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Is there a general formula for the canonical divisor $K_X$ of a smooth conic bundle $X$?

Motivation: for smooth hypersurfaces of degree $d$ in $\mathbb{P}^n$, $K_X = \mathcal{O}_X(d-n-1)$. But smooth conic bundles are not "naturally" embedded in some $\mathbb{P}^n$, so I was wondering if an analogue of the above formula holds for them.

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  • $\begingroup$ What is the base for your conic bundle morphism? Something smooth and projective? Also, when you say "smooth conic bundle $X$", do you mean the conic bundle morphism is smooth, or that $X$ is non-singular as an algebraic variety? $\endgroup$
    – Daniel Loughran
    Commented Oct 31, 2014 at 8:48
  • $\begingroup$ The base is $\mathbb{P}^1$. By smooth I mean $X$ is smooth. $\endgroup$
    – user61125
    Commented Oct 31, 2014 at 9:18

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Here is a method which should give you what you want, but I leave the details to you as I don't have time to work them out now.

Without loss of generality we may work over an algebraically closed field. Each singular fibre consists of a union of two $(-1)$-curves meeting at a single point. Contracting a choice of $(-1)$-curve in each single fibre, we obtain a ruled surface over $\mathbb{P}^1$. There is a formula for the canonical divisor of a ruled surface, which when combined with the formula for the canonical divisor of a blow-up, should give you what you want (all these tools can be found in Ch. V of Hartshorne).

Note that a useful check is given by Noether's formula, which implies that $$(K_X)^2 = 8 - \mbox{number singular of fibres}.$$

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  • $\begingroup$ Just to comment: for a general nice base curve of genus $g$, the above self-intersection should be $K_X^2 = 8(1-g) - \text{no of sing fibers}$. $\endgroup$
    – user61125
    Commented Oct 31, 2014 at 17:57

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