Closed felixwellen closed 6 months ago
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 $1$.
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).
Nice - thanks for refusing to see the problem non-existent problem ;-)
written down in "finite" -> closing.
Hugo suggested to me, it might be good to call an affine scheme $\mathrm{Spec} A$ finite, if A is a finite free R-module. So if $A=R^n$ is given, we can directly construct an embedding into $\mathrm{Spec}A\to \mathbb{P}^{n-1}$, by sending a homomorphism $R^n\to R$ to the vector given by the images of a standard basis $(e_i)_i$. The proposition $C$, to be in this subtype, is closed: For $[v]:\mathbb{P}^n$ let $\lambda_v=\sum \lambda_i v_i$, for $\lambda_i$ such that $\sum \lambda_i e_i=1$. Let $\varphi_v$ be the linear map sending $e_i$ to $v_i$. Then the conjunction of all $\lambda_v\cdot \varphi_v(e_i\cdot e_j)=\varphi_v(e_i)\cdot \varphi_v(e_j)$ is well-defined and closed and cuts out the $[v]$ such that $\varphi_v$ is an algebra homomorphism up to normalization.