When I was reading a paper, I came across a statement like "Since ince $M$ is affine, the trivial bundle is ample and …" I think that line bundle $L$ on a variety $M$ is ample if it the global sections of $L^{\otimes n}$ give an embedding $M$ to some projective space. Is this clear for the trivial line bundle of affine variety?
[Math] Is the trivial bundle ample on an affine variety
algebraic-geometryvector-bundles
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(Let me collect the comments into a CW answer. Others can then edit the answer to add their own favourites.)
Here are a couple of simple examples, and one non-simple one. Note that any line bundle of degree $\geq 2g+1$ on a curve of genus $g$ is very ample, so any line bundle of positive degree on a curve is ample.
The canonical bundle $K$ on a hyperelliptic curve of genus $\geq 2$. Sections of $K$ define a 2:1 cover, so $K$ is globally generated and ample, but not very ample. On the other hand $K^2$ is very ample: for $g \geq 3$ this is immediate by the above comment; for $g=2$ the argument is a little more involved (Hartshorne IV.3.1).
Any bundle $L=O_C(p)$ where $p$ is a point on an elliptic curve $C$. Riemann—Roch shows that such a bundle has a 1-dimensional space of global sections, so is not very ample, or even globally generated, but it has positive degree, so is ample. On the other hand $L^3$ is very ample, and embeds $C$ into $\mathbf{P}^2$ as a smoooth cubic, with $p$ mapping to a flex.
If $C$ is a curve of genus $2$, and $p,q,r$ are general points on $C$, then the bundle $L=\mathcal{O}_C(p+q-r)$$ is ample, but has no global sections at all.
For a trickier example, one could consider a so-called Godeaux surface. This is a particular kind of surface of general type constructed as a quotient of a quintic surface in $\mathbf{P}^3$. It has the property that the canonical bundle $K_S$ is ample, but has no global sections. For more details, see the excellent answer of Clay Cordova here. Sadly, in this case I don't know what power of $K$ is needed to obtain a very ample bundle.
If $X$ is not too bad (I believe normal should be enough) you have an equivalence between Cartier divisor, line bundle and invertible sheaf. A line bundle $L$ gives you a Cartier divisor (up to linear equivalence) as the zero set of a section $s : X \to L$.
(Edit : as Mohan said, one should really put Cartier here : in fact, the sheaf $O_X(D)$ is an invertible sheaf, i.e a line bundle if and only if $D$ is Cartier)
Now, in term of $\mathcal O(1)$ on $\Bbb P^n$, a section $s$ of this bundle is by definition an homogenous polynomial of degree $1$, and its zero set will be an hyperplane $H$. Now, the pullback of $\mathcal O(1)$ to $X$ is $\mathcal O(1)_{|X}$ and of course $Z(s_{|D}) = Z(s) \cap D = D \cap H$. This shows why the definition of Hartshorne also gives you an hyperplane section.
Best Answer
Quoting Hartshorne, Example II.7.4.2:
Hartshorne's definition of ampleness (equivalent to yours; see Theorem II.7.6) is that $L$ is ample if, for any coherent sheaf $F$ on $X$, there exists an integer $n_0$ such that $F \otimes L^n$ is generated by globals for all $n > n_0$. You get the example I quote above upon recalling (II.5.16.2) that coherent sheaves on an affine scheme are determined by their global sections, and you get the result you quote because the trivial bundle is invertible.