New answers tagged

1

In general the map might not be either open nor closed. For example, for $p \in Spec(R)$, $Spec(R_p)$ is the set of all prime ideals that are contained in $p$. So $(2) \in Spec(\mathbb{Z})$ is the set $\{(0), (2)\}$ which is neither open nor closed. if $f : X \rightarrow Y$ is a continuous bijection between topological spaces, then $f$ is a homeomorphism ...


0

Fellows of mine just answered my question, so I'm writing this answer to close this question. Since prevariety is irreducible, being both connected and a union of finite many irreducible spaces, any two (non-empty) open sets in the prevariety intersect. Therefore $U$ intersect $U_i$ for any open affine subvariety $U_i$. The proof of Proposition 1.54 tells ...


2

The first statement, i.e. that the codimension of the singular locus is at least two on a normal variety, is correct. It might be the case that someone said "the singular locus is codimension two" with an implied "or more" at the end - I know sometimes when I'm talking fast in person I can sometimes drop statements like that without ...


1

In case $I$ is a radical ideal which isn't the irrelevant ideal we can prove this as follows: The condition implies that for each $i$, $$x_i^kf=g_i$$ for some $k\in \mathbb{Z}$. If $k\leq 0$ for some $i$ we are done. If not then for any minimal prime ideal $p$ over $I$, take $i$ s.t. $x_i\notin p$. Then $x_i^kf=g_i\in p$ implies $f\in p$. So $f$ is in ...


2

The text says If $L(P_0)$ were generated by global sections, then $P_0$ would be linearly equivalent to some other point $Q \in X$. This goes as follows: Suppose $L(P_0)$ is generated by global sections. Then there is a global section $s \in \Gamma(X, L(P_0))$ which generated $L(P_0)$ at the point $P_0$. The effective Cartier divisor of zeroes $$D = (s)_0 =...


2

I do not believe the statement is true. Take for example $f=x_0$ and for $0\leq i\leq n$ let $g_i=x_0x_i$ and define $I=(g_0,\ldots ,g_n)$. Obviously $f\notin I$ since all the generators of $I$ are homogeneous of degree $2$ but $f$ is homogeneous of degree $1$. On the other hand, $$f(x_0,\ldots,x_{i-1},1,x_{i+1},\ldots,x_n)=g_i(x_0,\ldots,x_{i-1},1,x_{i+1},\...


1

If $\mathrm{Hom}(R/I,M)=0$, then there is no nonzero $a \in M$ whose vanishing ideal contains $I$. In other words, the (finite) set of associated primes of $M$ contains no prime ideal containing $I$. By the prime avoidance lemma, the reunion of the associated primes of $M$ does not contain $I$. But the reunion of the associated primes of $M$ is exactly the ...


0

Well every such ideal $I$ will be contained in a maximal ideal $\mathfrak{m}$ so that $A/\mathfrak{m}\simeq k$. Now take $a_i\in k$ such that $x_i$ corresponds to $a_i$ under this isomorphism. So $\mathfrak{m}=(x_1-a_2,\dots, x_n-a_n).$ Now $I \subset \mathfrak{m}=(x_1-a_1,\dots, x_n-a_n)$ says that for all $f\in I$, $f(a_1,\dots, a_n)=0$.


3

$\mathbb{P}(E) = \mathrm{Proj}\left( \bigoplus_{k=0}^\infty \mathrm{Sym}^kE \right)$.


0

After a while of thinking, I figured this out: $f$ is unramified, so $f^{-1}(P_0)$ is two points. Because $f$ is also a group homomorphism, we must have that one of those points is $P_0$, and the other point is $P_1$, a 2-torsion point. Letting $P_2$ and $P_3$ be the other 2-torsion points, we must have that $f(P_2)=f(P_3)$, since $P_2+P_1=P_3$, so we can ...


1

Your definition of $D$ as a sum of points is what happens in the case of curves. The general definition of a (Weil) divisor is a formal linear combination of codimension-one subvarieties, and Harris' quotation is how one would adapt this to the analytic setting.


2

An immersion of (locally Noetherian) schemes is flat if and only if it is an open immersion, or a closed immersion of connected components. Open immersions are flat since the induced maps on stalks are isomorphisms, as pointed out by Zhen Lin in the comments. Suppose $\phi\colon Z\to X$ is a closed immersion and assume it is flat. The question is local so we ...


3

The special fiber of a group scheme is always a group scheme. One way to think about this is follows. Fact (Yoneda's lemma): Let $S$ be a scheme and $G$ an $S$-scheme. Then, the map $$(m,i,e)\mapsto (R\mapsto (m(R),i(R),e(R))$$ is a bijection $$\left\{\begin{matrix}\text{group }S-\text{scheme}\\ \text{structures }(m,i,e)\end{matrix}\right\}\to \left\{\...


