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We work over the complex numbers. Fix a genus $g$. Does there exist a connected reduced base $ B $ and a flat projective family $ \pi : X \rightarrow B $ satisfying the following two conditions?

  1. its general fiber is a nonsingular curve of genus $ g $.
  2. it has a special fiber $ X_0 $ which is an irreducible, projective one-dimensional scheme such that $ (X_0)_\text{red} $ is isomorphic to $ \mathbb{P}^1$.

Toy cases: If $ g $ is a triangular number, write $ g = \frac{(d-1)(d-2)}{2} $ and we can take the universal family of degree $ d $ hypersurfaces in $ \mathbb{P}^2 $. If $ g = 4 $, we can take the family of complete intersections of type $ (2,3) $ in $ \mathbb{P}^3 $. This family has a fiber whose reduction is the twisted cubic. Similarly, if $ g=5 $, complete intersections of type $ (2,2,2) $ in $ \mathbb{P}^4 $ work. I don't know what happens for genera which can never occur as genera of complete intersection curves in projective space, say $ g=2,7,8 \dotsc $, although I think these occur as intersections in other varieties such as Grassmanians / products of projective spaces.

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    $\begingroup$ Embed your curve in projective space and use a torus action to specialize to a multiple line. $\endgroup$
    – Jason Starr
    Commented Dec 17, 2021 at 10:48
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    $\begingroup$ If I understand correctly, the OP is looking for some curve of genus g for each g not every curve of that genus. Another example is to take a type (2, g+1) divisor on $\mathbb{P}^1 \times \mathbb{P}^1$ and deform it to a double line $\endgroup$
    – Ben C
    Commented Dec 17, 2021 at 20:07
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    $\begingroup$ @BenC "... and deform it to a double line" If $g+1$ is odd, how can you deform a divisor of type $(2,g+1)$ to the double of a divisor? $\endgroup$
    – Jason Starr
    Commented Dec 18, 2021 at 1:14
  • $\begingroup$ @Jason Starr that's what I don't understand as well. Can you please explain what to do in your first comment? I thought of it but I don't know torus actions very well. $\endgroup$
    – Cranium Clamp
    Commented Dec 18, 2021 at 3:09
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    $\begingroup$ @JasonStarr you're right, what I said only works if $g$ is odd. In that case, I can choose a smooth divisor $D$ of type $(1, \tfrac{g + 1}{2})$ and then deform $2D$ to get a smooth divisor of type $(2,g)$ which then has genus $g$ while $D$ is isomorphic to $\mathbb{P}^1$. I guess this strategy should work if I can write $g = (a-1)(b-1)$ with $a$ dividing $b$ but I didn't think hard enough about the numerology when I said my example. $\endgroup$
    – Ben C
    Commented Dec 18, 2021 at 5:23

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