Given a base field $k$. There is a correspondence

$$ \{ \text{non-singular projective curves over } k \} \leftrightarrow \{ \text{algebraic function fields of trans. degree 1 over} k \}.$$

Here curves mean *integral scheme of finite type over $k$*. Let $C/k$ be a such curve and $k(C)/k$ be the corresponding algebraic function field, i.e the function filed of the curve $C$.

There may exist elements of $k(C)$ which are algebraic over $k$. The ** full constant field** $\tilde{k}$ is the collection of all such elements. When $k$ is perfect, $C$ is geometrically irreducible if and only if $\tilde{k} = k$.

If $\tilde{k} \neq k$, one can regard $k(C)$ as an algebraic function field over $\tilde{k}$, hence one has a new non-singular projective curve $\tilde{C}$ over $\tilde{k}$.

The points on $C$ (as a scheme) is the same as the *places* of the algebraic functioin field $k(C)/k$. One can easily show that the places of $k(C)/ \tilde{k}$ are exactly the same as those of $k(C)/k$. **Hence the points of $C/k$ are the same as those of $\tilde{C}/\tilde{k}$.**

* Quetion:*
But what is the geometrical meaning of $\tilde{C}/\tilde{k}$ and its relation with $C/k$?

$\tilde{C}/\tilde{k}$ **is not** the base change of $C/k$ to be over $\tilde{k}$.

And for the Riemann-Roch Theorem, one avoids the term $\[ \tilde{k} : k \]$ by assuming thatit equals to 1, i.e one assumes that $C$ is geometrically integral.

But they are more "arithmetic" aspects of $\tilde{C}$, not geometrical aspects.