# connection between absolute irreducibility and smooth+geometrically connected

Let $C_1$ and $C_2$ be two smooth, projective, and geometrically connected curves over a field $K$ of characteristic $0$. Assume there are two non-constant branch mapping of $K$-curves, $\phi_1\colon C_1\to P^1_K$ and $\phi_2\colon C_2\to P^1_K$ onto the projective line.

Taking function fields, one gets two finite extensions $F_1,F_2$ of $K(x)$ that are regular over $K$ (i.e. $K$ is algebraically closed in $F_1$ and in $F_2$).

Now assume that the compositum field $F_1F_2$ is also regular over $K$ and that $F_1, F_2$ are linearly disjoint over $K(x)$. This means that $C_1\times_{P^1} C_2$ is absolutely irreducible.

My question is whether in this setting it is also true that $C_1\times_{P^1} C_2$ is smooth, projective, and geometrically connected.

• How did you conclude that if $F_1F_2$ is regular over $K$ means $C_1\times_{\mathbb{P}^1} C_2$ is absolutely irreducible? For example, take $C$ to be an elliptic curve over the complex numbers and $\phi:C\to\mathbb{P}^1$ a degree 2 map. Let $C=C_1=C_2, \phi=\phi_1=\phi_2$. Then $F_1=F_2$, but the fiber product is not irreducible. Jun 30, 2015 at 15:33
• Mohan, I forgot to say that the fields are linearly disjoint. Fixed the question Jul 1, 2015 at 21:23

I do not think so. If you take $\phi_i:\mathbb{P}^1\to\mathbb{P}^1$ given by $\phi_i(x)=x^{i+1}$, then the fields are linearly disjoint, but the fiber product is singular.