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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.

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1 Answer 1

The factorization $C\to\text{Spec}(\tilde{k})\to\text{Spec}(k)$ is the Stein factorization of $C\to\text{Spec}(k)$. If you disregard base rings, we have $\tilde{C}=C$ as abstract schemes. The fact that $k\ne\tilde{k}$ accounts for the fact that $C$ may not remain connected after you extend scalars to an algebraic closure of $k$. You may find more on this relation with connectedness, as well as more general Stein factorizations (in nonproper cases), in this article, see for instance 2.5.2 and 3.2.5 there.

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