Re: like a definition question about sub-algebra

In article <20776843.1143221825835.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
eugene <jane1806@xxxxxxx> wrote:

Alas. The post you followed-up to is not in my server. I have not
seen it. I do not know if and when I will see it. So, I do not know
what was in the post you followed-up to. Why do I not know?

BECAUSE YOU DID NOT QUOTE ANYTHING IN IT.

Sorry. I'll repeat my question that was the first in this thread
http://mathforum.org/kb/message.jspa?messageID=4583614&tstart=0

I don't like reading the newsgroup through mathforum. I especially
don't like having to stop what I'm doing, firing up a browser, going
to mathforum, and trying to find the post in question in order to know
what the person is talking about. It is very bad practice,
unfortunately made all the easier by Mathforum and by Google, to
follow-up with no context. It wastes time, and it makes it harder for
people to help you (you are forcing people to do the work of
backtracking, which seems rather annoying when you are also asking for
free help).

Here is it's formulation

"I have an easy question about sub-algebras
If A -is algebra with the multiplicator domain(domain field) F, then
what is sub-algebra of A generated by for-example elements x and y
from A ?

I think that it is the set {Ax+Ay} which is also a vector space over F.

What is your definition of "algebra", and what is the "multiplicator
domain"?

I would interpret what you wrote by saying that an algebra over F is a
ring which contains F in its center. In particular, algebras would
necessarily be unital.

Certainly, the subalgebra generated by x and y will contain both Ax
and Ay; it will also contain Ax + Ay, so the real question is whether
Ax + Ay is a subalgebra of A. Unfortunately, it need not be: consider
the case of A=F[x,y]; then Ax + Ay is not a subalgebra of A, because
it does not have a 1, and does not contain F in the center; the
subalgebra generated by x and y turns out to be all of A in this
case. The problem is that Ax + Ay need not contain 1, hence need not
contain F in its center.

Of course, if your definition of algebra is something else, then the
answer might be something else.



--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

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