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| George Cox |
 Posted: Tue Nov 30, 2004 2:21 pm |
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To embed a ring in a field, the ring must have no divisors of zero since
fields have no divisors of zero. If the ring is commutative as well (=
integral domain) then it can always be embedded in a field.
What if the ring is non commutative? Can such a ring with no divisors
of zero always be embedded in a non commutative field (= skew field)? |
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| Arturo Magidin |
| Posted: Tue Nov 30, 2004 2:59 pm |
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In article <41AD0E82.FDB13560@spambtinternet.com.invalid>,
George Cox <george_coxanti@spambtinternet.com.invalid> wrote:
[quote:8303410347]To embed a ring in a field, the ring must have no divisors of zero since
fields have no divisors of zero. If the ring is commutative as well (=
integral domain) then it can always be embedded in a field.
What if the ring is non commutative? Can such a ring with no divisors
of zero always be embedded in a non commutative field (= skew field)?
[/quote:8303410347]
No. There is an example due to Mal'cev of a noncommutative ring with
no zero divisors that cannot be embedded in a division ring.
(I'm taking this from T.Y. Lam's "Lectures on Modules and Rings", GTM
no. 189; Chapter 4 is about rings of quotients).
You start with a cancellative semigroup H with elements
a,b,c,d,x,y,u,v that satisfy
ax = by
cx = dy
au = bv
cu not equal to dv.
Then let R be the semigroup algebra kH, where k is any domain (ring
with no zero divisors). Then R is a domain, and cannot be embedded
into any division ring.
I don't know if there are known necessary and sufficient
conditions. Here's what Lam has:
If a domain R is embeddable in a divisoin ring D, then R must satisfy
the following conditions:
(1) R must have Invariant Basis Number (i.e., R^n = R^m as right
modules implies n=m).
(2) R must be stably finite (the matrix rings over R are
Dedekind-finite, meaning for any c,d, cd=1 implies dc=1).
(3) (Klein's Nilpotence Condition) For any nilpotent n x n matrix
A over R, we have A^n = 0.
(4) The n x n matrix ring over R must satisfy ACC and DCC on left and
right annihilators.
It is known that (2)->(3); each of (3) and (4) implies (2). George
Bergman has shown that (3) is not sufficient for the
embeddability. The example in fact satisfies a very strong condition:
the multiplicative semigroup of nonzero elements of R is embeddable in
a group, but the ring is not embeddable in a division ring.
Some sufficient conditions known:
1. Any right noetherian domain can be embedded in a division ring.
2. If a domain R can be embedded in a direct product of division
rings, then it can be embedded in a division ring.
3. In particular, any ring which can be embedded into a strongly
regular ring (von Neumann regular and reduced) can be embedded
into a division ring (since strongly regular rings are subdirect
products of division rings).
4. Any right Ore domain has a division hull which is a division
ring.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu |
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| George Cox |
| Posted: Tue Nov 30, 2004 4:01 pm |
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Lee Rudolph wrote:
[quote:3709fe7cf1]
George Cox <george_coxanti@spambtinternet.com.invalid> writes:
To embed a ring in a field, the ring must have no divisors of zero since
fields have no divisors of zero. If the ring is commutative as well (=
integral domain) then it can always be embedded in a field.
What if the ring is non commutative? Can such a ring with no divisors
of zero always be embedded in a non commutative field (= skew field)?
I bet this is answered conclusively and exhaustively by P. M. Cohn
in _Free Rings and their Relations_ or one of his other books; but
I can't remember.
Lee Rudolph
[/quote:3709fe7cf1]
Thanks--I'll give him a ring. <groan> |
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| Lee Rudolph |
| Posted: Tue Nov 30, 2004 8:04 pm |
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George Cox <george_coxanti@spambtinternet.com.invalid> writes:
[quote:238e40f95a]To embed a ring in a field, the ring must have no divisors of zero since
fields have no divisors of zero. If the ring is commutative as well (=
integral domain) then it can always be embedded in a field.
What if the ring is non commutative? Can such a ring with no divisors
of zero always be embedded in a non commutative field (= skew field)?
[/quote:238e40f95a]
I bet this is answered conclusively and exhaustively by P. M. Cohn
in _Free Rings and their Relations_ or one of his other books; but
I can't remember.
Lee Rudolph |
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