On Rings

[Johnny5]

I want this thread to be about Rings.

 

I think the main issue for the thread should be, what is/are the purpose/purposes of defining a ring.

 

Let me start off with the definition Dave gave, of a ring, in a different thread.

 

A ring is a set R, with two operations defined on it, which are symbolized +, *.

 

The ring R, must be a group under at least one of the two operations. It doesn't have to be a group under both, but it could be.

 

Suppose that the ring is a group under +. That will mean the following four things are true about elements of R:

 

1. There is a zero element of R, denoted by 0, such that for any element x of the ring, 0+x=x.

2. Associativity: For any x,y,z elements of the ring R, we have:

(x+y)+z = x+(y+z)

3. (Inverse elements) For any element X of the ring R, there is an element (-x) of the ring, such that: x+(-x) = 0

4. Commutativity: For any x,y elements of the ring R, we must have:

x+y=y+x

 

(Thus R is a group under +).

 

Now, as for the other operation, namely *, we have to have associativity. That is:

 

1. Associativity of *

For any x,y,z elements of R: x*(y*z)=(x*y)*z

 

Lastly, there has to be two distributive laws:

 

a*(b+c) = a*b+a*c

(b+c)*a = b*a+c*a

 

A couple more questions.

 

Question: Using the ring axioms above, (assuming i got them right that is), can you prove that the element 0 is unique?

Question: Must the second operation *, be commutative too? If so, then the second distributive law above is redundant, and can be omitted from the set of ring axioms.

[Dave]

(Thus R is a group under +).

 

Well, this is true but it's not the whole picture. If a group is commutative then it's usually called abelian, so R is an abelian group under +.

 

Question: Using the ring axioms above, (assuming i got them right that is), can you prove that the element 0 is unique?

 

Yes. Assume you have two zero elements, [math]0[/math] and [math]\widetilde{0}[/math]. We know:

 

[math]0 = 0 + \widetilde{0}[/math]

[math]\widetilde{0} = 0 + \widetilde{0}[/math].

 

Combining these, we must have that [math]0 = \widetilde{0}[/math]. It's a standard kind of thing (you can use it for fields also).

 

Question: Must the second operation *, be commutative too?

 

No. Certain things are missing from the ring deliberately.

1) * commutative; then R is called a commutative ring.

2) There is an identity element [imath]1_R[/imath] such that [imath]1_R \cdot a = a \cdot 1_R = a[/imath] for all a in R; then R is called a ring with identity.

3) Every element a has an inverse under *; then the ring is called a division ring.

[Johnny5]

Well' date=' this is true but it's not the whole picture. If a group is commutative then it's usually called abelian, so R is an abelian group under +.

 

 

 

Yes. Assume you have two zero elements, [math']0[/math] and [math]\widetilde{0}[/math]. We know:

 

[math]0 = 0 + \widetilde{0}[/math]

[math]\widetilde{0} = 0 + \widetilde{0}[/math].

 

Combining these, we must have that [math]0 = \widetilde{0}[/math]. It's a standard kind of thing (you can use it for fields also).

 

 

 

No. Certain things are missing from the ring deliberately.

1) * commutative; then R is called a commutative ring.

2) There is an identity element [imath]1_R[/imath] such that [imath]1_R \cdot a = a \cdot 1_R = a[/imath] for all a in R; then R is called a ring with identity.

3) Every element a has an inverse under *; then the ring is called a division ring.

 

Ok, this is good.

 

A ring R is necessarily a group under +, but furthermore, it must be an abelian group under+.

 

(Now I have a side question... if ring R was a group under +, but not an abelian group under +, could you still have proved that 0 is unique?)

 

 

So any ring R is an abelian group under +.

Now, * has to be associative.

 

1) If a*b=b*a then the ring is commutive.

2) If 1*a=a*1 then the ring has an identity.

3) if a*b=b*a and 1*a=a*1 and (for any a in R, other than 0, there is 1/a in R such that a*(1/a) = 1 then R is a division ring???

 

Can you check my 3 above?

[Dave]

(Now I have a side question... if ring R was a group under +, but not an abelian group under +, could you still have proved that 0 is unique?)

 

Well, if it wasn't abelian then it wouldn't be a ring, so the question is null But that doesn't matter; definition of a zero element means that 0+a = a+0 = a, so it's commutative in that respect.

 

1) If a*b=b*a then the ring is commutive.

