Is there a Pick's Theorem for a general lattice in $\mathbb{R}^{2}$?

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Can't you just apply a linear transformation to convert the lattice to the standard lattice, and then apply the standard Pick's theorem to the image of the polygon? – Grumpy Parsnip Nov 14 '11 at 3:22
So if I do apply a linear transformation to convert the lattice to the standard lattice, why is it guaranteed that we have the same number of lattice points inside and on the boundary of the polygon before and after the transformation? – user4269 Nov 14 '11 at 3:57
Because the transformation maps lines to lines. Also, the intersection of two lines is mapped to the intersection of the mapped images of the lines. So parallel lines are mapped to parallel lines. Each lattice point is the intersection of two lines, parallel to the lines/vectors defining the lattice. This shows the number of lattice points in the boundary is invariant. – Andrés E. Caicedo Nov 14 '11 at 4:05
A similar geometric argument (perhaps a little bit harder to make rigorous; but this is just a general fact about invertible linear transformations) shows the interior of a polygon is mapped to the interior of the image of the polygon. So also the number of lattice points in the interior is invariant. – Andrés E. Caicedo Nov 14 '11 at 4:08
+1 I was just about to ask the same. – draks ... Mar 30 '12 at 17:48

Theorem (Pick's Theorem): Let $\Lambda$ be a two-dimensional lattice in $\mathbb{R}^k$ with 2-volume $\delta$. Let $P$ be a $\Lambda$-lattice polygon containing $h$ interior lattice points and $b$ boundary lattice points. Then the area $A(P)$ of $P$ is equal to $\delta \cdot (h + \frac{b}{2} - 1)$.
As a reference to the proof I give a 2003 book of Erdős and Surányi. But -- especially for the $k = 2$ case that you asked about -- Jim Conant's comment is right on: the proof consists simply of making a linear change of variables to get from $\Lambda$ to $\mathbb{Z}^2$.