Question

Need help with proof: Let R and S be commutative rings and let σ: R → S be a homomorphism. Prove that if σ is one to one and b is a zero divisor of R, then σ(b) is a zero divisor of S.

Answer 1

\(b\) is a zero divisor of \(R\) means there is a \(b^*\) in \(R\) with \(bb^*=0\)

Answer 2

the only natural thing to consider is \[0_S=\sigma (0_R)=\sigma(bb^*)=\sigma(b)\sigma(b^*)\]

Answer 3

i'll let you take it from there

Answer 4

thank you. i will ponder on how to form that into a paragraph or two

Answer 5

would i have to prove this under addition and multiplication?

Answer 6

or is this just showing that it's nonempty?