Finite affine schemes are projective about synthetic-zariski HOT 4 CLOSED

With Hugo, we just figured out that the above doesn't work like claimed. The problem is, that in the above, the closed subset of $\mathbb{P}^{n-1}$ does also contain linear maps that do not map $1$ to $1$.

from synthetic-zariski.

I don't see the problem. Phrased differently, we consider the projective space $\mathbb PA^\star$ associated with the R-linear dual of $A$. This is $\mathbb P^{n-1}$ after choosing a basis of $A$. Given $[\varphi] : \mathbb PA^\star$ we consider the proposition $C([\varphi])$ that $\varphi(1) \varphi(xy) = \varphi(x) \varphi(y)$ for all $x, y : A$. This is well-defined and a closed proposition because it suffices to check it for basis elements of $A$. $C([\varphi])$ implies $\varphi(1) \ne 0$ because otherwise $\varphi(x)^2 = 0$ for all $x : A$ and then $\varphi$ is not-not zero (projective space contains only non-zero vectors). So $x \mapsto \varphi(x) / \varphi(1)$ determines a point of Spec A for $[\varphi] : \mathbb PA^\star$ such that $C([\varphi])$ holds (and one can go in the reverse direction, and verify that the two maps are inverse to each other).

from synthetic-zariski.

Nice - thanks for refusing to see the problem non-existent problem ;-)

from synthetic-zariski.

written down in "finite" -> closing.

from synthetic-zariski.