Let $k \to k'$ a field extension / injective morphism and $A$ a $k$-algebra. Obviously $k \to k'$ is faithful flat (so flat and corresponding map $\operatorname{Spec}(k') \to \operatorname{Spec}(k)$ is surjective).
Why is $k \otimes_k A \to k' \otimes_k A$ also faithful flat.
I know that there is a theorem that $k \otimes_k A \to k' \otimes_k A$ is flat, but I don't see why the corresponding map $\operatorname{Spec}(k' \otimes_k A) \to \operatorname{Spec}(k \otimes_k A)$ is surjective.
Or more generally:
If $R \to S$ is a ring morphism such that the induced map $\operatorname{Spec}(S) \to \operatorname{Spec}(R)$ is surjective, why is then for every $R$-algebra $T$ the morphism $\operatorname{Spec}(S \otimes _R T) \to \operatorname{Spec}(R \otimes _R T)$ surjective?