We have the rings $R_1, R_2$, the elements $a,b\in R_1$, the ideal $I=(a)$ of $R_1$ and a ring homomorphism $f:R_1\rightarrow R_2$. It is given that $\ker f=I$. It holds in $R_1$ that $b\mid a$ but not that $a\mid b$.
If $b$ is not a unit in $R_1$ then $R_2$ is not a field, because since $b$ is not a unit then neither its image is, that's why $R_2$ cannot be a field.
Is this correct?
If $b$ is a prime in $R_1$ then is $R_2$ an integral domain or a field ?
For that we have to know if the image of a prime element is also prime or not? Or how can we check that?