Can we establish $f_0: X_0\to Y_0$ for given $f:X_0\times \operatorname {Spec}k\to Y_0\times \operatorname {Spec} k$ by taking finite extension?

Question

Sorry for my bad English.

Let $K/k_0$ be infinite field extension, and $X_0, Y_0$ be finite type schemes over $k_0$.

Now base change $X:=X_0\times_{k_0} \operatorname {Spec}K, Y:=Y_0\times _{k_0} \operatorname {Spec} K$, and $f: X\to Y$ is morphism over $K$.

Now is there intermediate field　$k_0\subset k_1\subset k$ such that $k_1/k_0$ is finite, and morphism $f_1: X_0\times_{k_0}\operatorname {Spec} k_1\to X_0\times_{k_0}\operatorname {Spec} k_1$ over $k_1$ such that $f_1\times_{k_1}k=f$?

If you need, you can assume various condition s.t. $k$ is algebraically closed, $k_0$ is finite field, or $X_0, Y_0$ is integral.