Section of conic bundle

Question

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).

Answer 1

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.

Answer 2

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.