Why is surjectivity stable under base change?

Solution 1:

Following your Idea 1 it only remains to show $X\times_S \mathrm{Spec} (k(t))$ is non-empty.

Let $s=\varphi(t)$, then the above fiber product is also $X_s\times_{\mathrm{Spec}(k(s))} \mathrm{Spec}(k(t))$. Take a non-empty affine open subset $U=\mathrm{Spec}(R)$ of $X_s$. It is enough to show $U\times_{\mathrm{Spec}(k(s))} \mathrm{Spec}(k(t))$ is non-empty.

The latter is an affine scheme given by $R\otimes_{k(s)} k(t)$. As $R\ne 0$, $R\otimes_{k(s)} k(t)\ne 0$ because we make a field extension (a vector basis of $R$ over $k(s)$ extends to a vector basis over $k(t)$). By Krull's theorem, $R\otimes_{k(s)} k(t)$ has a maximal (hence prime) ideal, so $U\times_{\mathrm{Spec}(k(s))} \mathrm{Spec}(k(t))\ne\emptyset$.