f is surjective ring homomorphism form (A,+, .) to (B, +, .)

Question

Let $f: (A,+,.) \rightarrow (B,+,.)$ is surjective ring homomorphism.

Prove or disprove:

(a) If B is commutative ring then A is also commutative.

(b) If B has a multiplicative identity then A has also multiplicative identity.

I know that the inverse implications are true and proof is not difficult. I think that these implications (a) and (b) are false (because it is true when $f$ is bijective, there is equivalence in these claims ). But I cannot find any contra example, or I cannot show that it is not true.

Thanks for any help.

Answer 1

(a) is false....Consider the (with-identity) free ring on two non-commuting variables $x$ and $y$, which has $\Bbb{Z}$ as a homomorphic image (just mod out the ideal generated by $x$ and $y$...i.e., map any expression to its constant term).

(b) I assume by "neutral element" you mean "multiplicative identity". This also fails. For example, the subring of 2x2 matrices over $\Bbb{Z}$ with the second row zero has no neutral element, but there is a surjective homomorphism onto $\Bbb{Z}$ given by taking the upper-left entry.

Answer 2

For (a), you can just take any non-commutative ring and map it onto the trivial ring $B=\{0\}$. The trivial ring is commutative.

For (b) You can do the same kind of thing by taking any ring without a neutral element and map it onto the trivial ring. It has a neutral element $0=1$, so this would work. The even integers are the easiest example of a ring without a neutral element.