3 added 23 characters in body

1) Let $C$ be a curve (of genues at least 2) with a $2:1$ morhpism on $\mathbb{P}^1$. Then $K_C$ has positive degree, hence it is ample. But you can check by Riemann-Hurwitz that $K_C$ is just the pullback of some multiple (depending on the genus) of the hyperplane section on $\mathbb{P}^1$. Moreover by Riemann-Roch you see that all sections of $K_C$ are pulled back by $\mathbb{P}^1$. It follows easily that the map given by sections of $K_C$ is just the map to $\mathbb{P}^1$ followed by a Veronese embedding; in particular it is not injective. So $K_C$ is ample but not very ample. Moreover by Serre duality you have $h^1(K_C) = h^0(\mathcal{O}) = 1$.

For a very ample line bundle with cohomology EDIT: removed nonsense

2) There is a bound depending only on the dimension of $X$ and the coefficients of the Hilbert polynomial of $L$; this is known as Matsusaka big theorem.

3) In general no, since $L$ will not give you a finite map. But it may happen that $L$ gives a finite map which is not an isomorphism; in this case you can then project to get a finite map to a projective space.

For instance take the double covering of $\mathbb{P}^2$ ramified over a sexitc, say this is $\pi \colon X \to \mathbb{P}^2$; then the line bundle $\pi^{*}(\mathcal{O}(1))$ gives the map $\pi$ itself, which is finite. But is it true that $\pi^{*}(\mathcal{O}(1))$ is ample, for instance by Kleiman's criterion. More generally the pullback of an ample line bundle under a finite map is still ample.

2 deleted 801 characters in body

1) Let $C$ be a curve with a $2:1$ morhpism on $\mathbb{P}^1$. Then $K_C$ has positive degree, hence it is ample. But you can check by Riemann-Hurwitz that $K_C$ is just the pullback of some multiple (depending on the genus) of the hyperplane section on $\mathbb{P}^1$. Moreover by Riemann-Roch you see that all sections of $K_C$ are pulled back by $\mathbb{P}^1$. It follows easily that the map given by sections of $K_C$ is just the map to $\mathbb{P}^1$ followed by a Veronese embedding; in particular it is not injective. So $K_C$ is ample but not very ample. Moreover by Serre duality you have $h^1(K_C) = h^0(\mathcal{O}) = 1$.

For a very ample line bundle with cohomology we can assume that $X$ is embedded in $\mathbb{P}^n$ and $L = \mathcal{O}_X(1)$. Then, since $h^1(\mathbb{P}^n, \mathcal{O}(1)) = 0$, having first cohomology for $L$ is equivalent to saying that the restriction map $H^0(\mathbb{P}^n, \mathcal{O}(1)) \to H^0(X, \mathcal{O}(1))$ is not surjective. And this can surely happenEDIT: it is enough that $X$ is embedded in $\mathbb{P}^n$ by a linear series which is not the whole $H^0(L)$. For instance you can embed $\mathbb{P}^1$ in $\mathbb{P}^2$ via sections of $\mathcal{O}(2)$ (so you obtain a conic) and then embed $\mathbb{P}^2$ in $\mathbb{P}^5$ via sections of $\mathcal{O}(2)$ (so you obtain the Veronese surface). The resulting $\mathbb{P}^1$ in $\mathbb{P}^5$ is embedded by a sub-vector space of all quartic polynomials, namely compositions of polynomials of degree two.removed nonsense

2) There is a bound depending only on the dimension of $X$ and the coefficients of the Hilbert polynomial of $L$; this is known as Matsusaka big theorem.

3) In general no, since $L$ will not give you a finite map. But it may happen that $L$ gives a finite map which is not an isomorphism; in this case you can then project to get a finite map to a projective space.

For instance take the double covering of $\mathbb{P}^2$ ramified over a sexitc, say this is $\pi \colon X \to \mathbb{P}^2$; then the line bundle $\pi^{*}(\mathcal{O}(1))$ gives the map $\pi$ itself, which is finite. But is it true that $\pi^{*}(\mathcal{O}(1))$ is ample, for instance by Kleiman's criterion. More generally the pullback of an ample line bundle under a finite map is still ample.

1

1) Let $C$ be a curve with a $2:1$ morhpism on $\mathbb{P}^1$. Then $K_C$ has positive degree, hence it is ample. But you can check by Riemann-Hurwitz that $K_C$ is just the pullback of some multiple (depending on the genus) of the hyperplane section on $\mathbb{P}^1$. Moreover by Riemann-Roch you see that all sections of $K_C$ are pulled back by $\mathbb{P}^1$. It follows easily that the map given by sections of $K_C$ is just the map to $\mathbb{P}^1$ followed by a Veronese embedding; in particular it is not injective. So $K_C$ is ample but not very ample. Moreover by Serre duality you have $h^1(K_C) = h^0(\mathcal{O}) = 1$.

For a very ample line bundle with cohomology we can assume that $X$ is embedded in $\mathbb{P}^n$ and $L = \mathcal{O}_X(1)$. Then, since $h^1(\mathbb{P}^n, \mathcal{O}(1)) = 0$, having first cohomology for $L$ is equivalent to saying that the restriction map $H^0(\mathbb{P}^n, \mathcal{O}(1)) \to H^0(X, \mathcal{O}(1))$ is not surjective. And this can surely happen: it is enough that $X$ is embedded in $\mathbb{P}^n$ by a linear series which is not the whole $H^0(L)$. For instance you can embed $\mathbb{P}^1$ in $\mathbb{P}^2$ via sections of $\mathcal{O}(2)$ (so you obtain a conic) and then embed $\mathbb{P}^2$ in $\mathbb{P}^5$ via sections of $\mathcal{O}(2)$ (so you obtain the Veronese surface). The resulting $\mathbb{P}^1$ in $\mathbb{P}^5$ is embedded by a sub-vector space of all quartic polynomials, namely compositions of polynomials of degree two.

2) There is a bound depending only on the dimension of $X$ and the coefficients of the Hilbert polynomial of $L$; this is known as Matsusaka big theorem.

3) In general no, since $L$ will not give you a finite map. But it may happen that $L$ gives a finite map which is not an isomorphism; in this case you can then project to get a finite map to a projective space.

For instance take the double covering of $\mathbb{P}^2$ ramified over a sexitc, say this is $\pi \colon X \to \mathbb{P}^2$; then the line bundle $\pi^{*}(\mathcal{O}(1))$ gives the map $\pi$ itself, which is finite. But is it true that $\pi^{*}(\mathcal{O}(1))$ is ample, for instance by Kleiman's criterion. More generally the pullback of an ample line bundle under a finite map is still ample.