# Regarding the definition of orders (as in subrings)

1. Sep 21, 2011

### Somewheresafe

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

Some things I'm not quite sure of:

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

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

3. I'm quite unsure about the notation $\mathbb{Q}R$. Is this equal to
$\left\{q r | q \in \mathbb{Q}, r \in R\right\}$? 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 $R$ can be expressed as a linear combination of elements in the generating set over $\mathbb{Z}$... but does the other way also hold? I mean, is it that any linear combination in the generating set over $\mathbb{Z}$ is also an element in $R$?

Thanks!

2. Sep 21, 2011

### micromass

I don't see any mention of group algebra's. So I think they just mean $\mathbb{Q}$-algebra as a $\mathbb{Q}$-module that is also a ring.

Every abelian group defines a $\mathbb{Z}$-module by

$$nx=x+x+x+x+...+x~~~~(n~times)$$

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

I guess it means a linear combination of such elements.

Yes, both implications hold.

3. Sep 22, 2011

### Somewheresafe

Thanks for the reply! It's a little clearer to me now! :) But just one little thing...

4. Sep 22, 2011

### micromass

No, certainly not every $\mathbb{Q}$-algebra is a group algebra. For example, each $\mathbb{Q}$ group algebra would have no zero divisors. However, $M_n(\mathbb{Q})$ (the matrices) do have zero divisors if n>1.

5. Sep 23, 2011

### Hurkyl

Staff Emeritus
What about (1 - [e])(1 + [e]) = 0 for any element e of order 2?

6. Sep 23, 2011

### micromass

Hmm, I should have known better than to post that...

7. Sep 24, 2011

### Somewheresafe

Hmm so they're the same?

Anyway thank you again for the replies! :)