## Q and Q^2 are not Isomorphic as Groups

**Point of Post: **In this post we prove that as abelian groups, and discuss a generalization.

*Motivation*

As I was trying to go to sleep last night, I was pondering the contents of my post the other day. In particular, I was thinking about the statement that I made that “one should always ask the obvious questions”, whatever that’s supposed to mean. It hit me then, that while there are many ‘obvious’ questions that could follow from the fact that as abelian groups, I didn’t ask perhaps the most obvious one. Namely, is as groups? I was pretty sure the answer was no, but it took me a little bit to come up with a proof. Of course, as one is supposed to, one sees that the argument that proves this really only has to do with the fact that as a group, is the localization of some commutative unital ring and the fact that we’re discussing the localization as a module over the original ring. So, in the spirit of quicker, simpler post I record this fact here.

*The Theorem*

Let’s get right down to it. Why is as abelian groups? Roughly because we shall show that any -linear map is actually -linear, from where the result will clearly follow. Indeed, the key is the following lemma:

**Lemma: ***Let be a -linear map, then is -linear.*

**Proof: **We must merely note that if then

Thus, if , then

Thus, for all , and so clearly is -linear.

Incidentally, this also proves:

**Corollary: *** as left -modules.*

Now onto the titular theorem:

**Theorem: ***Let be a cardinal. Then, as abelian groups if and only if .*

**Proof: **Suppose that is a group isomorphism. Note then that since are group homomorphisms for each , we may conclude by the previous lemma that for each and so, in particular, is a -map. Since then is a -isomorphism and commutative rings have the IBN property we may conclude that as desired.

Of course, a natural question then is to ask whether for . The answer for infinite cardinals is a resounding yes, since, as we have proven in the quest to prove that infinite dimensional vector spaces aren’t isomorphic to their dual since, since is countable and infinite we have that and . So, if the two are isomorphic as groups then they must be equipotent, and so . As for the finite case, I’m not quite sure as of now. I’ll give it a thought, anyone who has a lead/knows the answer, feel free to leave a comment saying as much!

**EDIT: **Let me now resolve the dispute about whether or not as abelian groups for any cardinals . The basic idea (secretly very similar to Yaghoub’s, as below) is to show that if they are isomorphic as -modules they are isomorphic as -modules. What we show, more generally, is that if is an integral domain then as -modules implies that . The key lemma is the following:

**Theorem: ***Let be an integral domain. Then, as -spaces.*

**Proof: **We know that is a -space from basic theory. It clearly is not zero since given by is a non-zero -bilinear map. That said, note that for any simple tensor is equal to and thus is singly generated as a -space. The conclusion follows.

We can now prove our desired theorem:

**Theorem: ***Let be an integral domain. Then, if and only if .*

**Proof: **Suppose that as -modules. Then,

as -spaces. But, then we see that

as -spaces, which implies that as desired.

It may seem, at first, that this implies that as -modules, for it seems the same argument applies. Let’s analyze where it breaks down. Basically we, using the same method, get to the point that as -spaces. Now, we know that is an -space an and so looks like for some cardinal . So, all we can really conclude is that . So, if is infinite this is no contradiction if, say, and . In fact, since we know it can’t be a contradiction for finite we MUST have that is infinite.

If as abelian groups, then

as -algebras. Now take dimensions of both sides, as vector spaces over to get 🙂

Comment by Yaghoub | February 1, 2012 |

Haha! You sneaky snake! That’s very clever! I gave up to easily! Now I see how what you wrote is the obvious generalization of what I wrote! That’s cool man. Thank you.

Comment by Alex Youcis | February 1, 2012 |