5

The definition of this sheaf can be found on p.68 of the book. In particular, for an open set $V\not\subseteq U$ by definition $j_!\mathcal{O}_U(V)=0.$ Hence, the global sections over your $V=\mathrm{spec} (A)$ are zero because $V$ is not contained in $U$. Edit: As red_trumpet mentions below, this answer is probably not quite right as written. Instead, we ...


1

First, as hm2020 notes in their comment, the canonical map $R\to \prod_{i = 1}^r R[g_i^{-1}]$ can be viewed as the composition $$ R\to\prod_{i = 1}^r R[X]\to\prod_{i = 1}^r R[X]/(Xg_i - 1). $$ This is because the localization map $R\to R[g_i^{-1}]$ is identified with the composition $R\to R[X]\to R[X]/(Xg_i - 1)\cong R[g_i^{-1}]$. Now, let $R$ and $I = (g_1,...


1

No, there's no need to suppose that $k$ is algebraically closed. In fact, if $\operatorname{char} k \neq 2,3$ we don't need any assumptions at all to reach the conclusions of this problem: in this situation, any elliptic curve can be put in short Weierstrass form with an (affine) equation of the form $y^2=x^3+ax+b$. This has a singularity iff the equations $...


0

I think you don't understand the argument, let me try to repeat it, he start with an effective divisor $D$ such that $L=O(D)$, having this $D$ is equivalent to having a global section $s\in \Gamma(L)$, then he prove that if $D$ has no multiplicity, $D$ should be a pullback of a divisor $D_1$, so the $L$ should be the pulback of $L'=O(D_1)$, and the section $...


2

You have that $\varepsilon$ is a Neron model of $E/K$. This means that $\varepsilon_K = E$ as schemes over $\operatorname{Spec} K$ (Note that $\varepsilon_K$ is the generic fiber of $\varepsilon \to \operatorname{Spec} R$). Therefore, by base change, an $R$-valued point of $\varepsilon$ gives a morphism $$\operatorname{Spec} K = (\operatorname{Spec} R)_K \to ...


1

$\operatorname{Spec}$ is a contravariant equivalence of categories between the categories of commutative rings and affine schemes. In your situation, this means that $K$-morphisms $$\operatorname{Spec} K\to X=\operatorname{Spec} K[x_1,\cdots,x_n]/(f_1,\cdots,f_r)$$ are in bijection with $K$-algebra homomorphisms $K[x_1,\cdots,x_n]/(f_1,\cdots,f_r)\to K$. You ...


1

As described in the comments, taking $\operatorname{Spec}(R[T,T^{-1}])$ will not work. Instead you have to do a glueing process, which is described in the reference you mentioned. Let me recall the construction, which should also address your point about being locally of finite type, but not of finite type (Recall finite type=locally of finite type + quasi-...


1

I suggest reading Guillemin, Victor; Pollack, Alan, Differential topology, Providence, RI: AMS Chelsea Publishing (ISBN 978-0-8218-5193-7/hbk). xviii, 222 p. (2010). ZBL1420.57001. The relevant result is that if $X, Y$ are smooth compact manifolds, $f: X\to Y$ is a smooth map and $B\subset Y$ is a smooth (compact) submanifold with boundary such that $f$ is ...


1

In the affine line $\mathbb{A}^1_k=\operatorname{Spec} k[t]$, the closure of the set $M$ will be $\mathbb{A}^1_k$ in the Zariski topology. Note the closed subsets in $\mathbb{A}^1_k$ are given by $V(\mathfrak a)$ for an ideal $\mathfrak a$ in $k[t]$. $k[t]$ being PID, any such ideal $\mathfrak a$ will be principal. So $\mathfrak a=(f)$ for some polynomial $...


2

Hartshorne, Proposition IV.5.2 has an answer, and I guess this is what Mohan hinted at in the comments: Proposition 5.2 Let $X$ be a curve of genus $g \geq 2$. Then $|K|$ is very ample if and only if $X$ is not hyperelliptic. As by Exercise IV 1.7. there are hyperelliptic curves of any genus $g$, the very-ampleness of $K$ is independent of the genus (save ...


1

Just to expand my comment that being flasque is local. If $\mathcal F$ is a locally flasque sheaf over (a topological space) $X$, let $U\subset X$ be an open subset, and $a\in\mathcal F(U)$. We want to show that $a$ can be extended to be a glocal section over $X$. By Zorn's lemma, we may pick a maximal extension $\tilde{a}$ of $a$. If the domain $V$ of $a$ ...


3

Tools from semi-algebraic and subanalytic geometry can help us solve this problem. We may obtain a locally finite triangulation of $U\cup V$ which has as sub-triangulations $U$, $V$, and $U\cap V$. Combining this with Mayer-Vietoris for simplicial complexes and simplicial homology plus the equivalence between simplicial homology and singular homology of the ...


