Re: Total quotient ring and integral closures

So here goes:

---------------------------------- Thereom: Let A be a commutative
unitary reduced Baer ring
(http://planetmath.org/?op=getobj&from=objects&id=7864). Let T(A) be
its total quotient ring. Then the following are equivalent

1. A is integrally closed in T(A) 2. For any prime ideal P in T(A),
A/(P/\A) is integrally closed in T(A)/P.

Proof: I write B:=T(A) for simplicity.

Note - For those unfamiliar with Baer rings, please take note of the
following facts. Baer reduced commutative ring have the property that
their minimal prime spectrum (i.e. set of minimal prime ideals with
topology relative to their prime spectrum) is extramally disconnected
and a Stone space. It has another property that their total quotient
ring is (von Neumann) regular and the canonical map from the prime
spectrums of the total quotient ring and the ring is a homeomorphism
ONTO the minimal prime ideals.

The second assertion essentially says that the image of the natural map
Spec(T(A)) -> Spec(A) *is* the set of minimal primes in A if A is von
Neumann regular and reduced. (*)


Well, it says a bit more does'nt it? There is no assumption that A is regular (when I say regular I mean von Neumann regular).
It says that th map is a homeomorphism on th minimal primes of A if >>> T(A) <<<< is regular (which automatically implies that A and T(A) are reduced).

Exercise: If A is a zero-dimensional commutative unitary ring. Then it is reduced and all quotients over prime are fields.


That this map is a homeomorphism on the image is clear, since in general
Spec(S^-1A) sits homeomorphically in Spec(A) if S is a multiplicative
subset of A.


Yes it does, what is not *clear* is that its a homeomorphism onto the minimal primes. But for the case T(A) is regular it is shown to be so.


It goes back to the fact (*) goes back to the equivalence: P is minimal
in A iff A_P is a field (note that A is assumed reduced here!).


We don't assume A is regular. We assume T(A) is regular! Its a more general case..


(I won't show this, but you can read
it probably in some books maybe Huckaba's "Rings with zero divisors"
has it. But these are known facts of the 60s or earlier). Another
fact is that a Baer ring contains all the idempotent of its total
quotient ring.

That sounds something like a Baer ring is integrally closed for elements
whose minimal polynomial has at most degree 2 - or so ...


This would be something I doubt. I think we can even do this with integral domains.. Integral domains are Baer.. so the question boils down to:
Are all domains integrally closed for elements of degree two in its quotient fields?

Ok, lets make the proof now! (an element e of a Ring,
is idempotent iff e=e^2)

"2=>1" Let f be in A[T] be monic and b be in B such that f(b)=0. For
any P in Spec B, consider a_P to be the element in A such that f(a_P)

rather: *an* element - since it is not uniquely determined


true :)

is in P (because of 2). Consider the clopen sets (because B is
regular!): V_P = {Q in Spec B : a_P = b mod Q} then P is in V_P for
all P in Spec B and the union of all V_P (P in Spec B) covers Spec B.
Thus there are finitely many a_1,...,a_n in A such that

\/_{i =1... n} V_i = Spec B

where V_i ={Q in Spec B: a_i = b mod Q}

Definite another family of clopen sets U_1 = V_1, U_i = V_i/U_{i-1}
for i>=2 then again.. U_i's cover Spec B.

Note that the U_i is a finite _disjoint_ cover of *opens*.


yes! clopen not only open.

now define the idempotents (look at the sheaf structure of Spec B if
this helps) such that

e_i mod P = 1 if P is in U_i and 0 if P is not in U_i

So e_i is the characteristic function (in B) of the clopen subset U_i.


we could call it so :)

clearly b= sum_{i=1..n} a_ie_i (look at the sheaf structure of Spec
B) and all a_i's and e_i's are in A (A contains all idempotents of B
because A is Baer).

Nice! I haven't seen a proof like this before. :-) This smells like a
generalization of the Chinese remainder theorem. Note that this is
fairly close to what is closed partition of unity in (differential)
topology which gives you the chance to patch local things together to
make up a global object. I am saying this since this analogy gives me a
better feeling what is happening here. ;)

?? How could this be a generalization of CRT? I don't quite see this... Yeah.. it is something similar to
partition of unity isn't it :)


Thus b is in A!

"1=>2" Suppose by contradiction there exists an f in A[T] monic, P in
Spec B and b in B such that i. f(b) is in P ii. (b+P)/\ A = emptyset

Case 1: b is in P, then b=0 mod P and this is a contradiction to ii

Case 2: b is not in P.

Then f(b) is in P which is a subset of T(A).

Initially P is a subset of A, I think you mean that f(b) is a subset of
P.B - but P = P.B /\ A.


No.. P is a prime ideal of B=T(A).. its a subset of B.. f(b) is an element of P (i.e. f(b) = 0 mod P)

Let c be the

Is c really unique? I am not sure here, but if not it does not matter
for the proof, existence is enough.


Exercise: Let A be a regular ring (not necessarily commutative I think). Then all quasi-inverses are unique.

quasi-inverse of f(b) in B (ie. c be the element in B such that
cf(b)^2= f(b), this can be done because B is regular). Then cf(b) is
an idempotent in B and so it must be in A (see Note). Now 1-cf(b) is
also an idempotent

in A

, denote e=1-cf(b) (clearly e is NOT in P because
1-e is in P) and observe that (1-e)f(b)=0

Now we can write f(T) as f(T) = T^n + sum_{i=0 .. n-1} a_i T^i for
some positive integer n and a_1,...,a_n in A. then e^nf(b) = 0 and so
eb is a zero of the monic polynomial g in A[T] defined by g(T) T^n +
sum_{i=0... n-1} a_ie^{n-i}T^i so because A is integrally closed in
B, we then know that eb is in A

But that implies that eb = b mod P (since 1-e is in P)

(1-e)b in P.B /\ A = P


P is a subset of B.. P.B=P

and so (b +
P)/\A is NOT emptyset. Again a contradiction.

The theorem is proven ^_^ --------------------------------

This is a really nice one !!! Actually I haven't found out yet what is
happening here from a geometric point of view. When I will have some
more time I might come back to that.


Thanks a lot =) .. its really always a pleasure discussing math with you :) ..
Well, honestly.. maybe the difficulty is in describing reduced rings with regular total quotients "geometrically"..
actually specifically speaking we want to describe here Baer rings geomtrically. Is there a way we can describe
annihilator ideals geometrically?

Thanks again.

Sincerely,
Jose Capco
.

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