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This last week I've been trying to prove that a certain ring is von Neumann regular. Here come the details.

Let $A$ be a commutative ring with unity. Put $A'=A[X_a]_{a\in A}/(a^2X_a-a,aX_a^2-X_a)_{a\in A}$ and let $\psi:A\to A'$ be the canonical homomorphism, defined by $\psi(b)=b+(a^2X_a-a,aX_a^2-X_a)_{a\in A}$. I want to prove that $A'$ is von Neumann regular.

Jean-Pierre Olivier seems to have proved in Anneaux absolument plats universels et épimorphismes à buts réduits. Séminaire Samuel. Algèbre commutative, 2 (1967-1968), Exp. No. 6, 12 p., that $A'_\mathfrak{p}$ is a field for every prime ideal $\mathfrak{p}\in\text{Spec}(A')$. (Proposition 5.) I don't follow his proof at all however. (It's very terse.)

I know from exercise 3.10 of Atiyah-Macdonald that $A'$ is von Neumann regular iff $A'_\mathfrak{m}$ is von Neumann regular for every maximal ideal $\mathfrak{m}\in\text{Spm}(A')$, and this is the approach I'd like to take.

It's equivalent to prove that for every maximal ideal $\mathfrak{m}$ we've that $A'_\mathfrak{m}$ is a field.

Update: I tried to prove $\mathfrak{m}A'_\mathfrak{m}=\{0\}$ immediately instead. Let $x\in\mathfrak{m}A'_\mathfrak{m}$, and write $x=\frac{a}{s}$. We've that $a\in\mathfrak{m}$ and $s\in A'_\mathfrak{m}$. Suppose that $x\neq 0$, i. e. that there exists no zero-divisors of $a$ in $A'\setminus\mathfrak{m}$.

If $a^{-w}$ (weak inverse of $a$) exists in $A'$, we've that $1-aa^{-w}\in A'\setminus\mathfrak{m}$ and $a(1-aa^{-w})=0$, which is a contradiction.

If $a^{-w}$ doesn't exist in $A'$ we've that for all $b\in A'$ it holds that $a(ab-1)\neq 0$ or $b(ab-1)\neq 0$. Moreover, by the structure of $A'$, there exists a $b_0\in\psi(A)$ such that $(ab_0)^{-w}$ exists in $A'$.

If $b_0\in A'\setminus\mathfrak{m}$ then since $1-ab_0(ab_0)^{-w}\in A'\setminus\mathfrak{m}$, we've that $ab_0(1-ab_0(ab_0)^{-w})=0$ which is a contradiction.

But if $b_0\in\mathfrak{m}$ there must be a contradiction, and since I've found no contradiction of the assumption that $a^{-w}$ doesn't exist, I guess $b_0\in\mathfrak{m}$ might imply that $a^{-w}$ can be constructed somehow.

Tobias Shuxue Laoshi
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  • That's right, $\psi$ is not needed. In the paper, Olivier uses $T(A)$ as his notation for $A'$, and writes in part 3 of prop. 5: ``for every $x\in\text{Spec}(T(A))$, $T(A)_x$ is isomorphic to the field of fractions of $A$ in $\psi^{-1}(A)$''. Where $\psi$ is the canonical map. – Tobias Shuxue Laoshi Dec 13 '16 at 08:39
  • Ok, so I've taken a different approach now. It does in fact use properties of $A'$, which is hopeful. There's just one detail to remedy, see update.

    Also ``of $A$ in $\psi^{-1}(x)$'' above.

     – Tobias Shuxue Laoshi Dec 13 '16 at 08:59
  • If $P$ is a prime ideal in $A[X_a]{a\in A}$ containing $(a^2X_a-a,aX_a^2-X_a){a\in A}$ then there are two cases: $aX_a-1\in P$ (and then $a\notin P$), or $a,X_a\in P$. This shows that the ideal $(a^2X_a-a,aX_a^2-X_a)_{a\in A}$ localized at $P$ equals the ideal $(bX_b-1,c, X_c)$ localized at $P$ (I've split the generators of the given ideal in two subsets accordingly to the cases mentioned before). Since $b\notin P$ it is invertible in the localization, so we get the ideal $(X_b-b^{-1},c, X_c)$ localized at $P$. Now, in the quotient ring $X_b$ and $X_c$ vanishes, so we get $A_p/pA_p$ – user26857 Dec 13 '16 at 23:50
  • where $p=P\cap A$ (or $p=\psi^{-1}(P)$ if you like it) since the set $c\in A$ with $c\in P$ is in fact $p$. – user26857 Dec 13 '16 at 23:53
  • I hope you understand my notation (and arguments). – user26857 Dec 14 '16 at 00:00
  • Sorry, I don't see what's going on in the end. How do you conclude that $A'_p$ is a field? – Tobias Shuxue Laoshi Dec 14 '16 at 07:53
  • Maybe you mean $A'_P$. Well, I've shown that $A'_P\simeq A_p/pA_p$. – user26857 Dec 14 '16 at 08:36
  • Ok. I'll take a deep look and get back to you. – Tobias Shuxue Laoshi Dec 14 '16 at 09:01