2) If 1*a=a*1 then the ring has an identity.

3) if a*b=b*a and 1*a=a*1 and (for any a in R, other than 0, there is 1/a in R such that a*(1/a) = 1 then R is a division ring???

 

They're almost correct. For 2, you want 1*a = a*1 = a. For 3, you're better off using the notation [imath]a^{-1}[/imath] instead of [imath]\frac{1}{a}[/imath]. Also, it's not necessary for a division ring to be commutative, although it is necessary for there to be an identity (obviously).

[Kygron]

Any good examples of rings/ring usage other than the real numbers?

[matt grime]

Rings:

 

Z, Q, R, C the integers rationals reals and complexes resp.

 

Let S denote any of those ie let S stand for some basic ring. then M_n(S) is a ring = the nxn matrices with elements in S, as is the set of all upper triangular matrices.

 

Let I be any interval on the real line then C^r(I) the r times differentiable functions from I to R are a ring. Sets of holomorphic, etc functions are all rings.

 

With S as above S[x,y,..,z] the polynomial ring in arbitrary many variables is a ring, as it is if we quotient out by any ideal.

 

Group rings are rings, as are hopf algebras, universal enveloping algebras, fields, field extensions, the algebraic numbers, the algebraic integers, finite fields, the p-adics, the p-locals.

 

I've barely touched the surface. Hopefully you get the idea that most anything algebraic in maths has something to do with a ring in there somewhere. Heck, even lots of mathematical physics is simply finding ways some rings can act ons spaces.

 

 

And, for Johnny, the identity element of *any* group is unique, as are inverses.

[Dave]

Another interesting ring that we touched upon in my Algebra course was the ring of differential operators:

 

[math]\mathcal{D} = \{ a_0 + a_1 \frac{d}{dt} + a_2 \frac{d^2}{dt^2} + \cdots + a_n \frac{d^n}{dt^n} \ | \ a_0, \dots, a_n \in C^{\infty}(\R,\R) \}[/math]

 

This isn't a ring under addition and multiplication, but under addition and composition. You can use the rule:

 

[math]\frac{d}{dt} \circ g = g' + g \frac{d}{dt}[/math]

 

You can then use this to solve a lot of nasty differential equations. For example, take a look at:

 

[math]\left[ \left(\frac{d}{dt}\right)^2 +(t + \cos t)\frac{d}{dt} + (t\cos t + 1)\right] f(t) = 0[/math]

 

That's a fairly nasty little equation. But using the rule above, we can see that [imath]1+t\frac{d}{dt} = \frac{d}{dt}\circ t[/imath]. Now we get (if we just consider the operator for a second):

 

[math]\frac{d}{dt} \circ \frac{d}{dt} + \frac{d}{dt} \circ t + \cos t \frac{d}{dt} + t \cos t[/math]

 

You can "factorize" this:

 

[math]\left(\frac{d}{dt} + \cos t\right) \circ \left( \frac{d}{dt} + t\right)[/math]

 

So to get solutions, you need only consider the two, easier differential equations given by:

 

[math]\frac{df}{dt} + f(t)\cos t = 0, \frac{df}{dt} + tf = 0[/math].

 

Of course, you can get the easiest solution using the second equation.

[Johnny5]

the identity element of *any* group is unique, as are inverses.

 

I guess my question would be, is it provable within the group axioms, that the identity is unique. I don't think that is true, so that you would have to add that as an axiom.

 

But i will try to answer my own question right here, in a moment.

[matt grime]

Of course it is provable from group axioms, and it is a VERY trivial proof too, indeed the proof already exits in this thread in Dave's post when he demonstrates it for any Ring. It is part of the axioms that inverses and identities are two sided, thus making them unique. Exercise find a unital ring which has elements L and R such that LR=1 but RL is not 1.

[Johnny5]

Another interesting ring that we touched upon in my Algebra course was the ring of differential operators:

 

[math]\mathcal{D} = \{ a_0 + a_1 \frac{d}{dt} + a_2 \frac{d^2}{dt^2} + \cdots + a_n \frac{d^n}{dt^n} \ | \ a_0' date=' \dots, a_n \in C^{\infty}(\R,\R) \}[/math']

 

This isn't a ring under addition and multiplication, but under addition and composition.

 

 

Dave this looks strangely familiar for some reason.

 

It resembles a power series, which hasn't operated on anything yet. Or a Maclaurin series expansion I should say.

