# Check my proof that the nilradical and the Jacobson radical are equal (A&M 1.6)

Introduction to Commutative Algebra, chapter 1, exercise 6 is this:

A ring $$A$$ is such that every ideal not contained in the nilradical contains a nonzero idempotent (that is, an element $$e$$ such that $$e^2 = e \neq 0$$). Prove that the nilradical and Jacobson radical of $$A$$ are equal.

As every prime ideal is contained in a maximal ideal, the nilradical must be contained in the Jacobson radical. Assume for contradiction that the Jacobson radical is not contained in the nilradical. Then there must be a nonzero idempotent $$x \in \mathfrak{R}$$.

Because $$x\in \mathfrak{R}$$, $$1 - xy$$ is a unit for all $$y\in A$$. So $$1 - x$$ is a unit. But we have

$$1 - x^2 = (1 + x)(1 - x) = 1 - x.$$

This implies $$x = 0$$, a contradiction.

Does my proof work? It appears to be much simpler than other proofs I have seen online.

• It seems correct to me. Jun 12 '18 at 7:46

The inclusion $$\sqrt{(0)}\subseteq\mathfrak{R}(A)$$ is always valid.
For the converse, suppose $$\mathfrak{R}(A)\not\subseteq\sqrt{(0)}$$, then there exists $$x\in\mathfrak{R}(A)$$ such that $$x^2=x\neq 0$$.
Since $$x\in\mathfrak{R}(A)$$, then $$1-xy$$ is a unit for every $$y\in A$$, in particular $$1-x$$ is a unit. Therefore we have $$(1-x)x=x-x^2=x-x=0,$$ so $$1-x$$ is a zero divisor, contradicting the fact it is a unit.
Note that since $$\mathfrak{R}(A)\subset A$$ (is a proper subset), then $$x$$ cannot be 1, hence $$1-x$$ cannot be zero.
PS: I wrote this answer because I could not see why you concluded $$x=0$$ in your proof.
• $(1 + x)(1 - x) = 1 - x$ and $1-x$ invertible implies $1+x=1$. Mar 28 '19 at 21:38