# Abstract Algebra - Rings and Subrings

• Feb 19th 2013, 06:52 PM
jll90
Abstract Algebra - Rings and Subrings
Let R be a finite commutative ring, that is if a, b are elements of R, then ab = ba.
Show that for any t $\in$ R, T = tR = {tr|r $\in$ R} is a subring of R

Further, if T $\neq$ R, there are r1 $\neq$ r2 such that tr1 = tr2. (Hint: if tr $\neq$ ts for all r, s $\in$ R,
then #tR = #R. )
• Feb 19th 2013, 07:28 PM
jakncoke
Re: Abstract Algebra - Rings and Subrings
Closed under multiplication? $tr_1 * tr_2 = t*(t*r_1*r_2)$ , $t*r_1*r_2 \in R$ so closed under multi.
Closed under subtraction? $tr_1 - tr_2 = t*(r_1 - r_2)$, $r_1 - r_2 \in R$ so closed under subtraction
$0*t = 0 \in T$

The third thing. If T = R, then they would have to be the same "size" no? meaning if $r_1t = r_2t$ and $r_1 \not= r_2$ then what were originally 2 elements in R collapsed into one element in T (thus making it smaller and thus not equal to R).