If $b$ is not a prime in $R_1$ then is $R_2$ an integral domain or a field?

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?

In the first part, you say ''because since $b$ is not a unit then neither its image is''. This is not generally true, consider a natural monomorphism $k[x] \to k[x, \frac 1 x]$: $x$ isn't an unit in $k[x]$, but it is in $k[x, \frac 1 x]$.
In your case, you have an extra condition: $b | a$, and $(a) = \ker f$, and you should use it. If you write $a = h \cdot b$, then you have $$0 = f(a) = f(h) \cdot f(b),$$
and that implies that $f(b)$ is not an unit. Because $a$ doesn't divide $b$, $f(b)$ is non zero. From this, you see that $R_2$ can't be a field. Write out the details if this doesn't convince you! I'll let you take care of the second part, it should be easy after you solved the first.