# Ring laws

1. Apr 14, 2005

### BicycleTree

I'm just starting rings and I don't think I really understand them. When an operation is shown as a grid where the entry i, j in the grid is the element (i op b), I can see immediately what the commutative property looks like (grid symmetric about diagonal from top left) and what a unit, a zero, or a unity looks like. But I don't understand what the associative or distributive properties look like, graphically. Is there a way to recognize visually when they are there besides just testing every single case?

2. Apr 14, 2005

### Hurkyl

Staff Emeritus
Associativity and distributativity are usually easier to check when you have some sort of formula or algorithm for the operations, rather than a messy table.

3. Apr 19, 2005

### BicycleTree

Well, the reason I was asking was because of a particular proof which I don't think I understand rings well enough to know how to do. Coming back to it now, I still can't do it. The problem is this:

Prove that for a ring (R, +, *), if S is a nonempty finite subset of R and for all a, b elements of S, a + b and a * b are elements of S, then S is a subring of R.

I need to show that there exist additive inverses of each element of S but I'm not making any progress. I can't get an intuitive idea of the situation.

4. Apr 19, 2005

### snoble

The trick here is to think about finiteness and the pigeon whole principal. The only thing you need to understand about distibutivity is that since S is a subset of a Ring distributivity still holds (or is said to be inherited) in S.

Sometimes it helps to think of counter examples. For this case consider the integers. If you take S to be the integers greater than 2 then you have a+b and a*b are in S but S does not have additive inverses. So finiteness seems to be important. Pigeon hole principal is a big hint. If you don't know what that is I think mathworld has a page on it.

I tend to find that when people try to explain how to think of things intuitively in mathematics they just get all flumuxed (I certainly do). It is easier to start with just rigor and if intuition eventually comes then that's also great.

Good Luck,
Steven

5. Apr 19, 2005

### BicycleTree

I know that the finiteness condition is necessary, and the book even gave an example of a ring proof that used the finiteness condition, but I am not having any insights.

Personally, I need intuitive understanding first. Otherwise it's pretty much a random search of all possible steps.

I also know that you need multiplication as well as addition to show that S is a subring. For addition, the relation {((1, 2), 2), ((2, 2), 2), ((2, 1), 2), ((1, 1), 2)} satisfies commutivity and associativity but it lacks a zero, so a proof based only on commutivity and associativity cannot be what I want. Multiplication must be used and the only thing that relates multiplication to addition is the distributive property.

6. Apr 19, 2005

### Hurkyl

Staff Emeritus
Yes, but if S had that addition law, it couldn't be a subring of R now, can it?

I claim that you can get additive inverses by addiiton alone.

7. Apr 19, 2005

### BicycleTree

It couldn't be a subring but it could be a subset, couldn't it? I mean that the laws of commutivity and associativity of addition are blind to that addition rule; they couldn't discover it not to be valid ring addition even though it isn't.

8. Apr 19, 2005

### Hurkyl

Staff Emeritus
That addition law clearly cannot be a subset of the addition law of a ring.

9. Apr 19, 2005

### BicycleTree

Aha! It took me quite a while, making examples and such, before I figured out why that is. Now I know how to do the proof--I was thinking I couldn't use cancellation because I don't know if S has any additive inverses, but then if R has additive inverses then I can use cancellation. Thanks.

10. Apr 20, 2005

### matt grime

I don't see why that helps, since it doesn't use the finiteness of S.

Use the advice you got earlier.

let a be in S, so is 2a, 3a, 4a and so on . S is finite, what does that imply?

11. Apr 20, 2005

### BicycleTree

What advice I got earlier? I only got one piece of advice.

The way I approached it was, assume that for some element m of S there is some element n of S so that there does not exist any element o of S such that m + o = n. Since there are |S| different elements x so that m + x = some y, and at most |S| - 1 possible values for y, there are some distinct elements a and b of S so that for some element q of S, m + a = q and m + b = q. So m + a = m + b and a = b by cancellation, violating the assumption. So for every element n in S the exists some element p so that m + p = n. So if S has a zero, then every element of S has an additive inverse.

(That is probably not the most concise way to write this)

Actually, I don't know that S has a zero. But I'll tackle that later.

Last edited: Apr 20, 2005
12. Apr 20, 2005

### matt grime

The earlier advice was to use the fact that you can work out -x from adding up x a lot.

You do know there is a 0 in S because you're told S is non-empty (or the question is vacuous) and that it is closed under addition. So it has x in it for some x, as well as x, 2x, 3x, etc. And that since R is finite that eventually this must hit zero!

13. Apr 20, 2005

### BicycleTree

Ah, that was a mistype. I meant "m + p = n" (edited it so it now reads that).

What do you mean by 2x, 3x, etc? x + x, x + x + x? Or multiplying x by arbitrary elements 2, 3, ...?

14. Apr 20, 2005

### matt grime

2x is defined to be x+x, and 3x is x+x+x, yes. This infinite family cannot be all distinct, so there must be integers m and n s( with m=/=n) uch that mx=nx, or equivalently, (m-n)x=0, or also equivalently (m-n-1)x = -x.

15. Apr 20, 2005

### snoble

I really like your argument Bicycletree, though I would have come up with one more like matt's. But yours is a great example of the pigeonhole principal. You get zero the same way. consider a fixed $$a \in S$$ and all the elements of the form $$(a + x)$$ with $$x \in S$$. Since each of those elements are distinct and there are |S| of them all in S one of them must be $$a$$ (this is almost exactly the pigeon hole principal). So there is an element x' in S s.t. a+x'=a in S. So x'=0. Then the same argument for inverses exept use a+x must equal 0 for some x.

Pigeon hole principal is one of those wheels that keep getting reinvented time and time again.

16. Apr 20, 2005

### BicycleTree

from mx=nx you get
mx + -nx = nx + -nx (in R)
(m - n)x = 0 (in R)

How do you go from that in R to S? Isn't that just saying that if -nx is in S, then 0 is in S?

Snoble, that's not a proof that there is a zero. That's a proof that something added to a equals a. A zero must be the additive identity for all elements of the ring, not just one.

17. Apr 20, 2005

### BicycleTree

Well, I think this is a proof (although you two may have already proved it and just not finished explaining it to me):
For a in S, you have an element b of S so that a + b = a (from Snoble's proof)
And also you have an element c so that b + c = b (same)
(a + b) + c = a + c
a + (b + c) = a + c
b + c = c
b = c

So now for any arbitrary element of S, x
x + b = x + c
x + (b + c) = x + c
(x + b) + c = x + c
x + b = x

So b is the additive identity for any element of S.

18. Apr 20, 2005

### snoble

Ah but you're still inside a ring remember. You know there's a zero out there you just need to show it is in S.
$$a+ x = a$$ so $$a+ x -a = a-a$$ so $$x =0$$.
This is the sort of intuition you want to develop when dealing with subsets of algebraic structures.

19. Apr 20, 2005

### BicycleTree

Ah, right.

Last edited: Apr 20, 2005
20. Apr 21, 2005

### matt grime

as far as i'm concerned I have finished explaining it, but you haven't taken enough care to try and understand it.

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