1

If there is a nonzero $f \in L(D)$, then the sum of $D$ and the divisor of $f$ has nonnegative entries and degree $0$, so it must be zero. Thus $D$ is principal (it’s the divisor of $1/f$). The reverse implication is easy.


2

It is not surjective in general. Indeed, the associated long exact sequence looks like $$ 0 \to H^0(Z,\mathcal{N}_{Z/X}) \to \mathrm{Ext}^1(i_*\mathcal{O}_Z,i_*\mathcal{O}_Z) \to H^1(Z,\mathcal{O}_Z) \to \mathrm{Ext}^1(\mathcal{I}_Z,i_*\mathcal{O}_Z) \to \dots $$ Now assume, for instance, $Z$ is a plane cubic curve. Then $\mathcal{I}_Z \cong \mathcal{O}(-...


1

First, one should assume that $l$ is not contained in $f$. Next, view $l$ as a copy of $\Bbb P^1$: then the restriction of $f$ to $\Bbb P^1$ is a homogeneous degree-$d$ polynomial of two variables. Over an algebraically closed field, such a polynomial decomposes in to a product of $d$ homogeneous linear factors. These are the roots, and the fact that some of ...


2

Question 1. It looks like you've understood correctly how to compute dimensions of hypersurfaces (i.e. varieties defined by a single equation). Here are some clarifying remarks. The height of a prime ideal is the supremum of the lengths of strictly ascending chains of prime ideals contained in $\mathfrak{p}$. Geometrically, this is the codimension of the ...


3

Let $K=\operatorname{Frac}R$ and $k=R/p$. There are natural homomorphisms $R\to K$ and $R\to k$. These induce morphisms on spectra in the usual way. In other words, $\operatorname{Spec} K$ and $\operatorname{Spec} k$ are $R$-schemes. Now, if $X$ is an $R$-scheme, the generic fiber is defined to be $$X\times_{\operatorname{Spec}R} \operatorname{Spec} K,$$ and ...


1

The statement is false. The reason is that the morphisms need not be isomorphisms onto their image, even though they are injective. That is, they need not be closed immersions. Here's an example of when it fails. Let $Y=\mathbb{A}^2_\mathbb{C}$. Then if $X_1=V(y^2-x^3)\subset Y$ and $X_2\simeq \mathbb{A}^1_\mathbb{C}$ is the normalisation of $X_1$, the ...


0

I think the statement has a counterexample. Suppose $A = \mathbb{F}_2$ and $B=A^3$. Then $C = B$, which cannot be expressed as a quotient of $A[X]$ as $A[X]$ only has two $A$-valued points. If $B$ is local, then the statement is true. In this case, $k'(\subseteq B\otimes_A k)$ is a field because $B\otimes_A k$ is local. $k'$ is finite separable over $k$ so ...


2

To get an action of $\mathrm{PGL}(n)$, you should twist the $\mathrm{GL}(n)$ action by (a power of) the determinant. For instance, in the example with the action of $\mathrm{PGL}(2)$ on $\mathrm{Sym}^2(\Bbbk^2)$, you should divide your 3-by-3 matrix by $ad - bc$ (and by the way, I think the middle row of your matrix should be divided by 2).


2

Well even classically it would be rather silly if the set of $K$-points was a variety. For example the elliptic curves 11a1 and 19a2 both have trivial Mordell-Weil group, but are clearly not isomorphic as varieties (e.g., they don't have the same conductor).


5

A scheme is a locally ringed space which is locally isomorphic as a locally ringed space to $\operatorname{Spec} A$. So the first reason that $E(K)$ is not a scheme is because you haven't provided a structure sheaf. The second and more important reason is that (assuming $K$ is algebraically closed) $E(K)$ is not the underlying topological space of a scheme ...


2

As I mentioned in the comments, Hartshorne's immersions are open immersions followed by closed immersions, which is different from the immersions of EGA, which are closed immersions followed by open immersions. The scenario you have here is that you have an immersion in the sense of EGA and would like to show that it's an immersion in the sense of Hartshorne....


3

$A$ finitely generated as a $\Bbb Z$-algebra exactly means the structure morphism $\operatorname{Spec} A\to\operatorname{Spec}\Bbb Z$ is of finite type by definition. Since $\Bbb Z$ is regular, proposition 14.103 applies. To check this, simply verify that every localization of $\Bbb Z$ at a prime ideal is a regular local ring: you just have to check $\Bbb ...


2

The point is that $n-1$ (general) homogeneous polynomials in $n+1$ variables will not intersect to give some points but rather to give a projective curve. There is a lot you can say about this curve (it's a complete intersection in projective space) for example you can compute its genus (see the adjunction formula) given the degrees of the polynomials but ...


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