Re: Rings and F-Algebras

From: Stuart M Newberger (smnewberger_at_comcast.net)
Date: 01/20/05 

Date: 20 Jan 2005 13:33:47 -0800

Porker899 wrote:
> Right? Wrong? Way off? I would appreciate help or a proof.
>
> Definition of F-Algebra: A ring R that is a vector space over F with
the same
> addition and ring multiplication and scalar multiplication are
related by
> (theta*a)*b=theta(a*b)=a(theta*b) for a,b in R and theta in F.
>
> Problem: Let R be a finite dimensional F-algebra where F is a field.
If Ris
> nontrivial and if r,s belong to R with rs=0 then either r=0 or s=0
IMPLIESS R
> is a division ring.
>
"Proof": For r not zero in R define the function s --> s*r is an
F-linear map from R to R. If
s*r=0 implies s=0 in R then there are no nonnzero elements in its
kernal.
Therefore the map is also surjective (onto R,since R is a finite
dimensional vector space over F). Thus it is a division ring.

Hold the phone.If you think that the axioms of a ring include a
multiplicative identity 1 then you have shown that given r not zero
that there is u with ur=1 and similarly (using the map s->rs) that
there is v with rv=1 .A well known one liner shows u=v and we are done.

What if you do not assume R has an identity .Then you must prove that
there is an identity in your ring with no zero divisors.Here is a hint
for this.f is any r in R not zero then by the above you can choose s
with rs=r s is not zero .Show s is an identity. 2nd hint: You can
choose t with st=1.Show using that R has no zero divisors that s=t
.Thus ss=s Now Every b in R is of the form b=sc (c->sc is onto) so that
sb=ssc=sc=b (similarly bs=b) and s is the identity for R.
Regards,Stuart M Newberger

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