Visualizing the Associative & Distributive Properties of Rings

In summary, Steven provides advice for understanding rings and proofs, including using the finiteness and pigeon hole principle to prove that a subset of a ring is a subring. The property of distributivity is also mentioned as a key factor in proving a subring. Cancellation and the existence of a zero element are also important in showing the existence of additive inverses in a subring.
  • #1
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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?
 
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  • #2
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. :frown:
 
  • #3
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
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
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
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.

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. :smile:
 
  • #7
Hurkyl said:
Yes, but if S had that addition law, it couldn't be a subring of R now, can it?
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
That addition law clearly cannot be a subset of the addition law of a ring.
 
  • #9
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
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
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.
 
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  • #12
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
BicycleTree said:
So for every element n in S the exists some element p so that m + p = r.
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
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
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 [tex]a \in S[/tex] and all the elements of the form [tex](a + x)[/tex] with [tex]x \in S[/tex]. Since each of those elements are distinct and there are |S| of them all in S one of them must be [tex]a[/tex] (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
matt grime said:
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.
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
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
BicycleTree said:
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.

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.
[tex] a+ x = a[/tex] so [tex] a+ x -a = a-a [/tex] so [tex] x =0[/tex].
This is the sort of intuition you want to develop when dealing with subsets of algebraic structures.
 
  • #19
Ah, right.
 
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  • #20
BicycleTree said:
Well, I think this is a proof (although you two may have already proved it and just not finished explaining it to me):

as far as I'm concerned I have finished explaining it, but you haven't taken enough care to try and understand it.
 
  • #21
I'm sorry. I have tried to understand it but maybe not enough.
matt grime said:
(m-n)x=0, or also equivalently (m-n-1)x = -x.
That's to show the existence of zero and additive inverses in S, right? But if you take it that way the first one says that if (m-n)x is in S, then 0 is in S. But starting here, -n is not known without other proof to be in S, so (m-n) is not known to be in S, so (m-n)x is not known to be in S, so 0 is not known to be in S by that expression. Am I reading this wrong?
 
  • #22
Did I say -n was in there? Perhaps I didn't make it clear you may assume wlog m>n os m-n is a strictly positive integer, k, say k>1, satisfying 0=kx=x+x+..+x (k times) which we know by hypothesis is in S. So (k-1)x is in S and is an inverse for x.

I mean, who even knows if -n is in R? R is not said to b unital is it? So it may make no sense to even talk about n or -n being in R.
 
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  • #23
Okay, I get it. From mx = nx you add the additive inverse of x to both sides n times to get (m-n)x=0. I was thinking mx being ring multiplication.
 
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1. What are the associative and distributive properties of rings?

The associative property of rings states that when performing operations on three or more elements, the order in which the operations are carried out does not change the result. The distributive property of rings states that the product of an element with the sum of two other elements is equal to the sum of the products of the element with each of the two other elements.

2. How do these properties apply to rings?

In a ring, these properties apply to the operations of addition and multiplication. This means that when adding or multiplying three or more elements in a ring, the order of the operations can be changed without affecting the result. Additionally, the distributive property allows for the expansion of expressions involving sums in a ring.

3. What is the importance of understanding these properties in ring theory?

The associative and distributive properties are fundamental to the study of rings and other algebraic structures. They provide a framework for understanding how operations behave in these structures and help to prove important theorems and properties.

4. How can the associative and distributive properties be visualized?

These properties can be represented visually through diagrams and illustrations. For example, a Venn diagram can be used to show how the associative property applies to three elements in a ring. A grid or matrix can also be used to demonstrate the distributive property in action.

5. Are there any real-world applications of these properties?

Yes, the associative and distributive properties have many real-world applications in fields such as computer science, physics, and engineering. For example, the associative property is used in parallel computing to optimize the order of operations, while the distributive property is applied in the distribution of electricity in power grids.

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