Let $ \phi : Y \to X $ and $ \psi : Z \to X $ be finite morphisms of integral algebraic curves over a field $ k $.
Let $ \phi^\* : K( X ) \to K( Y ) $ and $ \psi^\* : K( X ) \to K( Z ) $ be the pullbacks of $ \phi $ and $ \psi $.


Is the following true?


The fiber product $ Y \times\_X Z $ is an integral curve over $ k $ if and only if the tensor product $ K( Y ) \otimes\_{ K( X ) } K( Z ) $ is a field.


Thank you




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Edit: As shown in the comments, this is not true without the additional assumption that $ X $ is smooth.


Now, is it true with this assumption?



Fiber product of algebraic curves
This is ok when $X$ is smooth and $Y$ and $Z$ are integral. The point is that over a *smooth* curve, any finite morphism from an integral curve is automatically flat: over a Dedekind scheme, flat is the same as torsion-free [Tag [0AUW](https://stacks.math.columbia.edu/tag/0AUW)], and a finite integral extension is indeed torsion-free.


Then $Y \times\_X Z \to X$ is also finite flat, so satisfies both going up [Tag [00GU](https://stacks.math.columbia.edu/tag/00GU)] and going down [[00HS](https://stacks.math.columbia.edu/tag/00HS)]. This implies that all irreducible components of $Y \times\_X Z$ have dimension $1$, and for any point $p \in Y \times\_X Z$ the height of $p$ is the same as the height of its image in $X$. In particular, every generic point of an irreducible component maps to the generic point of $X$, so there is only one irreducible component if and only if $K(Y) \otimes\_{K(X)} K(Z)$ has only one irreducible component.


Finally we have to say something about reducedness. Recall that a scheme $S$ is reduced if and only if it is (R$\_0$) and (S$\_1$) [Tag [031R](https://stacks.math.columbia.edu/tag/031R)]. For $Y \times\_X Z$, the condition (R$\_0$) means exactly that $K(Y) \otimes\_{K(X)} K(Z)$ is reduced. Thus, we see that $K(Y) \otimes\_{K(X)} K(Z)$ is a field if and only if $Y \times\_X Z$ is (R$\_0$) and irreducible.


Recall also that a $d$-dimensional Noetherian scheme $S$ is Cohen–Macaulay if and only if it is (S$\_d$), hence a $1$-dimensional scheme $S$ is (S$\_1$) if and only if it is Cohen–Macaulay. So it remains to show that $Y \times\_X Z$ is always Cohen–Macaulay. But $\Delta\_X \colon X \hookrightarrow X \times X$ is a regular closed immersion [Tag [0E9J](https://stacks.math.columbia.edu/tag/0E9J)], hence so is $Y \times\_X Z \hookrightarrow Y \times Z$ since it is the pullback of $\Delta\_X$ along $Y \times Z \to X \times X$ [Tag [067P](https://stacks.math.columbia.edu/tag/067P)]. Since $Y$ and $Z$ are reduced curves, they are (S$\_1$), hence Cohen–Macaulay as they have dimension $1$. Then $Y \times Z$ is Cohen–Macaulay as well [Tag [045J](https://stacks.math.columbia.edu/tag/045J)], hence so is $Y \times\_X Z$ since it is a regular closed subscheme of $Y \times Z$ [Tag [02JN](https://stacks.math.columbia.edu/tag/02JN)].