 

I was wondering if there was some physical interpretation that we could give it. Like for example, suppose that it acts on position variable x (you can treat x as position vector r if you are so inclined, but I am just going to use x).

 

Then you have lets see...

 

a1x+

a2x`+

a3x``+

a4x```+

 

So I am wondering about units here, and how the coefficients manage to keep them the same.

 

For example, x` is interpreted as velocity, x`` as dv/dt which is acceleration, and then x```` would be da/dt, which can be nonzero, as for many many terms which follow. Yet, in physics we generally ignore the higher terms.

 

Can you add anything to this? (specifically I am wondering if i could do physics using all the higher terms)

 

(I thought about this long long ago, but forget what i did. I was thinking along these lines at the time, though this was very long ago... saying that the total motion of a particle was described by an infinite series of the form given by Dave, yet to avoid actual infinity, the formula for the coefficients of the series, eventually went to zero, so that there was a highest term, so that the series wasn't 'really' infinite. The main thing is, you would have more complex things than acceleration in the kinematics.)

[matt grime]

Yes, why do you think x is position? It isn't. Stop thinking physically. x is just a variable, and we're looking at the differential operators acting on the space of smooth functions in x, whatever. There is nothing that states x is position. It isn't.

[Johnny5]

Yes, why do you think x is position? It isn't. Stop thinking physically. x is just a variable, and we're looking at the differential operators acting on the space of smooth functions in x, whatever. There is nothing that states x is position. It isn't.

 

I didn't think it had to be position, i just wanted to choose an example, and that was the first that came to mind. I understand that x is a variable, very well i understand that.

 

The differential operator given by Dave, could act on any number of 'items' one of which is position. I am keenly aware of that, i assure you.

 

You don't have to interpret t as time, i understand that you are a 'maths' person, and you think mathematically, which is good. But it is also good to be able to apply mathematics to physical reality.

 

Much maths was developed in order to solve problems in physics. And i am sure the opposite has happened as well, that math was developed in a pure sense, and then applied to physics.

 

Regards

[matt grime]

Momentum etc are operators in this sense, and the you ought to at least know about solutions to the wave operator and so on.

[Johnny5]

Momentum etc are operators in this sense, and the you ought to at least know about solutions to the wave operator and so on.

 

I do, I studied d`Alemberts solution to the wave equation, done Fourier integrals, Fourier series in QM. I've seen a whole lot. I am not sure how much I would suddenly 'recall' but what exactly are you thinking of, when you say, "solution to the wave operator."

 

Also, i don't want the thread to stray too far from rings, so if the answer is long, then move it to another thread.

 

PS: For what it's worth, most of this stems from trying to understand quaternions. which i still dont.

[matt grime]

The space of operators on some vector space, eg the smooth functions on some compact subset of R^n is a ring, it has a spectral theory. It isnt' very complicated, but if you're still trying to understand quaternions, which are simply a 4 dimensional real division algebra, then talking about the spectral theory of operators is not going to be helpful. However, we can only second guess what it is you want to talk about, so why not formulate some questions?

[matt grime]

Back to rings.

 

The purpose of ring theory is simply to offer a self contained theory of certain objects that occur in algebra. There are many things we consider interesting and we have chosen to codify certain parts of them, ie pick out a set of rules common to all of them. We can then study what results follow simply from these axioms. Then we can add some more and study, say, commutative rings, which are important in geometry, or rings that are also vector spaces, ie algebras. Or perhaps we care about rings that have no zero divisors, which are called integral domains. As far as I'm concerned a Ring is simply the simplest algebraic object that is interesting in generality. More precisely it is the most basic case when the axioms it satisfies start to be interesting and also not too strong. Less restricting axioms are too general, they don't allow us to state and prove very many general results, and stronger axioms become a little too restrictive.

 

Of course, a group is arguably a simpler object, but I think they are in some sense too algebraic, that is the study of groups alone is purely algebraic, whereas rings are of more general interest to, say, number theorists. Ok a ring is a goup under addition, but it is frequently an infinite one.

[Johnny5]

However, we can only second guess what it is you want to talk about, so why not formulate some questions?

 

I do that very well.

 

Let's see...

 

Well I am trying to understand what a ring is, that's the point of this thread.

 

As I already know the field axioms, this shouldn't be a hard thing to do, since a field is more complex than a ring. Yet I am still fuzzy on a few things, so lets see just what. At least this way if I make any errors, you can point them out.

