Regarding the definition of orders (as in subrings)


by Somewheresafe
Tags: definition, orders, subrings
Somewheresafe
Somewheresafe is offline
#1
Sep21-11, 12:17 AM
P: 4
Hi everyone! I'm new. :) Anyway there's this textbook I found regarding the definition of orders (a type of subrings). I'm kinda having trouble with the notations and the phrasings used. If anyone knows about this your help would be greatly appreciated. :)

Anyway the definition goes like this:

Let [itex]A[/itex] be a [itex]\mathbb{Q}[/itex]-algebra. A subring [itex]R[/itex] of [itex]A[/itex] containing its unity is called a [itex]\mathbb{Z}[/itex]-order (or simply an order) in A if R is finitely generated as a [itex]\mathbb{Z}[/itex]-module and [itex]\mathbb{Q}R=A[/itex].

Some things I'm not quite sure of:

1. Does [itex]\mathbb{Q}[/itex]-algebra refer to any group algebra of [itex]\mathbb{Q}[/itex]? Ie, the group algebra of [itex]\mathbb{Q}[/itex] over any group?

2. [itex]R[/itex] is finitely generated as a [itex]\mathbb{Z}[/itex]-module = [itex]R[/itex] itself is a module over [itex]\mathbb{Z}[/itex] with a finite generating set? (Kinda confused here. @_@)

3. I'm quite unsure about the notation [itex]\mathbb{Q}R[/itex]. Is this equal to
[itex]\left\{q r | q \in \mathbb{Q}, r \in R\right\}[/itex]? Or a linear combination of elements from this set? The previous pages don't actually indicate anything about it. :( (Or maybe I've missed it.)

4. Also, now that I'm at it, if I'm correct in no. 2, it means that any element in [itex]R[/itex] can be expressed as a linear combination of elements in the generating set over [itex]\mathbb{Z}[/itex]... but does the other way also hold? I mean, is it that any linear combination in the generating set over [itex]\mathbb{Z}[/itex] is also an element in [itex]R[/itex]?

Thanks!
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
micromass
micromass is offline
#2
Sep21-11, 04:24 PM
Mentor
micromass's Avatar
P: 16,703
Quote Quote by Somewheresafe View Post
Hi everyone! I'm new. :) Anyway there's this textbook I found regarding the definition of orders (a type of subrings). I'm kinda having trouble with the notations and the phrasings used. If anyone knows about this your help would be greatly appreciated. :)

Anyway the definition goes like this:

Let [itex]A[/itex] be a [itex]\mathbb{Q}[/itex]-algebra. A subring [itex]R[/itex] of [itex]A[/itex] containing its unity is called a [itex]\mathbb{Z}[/itex]-order (or simply an order) in A if R is finitely generated as a [itex]\mathbb{Z}[/itex]-module and [itex]\mathbb{Q}R=A[/itex].

Some things I'm not quite sure of:

1. Does [itex]\mathbb{Q}[/itex]-algebra refer to any group algebra of [itex]\mathbb{Q}[/itex]? Ie, the group algebra of [itex]\mathbb{Q}[/itex] over any group?
I don't see any mention of group algebra's. So I think they just mean [itex]\mathbb{Q}[/itex]-algebra as a [itex]\mathbb{Q}[/itex]-module that is also a ring.

2. [itex]R[/itex] is finitely generated as a [itex]\mathbb{Z}[/itex]-module = [itex]R[/itex] itself is a module over [itex]\mathbb{Z}[/itex] with a finite generating set? (Kinda confused here. @_@)
Every abelian group defines a [itex]\mathbb{Z}[/itex]-module by

[tex]nx=x+x+x+x+...+x~~~~(n~times)[/tex]

what they mean is indeed that this group is finitely generated (which is equivalent to finitely generated as module). Thus there exists an epimorphism [itex]\mathbb{Z}[X_1,...,X_n]\rightarrow R[/itex].

3. I'm quite unsure about the notation [itex]\mathbb{Q}R[/itex]. Is this equal to
[itex]\left\{q r | q \in \mathbb{Q}, r \in R\right\}[/itex]? Or a linear combination of elements from this set? The previous pages don't actually indicate anything about it. :( (Or maybe I've missed it.)
I guess it means a linear combination of such elements.


4. Also, now that I'm at it, if I'm correct in no. 2, it means that any element in [itex]R[/itex] can be expressed as a linear combination of elements in the generating set over [itex]\mathbb{Z}[/itex]... but does the other way also hold? I mean, is it that any linear combination in the generating set over [itex]\mathbb{Z}[/itex] is also an element in [itex]R[/itex]?
Yes, both implications hold.
Somewheresafe
Somewheresafe is offline
#3
Sep22-11, 08:55 PM
P: 4
Thanks for the reply! It's a little clearer to me now! :) But just one little thing...
[QUOTE=micromass;3514575]I don't see any mention of group algebra's. So I think they just mean [itex]\mathbb{Q}[/itex]-algebra as a [itex]\mathbb{Q}[/itex]-module that is also a ring.[\quote]
Wouldn't that be the same thing as a group algebra of [itex]\mathbb{Q}[\itex] over any group?

micromass
micromass is offline
#4
Sep22-11, 09:03 PM
Mentor
micromass's Avatar
P: 16,703

Regarding the definition of orders (as in subrings)


Quote Quote by Somewheresafe View Post
Thanks for the reply! It's a little clearer to me now! :) But just one little thing...
I don't see any mention of group algebra's. So I think they just mean [itex]\mathbb{Q}[/itex]-algebra as a [itex]\mathbb{Q}[/itex]-module that is also a ring.
Wouldn't that be the same thing as a group algebra of [itex]\mathbb{Q}[\itex] over any group?
No, certainly not every [itex]\mathbb{Q}[/itex]-algebra is a group algebra. For example, each [itex]\mathbb{Q}[/itex] group algebra would have no zero divisors. However, [itex]M_n(\mathbb{Q})[/itex] (the matrices) do have zero divisors if n>1.
Hurkyl
Hurkyl is offline
#5
Sep23-11, 03:51 AM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
Quote Quote by micromass View Post
For example, each [itex]\mathbb{Q}[/itex] group algebra would have no zero divisors.
What about (1 - [e])(1 + [e]) = 0 for any element e of order 2?
micromass
micromass is offline
#6
Sep23-11, 09:04 AM
Mentor
micromass's Avatar
P: 16,703
Hmm, I should have known better than to post that...
Somewheresafe
Somewheresafe is offline
#7
Sep24-11, 11:55 PM
P: 4
Hmm so they're the same?

Anyway thank you again for the replies! :)


Register to reply

Related Discussions
All discrete subrings of R Calculus & Beyond Homework 3
Homomorphism and Subrings Calculus & Beyond Homework 1
Subrings Calculus & Beyond Homework 7
Ring homomorphism and subrings Calculus & Beyond Homework 3
Subrings and Subfields Linear & Abstract Algebra 14