## Understanding Baer Rings

I have been kind of inactive in mathematics, but hopefully not yet infertile. Recently there is a big thing that has been teasing me and I have to write about it. If not anywhere, at least here!

Here, we will always assume commutative unitary rings.

Definition: Suppose that $A$ is a ring, then $A$ is said to be Baer if for any subset $S\subset A$ theres is an idempotent $e\in A$ such that $eA$ is the set of annihilator of $S$.

It was in the 70’s that Mewborn showed that every reduced commutative ring has a Baer Hull, this is the smallest intermediate ring between the ring and it’s complete ring of quotients that is Baer. Mewborn showed a more detailed construction of the Baer hull of a reduced commutative ring. He showed that the Baer Hull of a reduced commutative ring is just the adjoint of the ring with all the idempotents of its complete ring of quotients.

But… there is a question, rather a conjecture, that lingers in my head…

Conjecture. Suppose that $A$ is a reduced ring and suppose that there are finitely many idempotents (in its complete ring of quotient) $e_1,e_2,\dots, e_n$ such that $A[e_1,\dots,e_n]$ is a Baer ring. Then $A$ is actually a Baer ring.

I think I can show/prove the following:

Conjecture. Let $A$ be a real Baer ring and let $T$ be the total integral closure of $A$ such that $T$ is in a natural way a finitely-generated $A$-module. Then $A$ is actually a real closed ring and

$T= A[\sqrt{-1}]$

and if I know that the first Conjecture holds true, then I can even remove the requirement of $A$ being Baer. This would be great!

But, why am I trying so hard to show that this second Conjecture holds?? Simply because the second Conjecture is beautiful 🙂 not to mention the fact that it is a generalization of the “classical” Artin Schreier Theorem (if you $A$ were a field, then its total integral closure is no other than its algebraic closure. As a note, I am afraid I cannot remove the requirement that $A$ should be real.).

Artin-Schreier Thereom. Let $F$ be a field and let $K$ be its algebraic closure. Suppose furthermore that $F$ and $K$ are unequal, but $K$ is a finite field extension of $F$. Then $F$ has actually characteristic 0. In fact, $F$ is a real closed field and

$K = F[\sqrt{-1}]$

Ancel C. Mewborn, “Regular Rings and Baer Rings”, Math. Z. 1971, vol. 121, p. 211-219