I am reviewing some old problems from Atiyah MacDonald Chapter 1, in particular problem 6. I was reviewing a proof that I found in this solution manual. I have some questions regarding minute details on the validity of this proof. I will reproduce the question and the solution below to keep the question self contained.
- A ring $A$ is such that every ideal not contained in the nilradical contains a non-zero idempotent (an element $e$ such that $e = e^2 \neq 0$). Prove that the nilradical and Jacobson radical are equal.
Solution 6 from online manual) It clearly suffices to show that every prime ideal in $A$ is maximal. Let $\mathfrak{p}$ be a prime ideal in $A$ and let $x$ be a non-zero element of $A-\mathfrak{p}$. Then the ideal $\langle x \rangle$ will contain an idempotent element $e \neq 0$, say $a_0x$. This implies that $a_0x(a_x-1) = 0 \in \mathfrak{p}$, so $a_0x(a_0x-1) = 0$ in $A/\mathfrak{p}$ too. However, $A/\mathfrak{p}$ is an integral domain, therefore $e = a_0x \neq 0$ implies $a_0x = 1$, thus $x$ is a unit, hence $A/\mathfrak{p}$ is a field.
My questions about the proof are:
- Does the $x$ we invoke need also be a non-unit since otherwise $\langle x \rangle = A$? However, if no such non-unit exists then $\mathfrak{p}$ would already be maximal?
- The proof relies on the $a_0x \neq 0$ in $A/\mathfrak{p}$ which is iff $a_0x \notin \mathfrak{p}$ but can we be sure $a_ox \notin \mathfrak{p}$ just because $x \notin \mathfrak{p}$?
- Assuming the logic questioned in 2 is valid, is the arbitrary $x \notin \mathfrak{p}$ being a unit in the quotient sufficient? It seems to me this would say that all the non units in $A$ correspond to units in the quotient, but then units always correspond to units also?