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Regarding the definition of orders (as in subrings) 
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#1
Sep2111, 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! 


#2
Sep2111, 04:24 PM

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P: 18,036

[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
Sep2211, 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? 


#4
Sep2211, 09:03 PM

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Regarding the definition of orders (as in subrings)



#5
Sep2311, 03:51 AM

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#7
Sep2411, 11:55 PM

P: 4

Hmm so they're the same?
Anyway thank you again for the replies! :) 


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