Hypersurfaces in Toric Varieties, Help understand a proof from Mikhalkin's paper

Question

Hello, in G. Mikhalkin's Papaer "DECOMPOSITION INTO PAIRS-OF-PANTS FOR COMPLEX ALGEBRAIC HYPERSURFACES": http://arxiv.org/pdf/math/0205011.pdf There is a lemma about the relation between intersection of a hypersurface with the boundary divisors in a toric variety and the truncation of the defining polynomial of a hypersurface to some face of the newton polytope. (Lemma 2.20. Page 15).

The lemma basically says that If we take a hypersurface with newton polytope P, and consider it's closure in the projective toric variery corresponding the the lattice polytope P, then to find it's intersection with the boundary divisors, it is enough to truncate the polynomial the the corresponding face of the newton polytope and take the zeroes of the truncated polynomial.

The proof given there is a one-liner about the order of vanishing of some monomials and I don't understand why it proves the claim.

I would appreciate if someone could explain this to me.

Thank you

Answer 1

Let's consider an example. Triangle $(0,0),(0,d),(d,0)\ $ gives us $\mathbb CP^2$ and each integer point $\{(k,l) |k,l\geq 0, k+l\leq d\}\ $ corresponds to monomial $x^ky^l$ (or $x^ky^lz^{d-k-l}$ in projective coordinates). There is the same situation for any toric variety - ring of function is generated by monomials corresponding to integer points inside the polytop $P$.

So, it is evident where is the zero set of each monomial -if a point $(k,l)$ belongs to interior of $P$ then zero set is just the union of boundary divisors. If no, for example the point is $(k,0) \to x^kz^{d-k}$, zero set is all boundary divisors without that where our point lies.

Therefore, if you truncate your polynomial, then its zero set doesn't change - throwed off monomials are equal zero on you boundry divisor.