Subring

**resident911** · August 26th 2012, 02:15 PM

Hi , i have this problem , i need help

Let R be a ring $1\neq 0$ and $G=\{g_1,g_2......g_n\}$ 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!!

**HallsofIvy** · August 26th 2012, 04:54 PM

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?

**resident911** · August 26th 2012, 05:04 PM

HI!

I have a problem, how is denoted an element of R (G)? , And not understand that part ,that says the coefficients add 0,

**Plato** · August 26th 2012, 05:25 PM

Originally Posted by resident911

I have a problem, how is denoted an element of R (G)? , And not understand that part ,that says the coefficients add 0,

I admit that it has been twenty years since I worked with ring theory.
That said, I think that you are at the mercy of the author of that question.
That is, that notation may not be standard. It may depend on a particular set of notes.

**Vlasev** · August 26th 2012, 07:06 PM

One way to think of elements in the group ring $R[G]$ (where $G = \{g_1,\dots,g_n\}$ ) is as follows. If $w \in R[G]$ , then

$w = \sum_{i=1}^n r_i g_i$ ,

where $r_i \in R$ , $g_i \in G$ for every $1 \leq i \leq n$ . This sum is formal. The $g_i$ 's act as placeholders (like the monomials in polynomials). You can only add elements with the same $g_i$ , that is

$u g_i + v g_i = (u+v) g_i$ ,

where the addition is done in the ring $R$ . It does not make sense to try to simplify something like $r_1 g_1 + r_2 g_2,$ no more than you can simplify something like $1+z$ among polynomials). To multiply two elements $r_i g_i$ and $r_j g_j$ you do

$(r_i g_i)(r_j g_j) = (r_i\cdot r_j) (g_i * g_j)$ ,

where $\cdot$ is multiplication in $R$ and $*$ is multiplication in $G$ . Finally, to multiply two $w$ 's together, you treat them like polynomials, as in

$(r_1g_1+r_2g_2)(p_1g_1 + p_2 g_2) =$
$=(r_1g_1)(p_1g_1) + (r_2g_2)(p_1g_1) + (r_1g_1)(p_2g_2) + (r_2g_2)(p_2g_2) =$
$= (r_1p_1)g_1^2+(r_2p_1)(g_2g_1)+(r_1p_2)(g_1g_2)+(r _2g_2)g_2^2$

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 $R[G]$ whose coefficients add up to $0_R$ " $S$ . It is indeed a subring (as you need to show) and can be written simply like this

$S = \{w = \sum_{i=1}^n r_i g_i : \sum_{i=1}^n r_i = 0_R\}$

Here is a simple example where $R = \mathbb{Z}$ and $G$ is the set of monomials with integer powers, i.e. $G = \{x^i : i\in \mathbb{Z}\}$ . This group $G$ is isomorphic to $\mathbb{Z}$ by the way. Now $R[G]$ is actually the ring of polynomials over $x$ . In this case $S$ is the set of polynomials whose coefficients sum up to 0. Here are some members of $S$

$1-x, 1+x-2x^2, 2-x+x^2-3x^3+x^4 \in S$

Now, for your question, you need to show that $S$ is a subring of $R[G]$ where $R$ is a ring and $G$ is a finite group. In general, to prove that $S$ is a subring you need to show that

1. $S$ contains the multiplicative identity of $R[G]$ (what is it?)
2. $S$ is closed under subtraction
3. $S$ 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.

**Vlasev** · August 26th 2012, 07:17 PM

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 $p,r \in S$ , you need $p-r \in S$ . You have

$p = \sum_{i=1}^n p_i g_i, \quad r = \sum_{i=1}^n r_i g_i,\quad p-r = \sum_{i=1}^n (p_i-r_i) g_i$

Given that $\sum_{i=1}^n p_i = 0_R, \sum_{i=1}^n r_i = 0_R$ you need to show that $\sum_{i=1}^n (p_i-r_i) = 0_R$ .

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.

**Deveno** · August 26th 2012, 09:37 PM

the group ring R[G] is all formal R-linear combinations of elements of G. so suppose we have:

$r = \sum_{i=1}^n a_ig_i$ and $s = \sum_{i=1}^n b_ig_i$ .

we need to show that if:

$S = \{ r \in R[G]: \sum_{i=1}^na_i = 0\}$

1) r - s in in S.
2) rs is in S.
3) S is non-empty.

directly from R-linearity we have:

$r - s = \sum_{i=1}^n a_ig_i - \sum_{i=1}^n b_ig_i = \sum_{i=1}^n(a_i - b_i)g_i$ .

and since r,s are in S:

$\sum_{i=1}^n (a_i - b_i) = \sum_{i=1}^n a_i - \sum_{i=1}^n b_i = 0 - 0 = 0$ . this proves (1).

(2) is a bit more complicated:

$rs = \left(\sum_{i=1}^n a_ig_i\right)\left(\sum_{i=1}^n b_ig_i\right) = \sum_{g_ig_j = g_k} a_ib_jg_k = \sum_{k=1}^n c_kg_k$ .

so we need to show that:

$\sum_{k=1}^n c_k = 0$ .

note that if a product $a_ib_j$ (for a specific i and j) occurs in the coefficient of $g_k$ 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 $c_k$ . hence:

$\sum_{k=1}^n c_k = \sum_{i,j = 1}^n a_ib_j = \sum_{j = 1}^n \left(\sum_{i=1}^n a_i \right)b_j = \sum_{j=1}^n (0)(b_j) = \sum_{j=1}^n 0 = 0$ this proves (2).

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

$0 = \sum_{i=1}^n 0g_i$ , which shows (3).

it is NOT true, that a subring S of a ring R must contain the multiplicative identity of R. for example, $2\mathbb{Z}$ is a subring of $\mathbb{Z}$ , 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:

$e_G = 1e_G = \sum_{g_i \neq e_g}(0)(g_i) + 1e_G$ , and $1+0+\dots+0 = 1 \neq 0$ , for a non-trivial ring with unity R.

**Vlasev** · August 26th 2012, 10:39 PM

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.

**Deveno** · August 27th 2012, 03:53 AM

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).

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