# Another basic ring question

1. Aug 21, 2014

### 1MileCrash

Is -(x^-1) = (-x)^-1 true for all nonzero x in any ring, where x^-1 denotes the multiplicative inverse of x?

2. Aug 21, 2014

### micromass

Staff Emeritus
Not all nonzero $x$ are invertible in a ring. So you must demand $x$ to be invertible.
In that case, why don't you try to prove it?

3. Aug 21, 2014

### 1MileCrash

I was only able to prove it for a commutative ring.

4. Aug 21, 2014

### economicsnerd

Small hint: What's $x*(-1) + x$? How about $(-1)*x + x$?

5. Aug 21, 2014

### micromass

Staff Emeritus
Please post your proof. Where do you use commutativity?

6. Aug 21, 2014

### 1MileCrash

-x • -(x)^-1 = -1 • x • -1 • (x)^-1 = -1 • -1 • x • (x)^-1 [with commutativity] = 1 • 1 = 1

Is the proof I did.

7. Aug 21, 2014

### micromass

Staff Emeritus
OK, so the step where you used commutativity is

$$(-1) * x = x* (-1)$$

But in fact, this is true in any ring, even noncommutative. So your goal should now be to prove this without using commutativity.

8. Aug 24, 2014

### 1MileCrash

Yeah, I haven't responded to this because I am just so embarrassed. I am so careful not to assume what I'm not given in proofs regarding algebraic structures that I often make mistakes like this. Thanks again.

9. Aug 25, 2014

### micromass

Staff Emeritus
So you found it?

10. Aug 25, 2014

### WWGD

AFAIK, the unity satisfies 1r=r1 for all r in the ring, by definition.

Last edited: Aug 25, 2014
11. Aug 25, 2014

### 1MileCrash

That's right, I just somehow chose not to see that this implied the same for -1.

mm, I just adjoined this as a lemma:
-1 • x = x • -1
Pf:
x + (-1 • x) = (1 • x) + (-1 • x) = (1-1)x = 0x = 0

x + (x • -1) = (x • 1) + (x • -1) = x(1•1 + 1•-1) = x0 = 0

Additive inverses are unique, so -1•x = x•-1

12. Aug 25, 2014

### micromass

Staff Emeritus
Seems good. In fact, if $n\in \mathbb{Z}$ (interpreted suitably), you can prove that $nx = xn$ for all $x\in R$.

13. Aug 25, 2014

### 1MileCrash

The proof is quick (I didn't write one, but it is obvious that (nx = (1+1+..+1)x = x+x+...x = (x•1+x•1+..+x•1) = x(1+1+..+1) = xn) - this seems even easier than the specific -1 case, but let me make sure I intuitively understand this..

If we have nx - even if these integers aren't in the ring R at all, this still makes sense because this can be considered to be "x to the (additive) power of n," x operated additively on itself n times, and that idea is commutative. It can't be considered a binary operation on members of R (because n is not in R here) but it is still well defined for any ring. Like, if I talk about the ring of mxm matrices in the usual way, and A is an mxm matrix, nA is just A added to itself n times, it doesn't matter that n isn't in the ring, nA makes sense.

If R DOES contain these integers n, then the above still applies, but we can additionally consider it to be exactly the same as a multiplicative operation on two members of the ring, n and x, which will be commutative since it is the same idea as the "x to the nth power" concept above.

Is that roughly correct, or am I spewing nonsense?

14. Aug 25, 2014

### micromass

Staff Emeritus
Yes, that is right. A more advanced way of seeing it is that there is a homomorphism of rings $\Phi:\mathbb{Z}\rightarrow R$ such that

1) $\Phi(1) = 1$ and $\Phi(0) = 0$
2) $\Phi(m+1) = \Phi(m) + 1$
3) $\Phi(-m) = -\Phi(m)$.

Of course you can now ask yourself if this is always injective/surjective.

All of this will be much clearer once you see the concept of $\mathbb{Z}$-module or $\mathbb{Z}$-algebra. Basically, a vector space over a field is an abelian group that allows you to multiply by scalars. A $\mathbb{Z}$-module allows you to multiply with integers. It is interesting to note that the $\mathbb{Z}$-modules correspond exactly with the abelian groups.

15. Aug 25, 2014

### disregardthat

Furthermore, rings are the monoid objects in the monoidal category of abelian groups together with the tensor product over $\mathbb{Z}$. I.e. a ring R is a abelian group together with a (multiplication) group homomorphism $R \otimes R \to R$ and a (unit) group homomorphism $\mathbb{Z} \to R$. This fact means that you can switch the order of any factor coming from $\mathbb{Z}$, including -1.

