3
votes
Accepted
$\operatorname{Spec}(R/I)\to \operatorname{Spec}(R)$ is closed immersion of schemes: why is the kernel locally generated by sections?
Well, that kernel is globally generated by sections: the kernel is $\widetilde{I}$, and you can find a surjective morphism from a free module by taking a surjective morphism $R^{\oplus S}\to I$ ...
- 69.1k
3
votes
Fibered product of schemes locally of finite type over a field is smooth if and only if its factors are smooth.
I like your idea of reducing to the case that $k$ is algebraically closed.
We can also reduce further to assume $z$ a closed point. Indeed, $X \times_k Y(k) \to X(k)$ is surjective, and to show $X$ is ...
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2
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Gortz, Wedhorn, Algebraic Geometry, Proof of Proposition 3.52 - why is there at most one reduced subscheme structure?
Suppose $Z_1,Z_2$ are two subscheme structures on $Z\subset X$. Let $Z_0=Z_1\times_X Z_2$, which is a common closed subscheme of $Z_1$ and $Z_2$ with the same underlying topological space (proof: ...
- 69.1k
2
votes
Accepted
On the openness of the map $\mathbb A^{n+1}_k\smallsetminus 0\rightarrow \mathbb P^n_k$.
Your idea to reduce to the situation over a standard affine open and prove this for the case of $D(g)$ is good. Let's continue from there. We observe that a point $p$ in $\Bbb A^n$ is not in the image ...
- 69.1k
2
votes
Accepted
if $Z\rightarrow X$ is closed immersion then $f:Y\rightarrow X$ factors through X iff $f^{\ast}\mathscr I\rightarrow f^{\ast}\mathscr O_X$ is zero
if $f$ factors as $Y\rightarrow Z\rightarrow X$ then the map $f^* \mathscr I\rightarrow \mathscr O_Y$ is zero.
Your logic is not correct. The pullback of $\mathcal{I}$ is not zero. For instance, ...
- 69.1k
1
vote
Accepted
Sheafification of function presheaves
I would start from the sheafification side. The idea is that you want to localize the information in your presheaf $\mathcal{F}$. Formally that means that the sheafification $\mathcal{F}^+$ is the ...
- 303
1
vote
Accepted
Help with R. Pink complete lecture notes of "Finite group schemes". Theorem 12.2
I will try clearing up the terminology you are unfamiliar with.
A Galois representation is a group representation of a Galois group. Group representation, in their naive form, are just group ...
- 16.3k
1
vote
Accepted
rational section non vanishing everywhere at any irreduicble component implies regular?
By working locally, we may assume $\mathcal{L}$ is $\mathcal{O}_X$ and that $X$ is affine. If $s$ were not regular, it would have to be contained in an associated prime of $X$ as the union of the ...
- 406
1
vote
Accepted
Questions about Corollary 5.23 of AG1: Schemes by Görtz & Wedhorn
Let $U \subset X$ be an open noetherian neighbourhood of $x$. Clearly every irreducible component of $X$ which contains $x$ intersects $U$. But as $U$ is noetherian, it has only finitely many ...
- 9,852
1
vote
Accepted
Example of a closed subscheme of $\operatorname{Proj} A$ which is not of the form $\operatorname{Proj}A/I$.
$\newcommand{\Spec}{\operatorname{Spec}}\newcommand{\Proj}{\operatorname{Proj}}\newcommand{\P}{\mathbb P}\newcommand{\V}{\mathbb V}$
For posterity, I answer this question with the one answered over at ...
- 4,029
1
vote
Accepted
Quasicompactness of $\operatorname{Proj}A$
I have figured this out.
By hypothesis we have that:
$$\sqrt{A_+}=\sqrt{\langle f_1,\dots, f_n\rangle}$$
hence:
$$V(A_+)=\bigcap_{i=1}^nV(f_i)$$
By taking compliments, it follows that:
$$\operatorname{...
- 4,029
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