Roots of line bundles

Let $k$ be a fixed algebraically closed field and $X/k$ an irreducible scheme smooth and proper over $k$. Can there exist a line bundles $\mathcal{L}, \mathcal{M}$ and an integer $m > 0$ so that

1.) $\dim_k \Gamma(\mathcal{L}) = 0$

2.) $\dim_k \Gamma(\mathcal{M}) > 0$

With $\mathcal{L}^m \cong \mathcal{M}^m$. If not, does an example exist if you drop smoothness (then replace $\mathcal{M}$ with the line bundle of an effective Weil divisor) or the condition that $k$ is algebraically closed?

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Yes. Take $\mathcal L$ the trivial line bundle, with a one-dimensional space of global sections, and $\mathcal M$ a nontrivial torsion line bundle, so $\mathcal M^k=\mathcal L$.
Then $\Gamma(\mathcal M)$ is certainly zero-dimensional, since otherwise $\mathcal M$ would have a nonvanishing section and be trivial or a somewhere vanishing section and then $\mathcal L$ would have a somewhere vanishing section and be nontrivial.
This provides a counterexample. Such examples exist on any smooth proper variety with a nontrivial Picard variety, meaning $\operatorname{dim}_kH^1(X,\mathcal O_X)>0$, such as curves of positive genus.
Yes. Take a curve of genus $2$ or higher, then most degree $1$ line bundles have a zero-dimensional space of global sections. Take the $n$th tensor power of this line bundle for $n\geq g$. Now it has degree $n$ and so a positive-dimensional space, by Riemann-Roch. – Will Sawin Oct 27 '12 at 18:37
Yes. Take a curve of genus $g\geq 2$. Then the space of line bundles of degree $1$ is a complex torus of dimension $g$. The space of line bundles which have a nontrivial global section is the image of the curve under the map sending a point to the line bundle corresponding to its divisor, so has dimension $1$. Fix one such line bundle. Now consider all the line bundles that are it tensor a torsion line bundle. These correspond to translations of the torus by torsion points on its Jacobian, so these points are dense, so at least one of them does not have a nontrivial global section. – Will Sawin Oct 27 '12 at 20:04
It is not too hard to see by counting that $m=2$ suffices on a hyperelliptic curve. If $f$ is the hyperelliptic projection to $\mathbb P^1$ there are $2g+2$ points such that $\mathcal O(P)^{\otimes 2}=f^*\mathcal O(1)$, but there $2^{2g}$ degree $1$ divisors such that $\mathcal L^{\otimes 2} =f^* \mathcal O(1)$, so there must be one line bundle of that form with a nontrivial global section and one line bundle of that form without a nontrivial global section. – Will Sawin Oct 27 '12 at 20:13
If there were any more global sections you would get a degree $2$ map to $\mathbb P^2$, which makes your curve a conic in $\mathbb P^2$, which makes it rational. I guess this is a special case of Clifford's Theorem (Hartshorne 4.5.4) + Riemann-Roch: By Clifford's theorem, if $f^* \mathcal O(1)$ is special then the dimension of the space of global sections is at most $2/2+1=2$. On the other hand if it's general then the dimension of the space of global sections is $2+1-g=3-g\leq 1$. So the dimension of the space of global sections is exactly $2$, so all of them come from the pullback. – Will Sawin Oct 28 '12 at 3:21
Another thing to generalize it from is Lemma 4.5.1, which says the canonical linear system has no base points, plus Riemann-Roch, to show the dimension of the space of global sections of divisor of degree $d>0$ on a curve of genus $g\geq 2$ is at most $d$. – Will Sawin Oct 28 '12 at 3:23