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.

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

i'll let you take it from there

anonymous · Answer

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

anonymous · Answer

would i have to prove this under addition and multiplication?

anonymous · Answer

or is this just showing that it's nonempty?