Re: like a definition question about sub-algebra

In article <21463759.1143224933304.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
eugene <jane1806@xxxxxxx> wrote:
What is your definition of "algebra", and what is the
"multiplicator
domain"?
Thank you for reply.
I used to the following definition of algebra:
Algebra is a ring which is a vector space over a field F.

And do rings have to have a 1?

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

Actually i was trying to solve the following problem: Let A be
sub-algebra of R[X], generated by X^2 and X^3. Show that A isn't
isomorphic with R[X].

Which means we are mixing up notation already, since here "A" is the
subalgebra, and in the question you asked it was the algebra... Sigh.

Unfortunalety there wasn't a definition of the sub-algebra,generated
by it's elements.Maybe you can deduce from the problem condition what
was supposed by "sub-algebra of R[X], generated by X^2 and X^3".

Depending on your definition of ->RING<- (like pulling teeth...), the
answer will change.

Suppose first you do not require your rings to have a 1. Then
The subalgebra will be the smallest subspace which is closed under
products.

So take R[X], which is an algebra over R. If a subset of R[X] contains
X^2, contains X^3, and is closed under products must contain all
powers X^i with i>=2, since every number greater than or equal to 2
can be written as 2a+3b, with a and b nonnegative integers, not both
zero. If it is a vector space it must also contain aX^i for all a in R
and all i>=2. And it must contain all the sums. So it must contain all
polynomials which have zero linear and zero constant term. Since this
set is a linear subspace, and is closed under products, this is an
algebra. It is the smallest such set, as discussed, so the subalgebra
A is the conllection of all polynomials with coefficients in R which
have zero constant and zero linear term.

Why is this not isomorphic to R[X]? Well, for one thing, it does not
have a 1, so we are done.

Now suppose that your definition or ring ->requires<- that it have a
1. Then the subalgebra will be the smallest subspace which contains
the elements given, contains 1, and is closed under products. In this
case, we already know it contains all polynomials with zero linear and
constant term. If we throw in 1, it is easy to verify that the
subalgebra A will be the collection of all polynomials that have zero
linear term.

I am guessing that this is the actual definition you have, as
otherwise the result is too easy...

You want to show that it is not isomorphic to R[Y] (I'm changing
notation to make the argument clearer). That is, you want to show that
there is no ring isomorphism R[Y]->A which respects scalar
multiplication. Suppose that h:R[Y]->A is an algebra
homomorphism. Then h(pq)=h(p)h(q), h(p+q)=h(p)+h(q) for all p,q in
R[Y], and for all p in R[y] and a in R, h(ap)=ah(p).

Now, h(1) must be either 1 or 0, since h(p) = h(1p) = h(1)h(p); if
h(1)=0, then h is zero and it is not an isomorphism. If h(1)=1, then
h(a)=a for all a in R, since h(a) = h(a1) = ah(1) = a. Therefore,

h(a_0 + a_1Y + ... + a_nY^n) = a_0 + a_1h(Y) + a_2[h(Y)]^2 + ... +a_n[h(Y)]^n.

Let h(Y) = b_0 + b_2X^2 + ... + b_mX^m. If b_3 is nonzero, then it
should be easy to verify that no polynomial maps to X^2, so h is not
surjective hence not an isomorphism. And if b_3=0, then you should be
able to check that no element maps to X^3, so h is not surjective.

--
======================================================================
"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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