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If $P \subset \Bbb R^2$ is a lattice polygon (i.e vertices with integer coordinates), we know that blowing-up the surface $X_P$ at a $T$-invariant point is the same as cutting a corner of $P$ and taking the corresponding toric variety. I understand this using the fan construction.

This is equivalent to take the closure of the embedding given by the monomial inside $P$ cutted at one corner but I have no intuition why this works. For example, the convex hull of $(0,0), (6,0)$ and $(0,6)$ gives the usual Veronese embedding so we get $X_P \cong \Bbb CP^2$. Now if I delete the monomials $x^ky^j$ for $k \geq 4$, the corresponding embedding should gives the blow-up of the projective plane $\Bbb CP^2$. Why ?

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I think I understood why, I'll just sketch how it works :

By definition $X_P$ is the closure of the graph $f : (\Bbb C^*)^2 \to \Bbb P^k, (x,y) \to (x^6, \dots, y^2, xy, x,y,1)$. I claim that the boundary of the image is composed of three axis intersecting each other in one point (so indeed we have $\Bbb P^2$). For this, this is enough to look at the limit when $t \to 0$ of any non-zero loop $\gamma_{\alpha, \beta, a,b}(t) = (\alpha t^a, \beta t^b)$. for $a,b \in \Bbb Z$ and $\alpha, \beta \in \Bbb C^*$.

But this time, $X_Q \backslash (\Bbb C^*)^2$ is the union of $4$ axis (again this can be see using the limits method), and the natural map $X_Q \to X_P$ is contracting one of the axis to a point so indeed, $X_Q$ is the blow-up of $X_P$.

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