Hi , i have this problem , i need help

Let R be a ring and is a finite group, let R [G] group ring.

Prove that the set of elements of R [G] whose coefficients add 0, is a subring of R [G]

Thanks!!

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- August 26th 2012, 03:15 PMresident911Subring
Hi , i have this problem , i need help

Let R be a ring and is a finite group, let R [G] group ring.

Prove that the set of elements of R [G] whose coefficients add 0, is a subring of R [G]

Thanks!! - August 26th 2012, 05:54 PMHallsofIvyRe: Subring
The only conditions for a subset of a ring to be a subring is that it be "closed" under addition and multiplication. Have you tried to show that?

- August 26th 2012, 06:04 PMresident911Re: Subring
HI!

I have a problem, how is denoted an element of R (G)? , And not understand that part ,that says the coefficients add 0, - August 26th 2012, 06:25 PMPlatoRe: Subring
- August 26th 2012, 08:06 PMVlasevRe: Subring
One way to think of elements in the group ring (where ) is as follows. If , then

,

where , for every . This sum is**formal**. The 's act as placeholders (like the monomials in polynomials). You can only add elements with the same , that is

,

where the addition is done in the ring . It does not make sense to try to simplify something like no more than you can simplify something like among polynomials). To multiply two elements and you do

,

where is multiplication in and is multiplication in . Finally, to multiply two 's together, you treat them like polynomials, as in

You can read more about group rings on Group ring - Wikipedia, the free encyclopedia.

Now, pertaining to your question, call "The set of elements of whose coefficients add up to " . It is indeed a subring (as you need to show) and can be written simply like this

Here is a simple example where and is the set of monomials with integer powers, i.e. . This group is isomorphic to by the way. Now is actually the ring of polynomials over . In this case is the set of polynomials whose coefficients sum up to 0. Here are some members of

Now, for your question, you need to show that is a subring of where is a ring and is a finite group. In general, to prove that is a subring you need to show that

1. contains the multiplicative identity of (what is it?)

2. is closed under subtraction

3. is closed under multiplication

The first two should be straight-forward (if you keep the analogy of polynomials rings). The last one can be tricky. I hope this helps. - August 26th 2012, 08:17 PMVlasevRe: Subring
Here is a bit of follow up on points 1, 2 and 3.

1. There are several ways in which you can find what the identity is. The simplest way would be to just think in a minimal sense. That is, try some small things.

2. The way to do this is straight-forward. Given , you need . You have

Given that you need to show that .

3. This is the same way as part 2 but with multiplication. It's a little more difficult. You might want to try small cases. - August 26th 2012, 10:37 PMDevenoRe: Subring
the group ring R[G] is all formal R-linear combinations of elements of G. so suppose we have:

and .

we need to show that if:

1) r - s in in S.

2) rs is in S.

3) S is non-empty.

directly from R-linearity we have:

.

and since r,s are in S:

. this proves (1).

(2) is a bit more complicated:

.

so we need to show that:

.

note that if a product (for a specific i and j) occurs in the coefficient of for some k, it does not occur in any other coefficient, and every possible product occurs as a term in the sum that is one of the . hence:

this proves (2).

finally, R[G] is non-empty, as it contains the element:

, which shows (3).

it is NOT true, that a subring S of a ring R must contain the multiplicative identity of R. for example, is a subring of , but clearly 1 is not an element. in this case, in fact, the identity of R[G] is not an element of S (unless R is the 0-ring), because the identity of R[G] is:

, and , for a non-trivial ring with unity R. - August 26th 2012, 11:39 PMVlasevRe: Subring
Deveno, it all the depends on the definition of subring your are using. For some definitions, the identity elements must coincide. But yea, in this case you're working with the definition of a subring where it just has to be a subset that is a ring with the given operations.

- August 27th 2012, 04:53 AMDevenoRe: Subring
unfortunately, this is true. some authors only call rings with unity "rings" (and the term rng for rings without identity is used).

however, in context, it appears the OP is being asked to prove something which is contingent upon the broader definition of ring. and if that is *not* the definition of ring then as we see S is not a sub-ring of R[G], as it fails to contain the multiplicative identity of R[G].

interesting digression: a subring S of a ring R can contain a different multiplicative identity. for example: let R = Z x Z, which has identity (1,1). let S = Z x {0}, which has identity (1,0).