Modern Algebra: Subring Test Proof Help?

Question

I've posted my question in detail at Cramster.com:

http://answerboard.cramster.com/math-topic-5-140229-0.aspx

I got stuck about halfway through the proof. Any help is greatly appreciated. Thanks!

Answer 1

note that we have to assume that either a=0, in which case S=R is trivially a subring, or else we don't include multiplicative identities in the definition of ring: if subrings need to have a multiplicative identity, then S has no chance, since 1a=a is nonzero.

let a in R and define S by {x in R | ax=0}. if x and y are in S, then ax=ay=0. we want to show that x-y and xy are in S. well, a(x-y)=ax-ay by the distributive property, and ax=ay=0 by hypothesis, so a(x-y)=0, which means x-y is in S. next, we see that a(xy)=(ax)y by associativity of multiplication, and ax=0, so (ax)y=0y=0. so by the subring test, S is a subring of R.

Answer 2

You want to show that if a and b are in S, so are a-b and ab.
Look at (a-b)x. By the distributive law in R
this is ax-bx = 0-0 = 0. So a-b is in S.
Now look at (ab)x. By the associative law in R
this is a(bx) = a*0 = 0, so ab is in S
and S is a subring of R.
Hope that helps!