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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.

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    $\begingroup$ It seems correct to me. $\endgroup$
    – Louis
    Jun 12 '18 at 7:46
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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.

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    $\begingroup$ $(1 + x)(1 - x) = 1 - x$ 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

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