(The polytope)
In \(M_\RR = \RR^3\) take \[ P = \operatorname{Conv}\big( (0,0,0),\ (0,1,1),\ (1,0,1),\ (1,1,0) \big) , \] a lattice \(3\)-simplex, and let \(D = D_P\) be the corresponding divisor on \(X_P\). Since \(P\) is a full-dimensional simplex, its normal fan has four maximal cones matched with the four vertices, so \(\varphi_D\) is strictly convex and \(D\) is ample.
(Why it is not very ample)
The only lattice points of \(P\) are its four vertices. At the vertex \(v = (0,0,0)\) the edge vectors are \[ (0,1,1), \quad (1,0,1), \quad (1,1,0) , \] and \[ \det \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} = 2 , \] so they generate a sublattice of index \(2\) in \(M = \ZZ^3\). The semigroup \(\ts{m - v \st m \in P \intersect M}\) is therefore not saturated, which is exactly the clause very ampleness requires, and \(D\) fails it.
(What the map does)
The four lattice points give a morphism \(X_P \to \PP^3\), and \[ \deg X_P = 3! \cdot \vol(P) = 3! \cdot \tfrac{1}{3} = 2 , \] so the map is finite of degree \(2\) rather than an embedding: \(X_P\) is a double cover of \(\PP^3\) branched over the four coordinate hyperplanes. Doubling repairs it: \(2P\) is normal and hence very ample, so \(2D\) is very ample while \(D\) is ample but not.
Three structural features make the example minimal. It must be three-dimensional, because ample and very ample agree on complete toric surfaces. It must be singular, because they also agree on smooth complete toric varieties. And the obstruction is a single determinant, which is the same index computation that detects a quotient singularity at the corresponding fixed point.
Dependencies
PR-TORPOSBase-point freeness, ampleness and very ampleness through support functionsusesD-FULNORMPOLYNormal polytopes and very ample dilationsusesT-FULDEGThe degree of a projective toric variety is the normalised volume of its polytopeuses