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Let $\pi: Y \to X$ be a finite morphism of projective varieties over a field $k$, ramified over a divisor $D \subset X$. Let {$\{ X_t \}$} be a pencil of hyperplane section with respect to some embedding $X \hookrightarrow \mathbb{P}^N$.

Question. Is $\pi^{-1}(X_t)$ a pencil of hyperplane sections on $Y$?

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Not in general.

In fact, the pullback of an ample divisor under a finite morphism is ample, but the pullback of a very ample divisor is not necessarily very ample.

The easiest couterexample is probably the following. Take $ \pi \colon Y \to X$, where $X \cong \mathbb{P}^1$, $Y$ is a genus $2$ curve and $\pi$ is induced by the hyperelliptic involution.

If $X_t$ is a point in $X$ (which is a very ample divisor), then $\pi^{-1}(X_t)$ is an element of the complete $g^1_2$ on $Y$, which is ample but not very ample.

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Thanks Francesco! In general the pullback of a very ample divisor will only be ample, so it would be necessary to pass to a power to have something very ample. Does it mean that if I consider a suitable multiple of my embedding $X \hookrightarrow \mathbb{P}^N$ and I replace hyperplane sections by hypersurface sections the condition I want will be verified? – leffe Apr 1 '13 at 12:31
As a comment, another example in dimension $2$ (which is almost the same as the one of Francesco): let $X\to \mathbb{P}^2$ be the double covering ramified over a smooth quartic. Then, $X$ is a smooth del Pezzo surface of degree $2$ and the pull-back of an hyperplane section is the anticanonical divisor of $X$, which is ample but not very ample (the morphism associated is exactly the double covering). – Jérémy Blanc Apr 1 '13 at 12:32
@ leffe: yes, this is exactly this. – Jérémy Blanc Apr 1 '13 at 12:33

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