Definition of degree of invertible sheaf Sorry for my bad English.
Let $X$ be regular curve over algebraic closed field $k$, and $T$ be scheme over $k$.
In Hartshorne algebraic geometry Section Iv, he define $Pic^0(X\times T)$ as follows:
invertible sheaf ${\scr L}\in Pic(X\times T)$ s.t. for any $t\in T$, degree of ${\scr L}$ is 0 over fiber $X_t$.
Now, I trouble about what is definition of degree of ${\scr L} $over fiber $X_t$.
If  $X_t$ is regular curve, I know this definition, but I don't know how $X_t$ is curve.
 A: If $p_T: T\times_k X \rightarrow T$ is the projection morphism, you define for any point $t\in T$
$$X_t:= p_t^{-1}(t)\cong  Spec(\kappa(t))\times_T (T\times_k X) \cong Spec(\kappa(t))\times_k X.$$
Here $k \subseteq \kappa(t)$ is a field extension and need not be an eqality in general. If $X$ is an integral scheme  over $k$ (let us assume $X$ is irreducible, then $X$ is integral since it is regular) it follows from Hartshorne, Ex II.3.20 that every irreducible component $X_i$ of $X_t$  has $dim(X_i)=dim(X)=1$ hence $X_t$ is a "curve" for any point $t \in T$.
Definition: We say a scheme $X$ of finite type over a field $k$ is a "curve" if the irreducible components $X_i$ of $X$ have $dim(X_i)=1$.
Question: "If $X_t$ is regular curve, I know this definition, but I don't know how $X_t$ is curve."
Answer: This follows from HH.Ex.3.20.
Example. Let $k$ be the field of real numbers and $K$ the field of complex numbers. Let $A:=k[x,y]/(x^2+y^2)$ and let $B:=K\otimes_k A$ with $C:=Spec(A)$. It follows $A$ is an integral domain and hence $C$ is an integral scheme of dimension one, but
$$K\otimes_k B \cong K[x,y]/(x+iy)(x-iy)$$
is no longer an integral domain. Hence $Spec(K)\times_k C:=C_K$ is no longer an irreducible scheme. The dimension of the irreducible components of $C_K$ is as you can see one.
Note: Since your interest is in the Picard scheme you cannot restrict to closed points as claimed in the comments. Hence the field extension $k \subseteq \kappa(t)$ will not be an isomorphism in general.
