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Let $R$ be a commutative ring. Let $ f : R \to A$ be a smooth morphism of relative dimension 1. Suppose $f$ admits a retraction $r : A \to R$ i.e $r f = Id_R$. Is it true that the kernel of $r$ is a principal ideal ?

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up vote 2 down vote accepted

The kernel is locally principal, but need not be principal.

Let $R$ be an algebraically closed field, let $A$ be the ring of regular functions of a smooth affine curve $C$, then any closed point of $C$ defines a section (=rectraction), but the corresponding maximal ideal is not principal in general because otherwise $A$ would be principal. Concrete example: $C=$ a projective smooth curve of positive genus minus one point. Then no maximal ideal of $A$ is principal.

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How do you prove that it is locally principal ? – Damien L Feb 7 '13 at 10:00
@DamienL: this is true at the fibers of $\mathrm{Spec}(A)\to\mathrm{Spec}(R)$, then use Nakayama. – user18119 Feb 7 '13 at 16:24

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