16. Aug 25, 2014

### 1MileCrash

Of course, this is all beyond my current studies, but I do plan to learn more about this eventually. Very cool stuff. So it appears by what you are saying is that by "a vector space being an abelian group with scalar multiplication" that this is actually not so different from a normal abelian group, it's not that scalar multiplication is a completely new operation, it is that scalar multiplication allows for all elements from an arbitrary field to act just as the integers already do for any abelian group. So the only thing keeping all abelian groups from being vector spaces is the fact that Z is not a field. That is quite an eye opener to me.

17. Aug 25, 2014

### micromass

Staff Emeritus
Yes, in that sense abelian groups and vector spaces are not different. They both have a scalar multiplication. An abelian group always has $\mathbb{Z}$ as multiplication, while a vector space has an entire field. This is the difference between the $\mathbb{Z}$-module and a vector space.

So in one sense they are very alike. But in another sense, the different scalars mean a world of difference. For example, any vector space has a basis, but the same is not true for a $\mathbb{Z}$-module. However, we can adapt some of the results. For example, you are undoubtly aware that finite-dimensional vector spaces are isomorphic to $\mathbb{K}^n$. It is interesting to notice that a similar theorem also holds for $\mathbb{Z}$-modules. You probably know the result too (but maybe didn't see it as similar to the vector space version). It is simply the fundamental theorem for abelian groups. http://en.wikipedia.org/wiki/Finitely-generated_abelian_group#Classification

Once you see modules, it is interesting to search for analogs like this.

18. Aug 26, 2014

### 1MileCrash

Now hold on, what about vector spaces whose underlying field F do not contain integers? Since they are abelian groups with additional features, scalar multiplication by integers in a vector space must always be possible, and so the set of scalars for the vector space would actually be F union Z, which could never be a field.

So what's the deal with that? I can't think of an example off hand, but I have no immediate objections to a field that does not contain integers.

19. Aug 26, 2014

### disregardthat

Every ring, and therefore field, has a distinguished homomorphism $\mathbb{Z} \to R$ picking out its unit, but it is not required to be injective. So it does have multiplication by $\mathbb{Z}$, and not-injectivity amounts to finite characteristic.

20. Aug 26, 2014

### micromass

Staff Emeritus
Certainly, some fields do not contain the integers. For example, $\mathbb{Z}_2$ (the integers mod 2) are a field that do not contain the integers. However, we see that there is still always a standard map $\Phi:\mathbb{Z}\rightarrow F$ as described in one of my previous posts. The issue is that this map isn't always injective. (by the way, the kernel of this map is always of the form $\alpha\mathbb{Z}$ and $\alpha$ is called the characteristic of the field, for example $\mathbb{Z}_2$ has characteristic 2, while $\mathbb{R}$ has characteristic 0. Exercise: prove that the characteristic of a field is either 0 or a prime number).

Now, take a vector field $V$ over $\mathbb{Z}_2$, for example. We can always multiply a vector $\mathbf{v}$ by an integer $m\in \mathbb{Z}$. But it turns out (as you can easily check) that $m\mathbf{v} = \Phi(m) \mathbf{v}$. So there is no real difference between multiplying by an element of $\mathbb{Z}$ and multiplying by a suitable element of $\mathbb{Z}_2$.

You can generalize this. If $M$ is an $R$-module with $R$ a ring (again an $R$-module is just a vector space where the scalars are in a ring $R$ and not a vector space) and if $f:R^\prime\rightarrow R$ is a morphism of rings, then $M$ is an $R^\prime$-module in a very canonical way: $\alpha\mathbf{m} := f(\alpha)\mathbf{m}$.
You of course know about every $\mathbb{C}$-vector space being a $\mathbb{R}$-vector space since if you can multiply by a complex number, then you can multiply with a real. This is just a special case of the previous generalization with the morphism $f:\mathbb{R}\rightarrow \mathbb{C}:x\rightarrow x$.

I guess what I want to say is that every module (and thus every vector space) determines in a canonical way many other modules and vector spaces. Indeed, any $R$-module determines an $R^\prime$-module for each morphism $\Phi:R^\prime\rightarrow R$. It is in this light that you should see every vector space being a $\mathbb{Z}$-module. That $\mathbb{Z}$ is actually included in the field is not necessary. Also note that for every ring $R$ (and thus field), there is always a unique morphism $f:\mathbb{Z}\rightarrow R$. So the result is even better: every $R$-module determines by this a unique $\mathbb{Z}$-module.

Does this make a bit of sense? I'm sorry to use so much module-specific language that I think you haven't seen yet, but I found no other way of discussing this.