# Onto homomorphism keeps integral domain and field?

Let $R,S\neq \{0\}$ be a rings and let $\varphi:R\to S$ be a rings homomorphism onto on $S$. Prove or disprove with counter example:

A). if $R$ is a integral domain then also $S$ is a integral domain.

B). if $R$ is a field than $S$ is a field

Attempt:

A). False, consider the map $\varphi:\mathbb Z\to \mathbb Z_6$,

$\mathbb Z$ is a integral domain and $\mathbb Z_6$ is not because $3,2$ are zero divisors

B). False, consider the map $\varphi:\mathbb R\to \mathbb Z$

$\mathbb R$ is a field but $\mathbb Z$ not

• What map $\varphi:\mathbb R\to \mathbb Z$? – Slade Apr 17 '17 at 13:03
• I don't know, to define some map, maybe something like this: $[0,1]_R\to 1_S,(1,2]_R\to 2_S$ – Error 404 Apr 17 '17 at 13:05
• Is that a homomorphism? – Slade Apr 17 '17 at 13:06

Hint for B: A field has exactly two ideals. Consider $\ker \varphi$.
• @Error404 If you have a homomorphism out of a ring, its kernel has to be an ideal. Fields only have two ideals, $\{0\}$ and the whole field. What would a homomophism look like if its kernel was $\{0\}$? What would it look like if its kernel was the whole field? – Oscar Cunningham Apr 17 '17 at 14:36
I'll assume the rings are commutative and have $1$.
Indeed $S$ can have zero divisors even if $R$ does not (i.e., is an integral domain), as in the case of the (unique) morphism $\Bbb Z\to\Bbb Z/n\Bbb Z$ given by reduction modulo a composite number$~n$. Or in the very similar case of $K[X]\to K[X]/(P)$ of reduction of polynomials modulo a reducible polynomial$~P$. In both cases the witnesses for the fact that $n$ is composite respectively that $P$ is reducible map to zero divisors.
In the case where $R$ is a field and $S\neq\{0\}$ (i.e., $0_S\neq1_S$) it does follow that $S$ is a field. A morphism of rings maps $0_R$ to $0_S$ and it maps invertible elements of $R$ to invertible elements of $S$ (because relations $ab=1$ are preserved). If all nonzero elements of $R$ are invertible and $\varphi$ is surjective, then each nonzero element of $S$ (of which at least one exist) is the image of a nonzero element of $R$, and therefore invertible; this ensures that $S$ is a field. Moreover $~\varphi:R\to S$ is an isomorphism since $\ker\varphi=\{0_R\}$.