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Aug 22, 2023 at 19:54 comment added Tom Ducat The reason it is an embedding is that $-K_X$ is a very ample divisor class for $X$. (Not true if $k>6$.) The reason you can express it in terms of cubics like that follows from the standard formula for pulling back divisors after blowing up a smooth point on a surface.
Aug 21, 2023 at 9:04 comment added woolly-minded What I don't understand is why this is true: why knowing that $-K_X=\pi^*\mathcal O_{\mathbb P^2}(3)-E_1-\dots-E_k$ (which I understand) gives us the fact that we can write down the anticanonical embedding in that way.
Aug 19, 2023 at 17:41 comment added Tom Ducat In which direction do you want more explanation? Why it is true? Or how to do it?
Aug 18, 2023 at 18:19 comment added woolly-minded "Therefore you can write down the anticanonical embedding of 𝑋 by writing down a basis for the linear system of cubic polynomials in 𝑥,𝑦,𝑧 that vanish at 𝑃1,…,𝑃𝑘." Can you explain this in more detail?
Nov 10, 2021 at 18:07 comment added Nicolas Banks @JasonStarr I am indeed interested in how the construction would generalize to, say, $\mathbb{Q}$ instead of $\mathbb{C}$. In particular, I want to take the points $P_1,...,P_k$ to be Galois conjugates over some number field of degree $k$. But this answer is a great starting point, especially the comments about the anticanonical embedding. Also to add, a discussion on the aforementioned automorphism group can be found here: mathoverflow.net/questions/212015/…
Nov 10, 2021 at 10:44 comment added Jason Starr Possibly the OP wants to work over a non-algebraically closed field (the degree $6$ del Pezzo has a large automorphism group).
Nov 10, 2021 at 10:00 history answered Tom Ducat CC BY-SA 4.0