# Commutative ring theorem

1. Dec 1, 2009

### j9mom

1. The problem statement, all variables and given/known data
I need to prove this theorem Let R be a commutative ring with identity and c1,c2,...cn E (element of) R Then the set I={r1c1+r2c2+....+rncn|r1,r2,...,rn E R} is an ideal in R

2. Relevant equations

Well I do know I need to prove closure under subtraction, closure under multiplication and absorption r E R and a E I show ra E I and ar E I I know because R is commutative I do not have to prove both ways. and I know that just proving absorption will prove multiplication.

3. The attempt at a solution

so, I guess I am just confused on what the set I looks like.

to prove closure under subraction I need to elements from I. What would these two elements look like?

2. Dec 2, 2009

### HallsofIvy

The set "looks like" exactly what it says: All things of the form $r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n$ where $c_1, c_2, \cdot\cdot\cdot, c_n$ are given (fixed) members of ring R and $r_1, r_2, \cdot\cdot\cdot, r_n$ can be any members of the ring.

So you want to prove:
(1) Closed under addition: If $r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n$ and $s_1c_1+ s_2c_2+ \cdot\cdot\cdot+ s_nc_n$ are members of this set, it their sum also a member? That is can it be written in the same way: find $t_1, t_2, \cdot\cdot\cdot, t_n$ such that the sum of those two is $t_1c_1+ t_2c_2+ \cdot\cdot\cdot+ t_nc_n$.

(2) "Absorption" of the entire ring: If $r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n$ is in the set, show that $r(r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n)$, for r any member of the ring, is of that same form.

Those are both just as easy as they look.

3. Dec 2, 2009

### j9mom

Ok that is easy.

So under addition, because R is a ring then r1+s1 will be in R so the (r1+s1)c1+(r2+s2)c2... is in I

And because R is a Ring and closed under multiplication r*r1, r*r2,.... are in R so r*r1c1+r*r2c2+..... is in I

So it is a ideal in R

Thanks