 

Lets start from Dave's definition of a Ring:

 

A ring [math] \mathcal{R} [/math] is a set of elements which is defined axiomatically, as follows:

 

There are two binary operations +,* on any arbitrary ring, which obey the following axioms:

 

Let x,y,z denote arbitrary elments of [math] \mathcal{R} [/math].

 

[math] \text{Axiom I:} x+y=y+x [/math]

[math] \text{Axiom II:} (x+y)+z=x+(y+z) [/math]

[math] \text{Axiom III:} \exists 0[0+x=0] [/math]

[math] \text{Axiom IV:} \exists (-x) [x+(-x)=0] [/math]

 

(Thus, for those who know what an abelian group is, a ring is an abelian group on at least one of the ring operations, possibly both, but necessarily at least one.)

 

[math] \text{Axiom V:} (x*y)*z=x*(y*z) [/math]

[math] \text{Axiom VI:} x*(y+z)=x*y+y*z [/math]

 

The above axioms are all necessary axioms of any ring [math] \mathcal{R} [/math].

 

Now, the first question is, are the axioms above sufficient to prove that 0 is the unique element of the ring, with the property listed above? If not, we have to add another axiom. The following proof was provided by Dave

 

Proof that 0 is unique (Post 2 above)

 

Assume you have two zero elements, [math]0[/math] and [math]

\widetilde{0}

[/math]. We know:

 

[math]

0 = 0 + \widetilde{0}

[/math]

 

Sending [math] \widetilde{0} [/math] through the ring axioms provides:

 

[math] \text{Axiom III:} \exists 0[0+\widetilde{0}=0] [/math]

 

And we have assumed that [math]\widetilde{0}[/math] satisfies the same property that 0 satisfies in the axioms so also we have:

 

[math] \exists \widetilde{0}[\widetilde{0}+x=\widetilde{0}] [/math]

 

And since 0 is a necessarily a ring element we have:

 

[math] \exists \widetilde{0}[\widetilde{0}+0=\widetilde{0}] [/math]

 

So we have:

[math] 0+\widetilde{0}=0 [/math]

[math] \widetilde{0}+0=\widetilde{0} [/math]

 

Now ring [math] \mathcal{R} [/math] is an abelian group under +, which means that its a group under +, with the additional axiom that + is commutative. Commutativity of + is axiom 1. So the following is true:

 

[math] 0+\widetilde{0}=\widetilde{0}+0 [/math]

 

Thus, by the transitive property of equality we have:

 

[math] 0=0+\widetilde{0}=\widetilde{0}+0=\widetilde{0} [/math]

 

From which it follows that:

 

[math] 0=\widetilde{0} [/math]

 

Contrary to the assumption that they were different elements. Hence the ring element 0, asserted in axiom 3 is unique.QED

 

(Without commutativity of +, this would not have been provable)

 

Thus, the following is a theorem of any Ring theory:

 

[math] \text{Theorem:} \exists! 0[0+x=0] [/math]

[Johnny5]

You asked for a question, well I really have one.

 

Must both of the following statements be taken as axioms of any ring theory, or can we get away with just one?

 

a*(b+c) = a*b+a*c

(b+c)*a = b*a+c*a

 

PS: I think the answer is we have to adopt both, unless the ring is commutative. Anyone?

[matt grime]

Firstly you must sepcify distributivity on both sides; i think there are examples to show this, try googling for them.

 

Secondly yes 0 is unique as is the identity in *any* group. That has been shown to you, explained to you at least three times now and still remains a trivial exercise, and the commutativity of the group is not important since identities are declared to be two sided. You can either assume + is commutative and show the unique left identity is also a right identity, or you can simply say R,+ is an abelian group. Your choice. There is redundancy in the group axioms when we restrict to it being abelian, that is all, and thus there are several equivalent ways to give the group axioms, just as there several equivalent ways to axiomatixe the real numbers, and just as we can show there to be lots of redundancy in other axiomatizations of other things.

[Johnny5]

Firstly you must sepcify distributivity on both sides; i think there are examples to show this

 

 

Ok thanks.

 

As for your comments about axiomatization, i liked them. What i just want to do right now, is be able to precisely formulate the axioms of a ring, and be able to understand what theorems emerge from what axioms.

 

Again thank you, your patience is appreciated.

 

Regards