Kernel of a ring homomorphism is prime ideal $\varphi: A\to B$ be a ring homomorphism.
We see that $\ker(\varphi)=\varphi^{-1}(0)$ and we know that if $(0)$ is prime ideal (equivalently $B$ is an integral domain) then $\ker(\varphi)$ is also prime ideal.
I am searching for the reverse direction. Can we say $\ker(\varphi)$ is prime then $B$ is an integral domain? I feel some contradiction, yet could not find.
Is reverse direction is correct, or can you give me a hint about it.
How can we weaken the conditions without imposing surjectivity and injectivity that double direction is satisfied with additional conditions.
 A: I assume every ring considered is commutative
If the kernel is a PRIME IDEAL
we can only say that the image of $\varphi$ is an integral domain. This image is a subring of $B$. But a subring being an integral domain says nothing about the whole ring: it can very well ave zero divisors.
We can take the polynomial ring with real coefficients, and then quotient it be an ideal generated by any degree 3 polynomial(they always have one real root). This quotient ring $B$, is not an integral domain, and the set of real numbers will be a subring ( actually a field.)
Now define $\varphi: \mathbf{R}\to B$ sending  every real number  to itself. The kernel now is (0) a prime ideal, but $B$ is not an integral domain.
A: If $\varphi\colon R\to S$ is a surjective homomorphism of commutative (unital) rings and $I=\ker \varphi$ is a prime ideal, then by the isomorphism theorem, $S\cong R/I$, and the latter is clearly a domain (this is easy to check directly).
If $\varphi\colon R\to S$ is not surjective, this is not necessarily true. For example, if $R$ is any domain, then $R[x]/(x^2)$ clearly isn't a domain, but the kernel of the natural embedding $R\to R[x]/(x^2)$ is $(0)$, which is a prime ideal (since $R$ is a domain).
