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This example is a test of several quantities invariant in a flat family.

Let $X$ be a projective variety, let $D$ be a divisor on $X$ with $\dim|D|\ge 1$. Choose a pencil $\{X_t\}_{t\in\mathbb P^1}$ inside $|D|$. This is in particular an algebraical family, therefore a flat family. We know Hilbert polynomial is invariant in a flat family, therefore all quantities arising from Hilbert polynomial, including dimension, degree and arithmetic genus, should be invariant in a such a family.

Here is an example seems contradicting to me: Let $p:X\to \mathbb P^1\times \mathbb P^1$ be the surface blowing up at a point, and let $\pi:X\to \mathbb P^1$ be the composite $p_1\circ p$ with $p_1$ the projection to the first coordinate. ($X$ is indeed a Hirzebruch surface $\Sigma_1$, but we are considering its $\mathbb P^1$-fibration in a different way.)

Then the map $\pi$ gives us a pencil family of divisors on $X$, with general fiber $X_t=\mathbb P^1$, and special fiber $X_0=\mathbb P^1\cup \mathbb P^1$, two $\mathbb P^1$ intersecting at a point. Now, let's check the those quantities mentioned above.

Obviously, the dimension is always 1 which doesn't change. The arithmetic genus is also the same by the answer of this post. However, the degree of $X_t$ is $1$, while the degree of $X_0$ is $2$ (The degree is additive on irreducible components, I believe.)

So my question is:

(1) Does $\pi: X\to \mathbb P^1$ gives a flat family of divisors?

(2) Is it true that $\deg(X_0)=2$?

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The map is flat. In fact, a map from a variety to a smooth curve is flat if and only if it is dominant (see Hartshorne).

The degree is computed with respect to some (ample) line bundle. For instance, if $h_1$ and $h_2$ are the hyperplane classes of factors of $\mathbb{P}^1 \times \mathbb{P}^1$ and $e$ is the exceptional divisor class of the blowup $p \colon X \to \mathbb{P}^1 \times \mathbb{P}^1$, then $2h_2 - e$ is ample on fibers of $p_1$ and with respect to it both general and special fibers have degree 2.

Finally, $X$ is NOT a Hirzebruch surface, since its Picard number is 3. It is the del Pezzo surface of (anticanonical) degree 7.

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  • $\begingroup$ Dear Sasha, thanks for your correction and explanation. Indeed, the Hirzebruch surface is blowing down a $(-1)$ curve on $X$. And I can now see that the map $\pi:X\to \mathbb P^1$ gives a family of conics degenerating to two lines. But here is a question in your answer: $2h_2-e$ is divisor on $X$ whose restriction on each fiber of $\pi$ is ample, but itself is not ample on $X$ (as its paring with $h_2$ is zero). It seems to be a notion "being ample with respect to a morphism" if I interpret correctly. It is new to me. Do you have a reference to this notion? $\endgroup$ – Yilong Zhang Jan 20 at 20:08
  • $\begingroup$ For instance, see stacks.math.columbia.edu/tag/01VG. $\endgroup$ – Sasha Jan 20 at 20:20

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