ac.commutative algebra - When is the radical of the extension of a prime ideal prime?

If you look at the situation geometrically, you have a morphism of affine schemes $Spec(S)to Spec(R)$ and you are asking if the inverse image of the integral subscheme
$V(mathcal P) subset Spec(R)$ is irreducible.
My intuition is that this happens only rarely.

For example if $k$ is a field, consider the diagonal embedding $k to ktimes k$. You get an algebra which has all the properties you require:injectivity of structure map, integrality, finite presentation (after all it is a 2-dimensional vector space...) Yet the zero ideal of $k$ extends to the zero ideal in $k times k$ , which is reduced but not prime.

Still, in order to end on an optimistic note, here is a positive result. If you take for $S$ the polynomial ring $R[X_1,ldots ,X_n]$ over $R$, then the extension of any prime ideal of $R$ will remain prime. This corresponds geometrically to
the product of the subscheme $V(mathcal P)$ with affine space $mathbb A^n_R$.