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.

  • 1
    $\begingroup$ It seems correct to me. $\endgroup$
    – Louis
    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
    $\begingroup$ $(1 + x)(1 - x) = 1 - x$ and $1-x$ invertible implies $1+x=1$. $\endgroup$
    – user26857
    Mar 28 '19 at 21:38
  • $\begingroup$ Ooh, thank you. It was very clear. $\endgroup$ Mar 29 '19 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.