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Suppose $X$ is a smooth projective surface with a dominant morphism $\pi:X \rightarrow \mathbb{P}^{1}$ over a field $k$, where all the fibres of $\pi$ are conics (i.e. a conic bundle). If $\pi$ admits a section $s$ over $k$ (i.e. there exists $s:\mathbb{P}^{1}_{k}\rightarrow X$ such that $\pi \circ s=\text{Id}_{\mathbb{P}^{1}}$) then why can't $s$ meet a singular fibre at it's singular point (the singular fibres are two traversal lines, where the intersection of these two lines is the singular point).

Additional question: Why does the singular point on one of the fibres of $\pi$ not define a singular point on $X$. My (very non-rigourus) guess was that as the singular point on this fibre would be 2 dimensional it defines something singular on a conic (which is 1 dimensional) but not on the surface (which is 2 dimensional).

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  • $\begingroup$ Are you assuming that the relative Picard number of $\pi$ is one? $\endgroup$
    – Puzzled
    Jun 16, 2022 at 22:37
  • $\begingroup$ @BlaCa No I wasn’t assuming my conic bundle is minimal. $\endgroup$
    – H U
    Jun 17, 2022 at 23:06

2 Answers 2

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The differential of a section is right inverse to the differential of $\pi$, hence $d\pi$ is surjective and $\pi$ is smooth along the section.

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    $\begingroup$ Would you mind spelling out what you mean by the differential is a right inverse? $\endgroup$ Jun 14, 2022 at 17:46
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    $\begingroup$ If you differentiate the equation $\pi \circ s = \mathrm{id}$, you obtain $$d\pi \circ ds = \mathrm{id}$$, which means that $ds$ is right inverse to $d\pi$. $\endgroup$
    – Sasha
    Jun 14, 2022 at 19:52
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Intersection theory provides another way of seeing why a section cannot pass through a double point.

If $D$ is a horizontal divisor corresponding to a section of $\pi$, then $D$ will have intersection multiplicity 1 with each vertical divisor $V$: $$D\cdot V =1$$

But if the vertical divisor has two irreducible components $V = E_1 + E_2$, and $D$ passes through their intersection point, transversely to each, then $$D\cdot V = D\cdot E_1+D\cdot E_2 = 1 + 1 \neq 1$$

See chapter 9 of Qing Liu's book for more deets.

His corollary 4.2.12 may also be useful to you in explaining why the singular point can be regular. The quotient $A/fA$ of a regular Noetherian local ring by $f\in \mathfrak{m} \setminus 0$ will again be regular if and only if $f \notin \mathfrak{m}^2$. If your total space is given locally around the singular point by an equation like $xy-t$, where $t$ is a function on the base, then the singular point will be regular as long as $t$ has valuation 1.

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