The independence of the degree of morphisms between two curves on field extension. Suppose $C_1$ and $C_2$ are two regular projective geometrically irreducible curves over $k$ and $F$ is a surjective morphism between them, then the degree of $F$ is the degree of the field extension 
\begin{equation}
\text{deg}~F=[K(C_1):K(C_2)]
\end{equation} 
If $k_1$ is a larger field, since $C_i$ is geometrically irreducible, $C_i \times_k k_1$ is also irreducible and $F$ induces a morphism $F_1$
\begin{equation}
F_1:C_1 \times_k k_1 \rightarrow C_2 \times_k k_1
\end{equation}
$F_1$ is also surjective since which is preserved by fibered product. So how to show the degree of $F_1$ is the same as $F$?
 A: Thanks to Brian Ng for helpful discussion on this.
With the hypotheses you write, it is not always true that if $C$ is regular then also $C \times_{\text{Spec}\,k} \text{Spec}\,k_1$ is regular. You should either specify that $C$ is generically smooth, or you should specify that $k_1/k$ is separable. Once you add these hypotheses, then the degree of $F_1$ (as you have defined it) equals the degree of $F$.
Since $C$ is geometrically irreducible, $k$ is algebraically closed in the fraction field of $C_i$. The hypothesis guarantees that $C \times_{\text{Spec}\,k} \text{Spec}\,k_1$ is integral. The fraction field of this integral scheme is $K \otimes_k k_1$. Thus, the equality is just preservation of dimension under field extension: for a $k$-vector space $V$, $\text{dim}_{k_1}(V \otimes _k k_1)$ equals $\text{dim}_k(V)$.
More pertinently, $K_1 = K \otimes k_1$ is the fraction field of $C \times_{\text{Spec}\,k} \text{Spec}\,k_1$. So for every $K$-vector space $W$, $\text{dim}_K(W)$ equals $\text{dim}_{K_1}(W \otimes_K K_1)$.
