $x$, namely the principal ideal $x=\{xa \ : \ a\in A\}$ is not contained in the nilradical. By hypothesis, this means that there exists a nonzero idempotent element in $(x)$. Hence, there exists $a\in A$ such that $(xa)^2=xa\neq 0$. Now since $x\in\mathfrak{R}$, using Proposition 1.9 we get that $1-xa$ is a unit. But $1-xa=1-(xa)^2=(1-xa)(1+xa)$. Since $1-xa$ is a unit, we can multiply this equation from the left by its inverse $(1-xa)^{-1}$ to get $1=1+xa$, or equivalently, $xa=0$. This contradicts $xa\neq 0$, and so we conclude that $x\in \mathfrak{N}$, as desired. This completes the proof that the nilradical and Jacobson radical of ring $A$ coincide.