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2

$\DeclareMathOperator{\Spec}{\operatorname{Spec}}$$\DeclareMathOperator{\codim}{codim}$$\Spec(\mathbb{Z})$ cannot be realized as $\Spec(A) \setminus \{m\}$ for a local ring $(A,m)$ - the argument given here can essentially be adapted verbatim. Set $X := \Spec(A)$, $Y := \{m\}$, $U := X \setminus Y \ne \emptyset$. Notice that $\dim U < \infty \iff \dim X ...


0

For your highlighted questions: a topological space has a nonempty proper subset which is both open and closed iff the space is disconnected. Thus in a (Noetherian) ring with no idempotents $\ne 0, 1$ (e.g. a local ring or a domain), no proper nonempty closed set is open. The reason for the confusion (both in your post and in the comments under Martin's ...


0

A good reference for your question is Ravi Vakil's notes on Foundations of Algebraic Geometry. There is a very nice chapter on Blowups. To answer your question, I would like to edit your statements a bit. Define $\phi: \mathbb{A}^n \rightarrow \mathbb{A}^s$, by mapping $(x_1, ..., x_n)$ to $(f_1(x), ... ,f_s(x))$, where $x = (x_1, ..., x_n).$ Under $\phi$ ...


3

No. Assume that $R$ is a field and $(A,\mathfrak{m})$ is a local ring such that $\mathrm{Spec}(R) \cong \mathrm{Spec}(A) \setminus \{\mathfrak{m}\}$. Then $A$ has exactly two prime ideals $\mathfrak{p} \subset \mathfrak{m}$ and $R \cong A_{\mathfrak{p}}$. The general fact $Q(A/\mathfrak{p}) \cong A_{\mathfrak{p}} / \mathfrak{p} A_\mathfrak{p}$ implies here ...


5

If $f : X \to Y$ is a continuous map and $\mathscr{F}$ is a sheaf on $X$, then $f_* \mathscr{F}$ is the sheaf on $Y$ defined by the following formula: $$f_* \mathscr{F} (V) = \mathscr{F} (f^{-1} V)$$ This is called the direct image sheaf. You should verify that the formula indeed defines a sheaf.


1

Assume that a root $w$ of your polynomial lies outside the disk $\|z\|\leq \frac{n}{n-1}$. Then: $$ \|w^n\| = \|w+1\|, \tag{1}$$ but, due to the following stronger version of the Bernoulli inequality, the LHS is greater than: $$\left(1+\frac{1}{n-1}\right)^n \geq e\left(1+\frac{1}{2n-1}\right),$$ while the RHS is at most $2+\frac{1}{n-1}$, so the equality in ...


5

For real $x>1+\frac{1}{n-1}$ then $$x^n = (1+(x-1))^n> 1+n(x-1) = x+(n-1)(x-1)> x+1$$ Now if $z^n+z+1=0$ then $\|z\|^n = \|z+1\| \leq \|z\|+1$. So, with $x=\|z\|$, we have $x^n\leq x+1$, so $\|z\|=x\leq 1+\frac 1{n-1}$.


2

Presumably, $\pi(z,\ell)=z$? It's not bijective, at least for $n>1$. $$\pi^{-1}(0)=\{(0,\ell)\mid\ell\in\mathbb CP^{n-1}\}$$ It is a bijection when excluding the $0\in \mathbb C^n$ and $\pi^{-1}(0)$ from the sets, because when $z\in\mathbb C^n\setminus\{0\}$, there is exactly one line $\ell$ through zero containing $z$.


4

This is indeed quite an interesting and subtle question! The point is that when $\mathcal{L}(-P)\subset \mathcal{L}$, every section $s$ of $\mathcal{L}(-P)$ is a section of $\mathcal{L}$ but its order of vanishing at $P$ seen in $\mathcal{L}(-P)$ is one less than its order of vanishing seen in $\mathcal{L}$. Proof: Since the problem is local, we may ...


1

A $2 \times 2$ symmetric matrix $M = \begin{bmatrix} a& b\\ b& c \end{bmatrix}$ defines a quadratic form by $f(x,y) = \begin{bmatrix} x & y \end{bmatrix} M \begin{bmatrix} x \\ y \end{bmatrix} = ax^2 + 2bxy + cy^2$. Thus a conic section (centered at the origin) can be regarded as a $2 \times 2$ symmetric matrix and vice-versa. An ellipse ...


3

For a smooth affine scheme $X$ of finite type over $\mathbb C$ the de Rham cohomology $H^*_{dR}(X)$ is just the cohomology of the complex of global differential forms: $$ H^*_{dR}(X)=H^*(\Gamma(X,\Omega^*_X)) $$ In your case you have to compute the cohomology of the complex $$0 \longrightarrow \mathbb{C}[x,x^{-1}] \stackrel {d}{\longrightarrow} ...


2

The idea underlying these questions is the structure group of the fiber bundle, and the issues raised here are related to the fact that the structure group is somehow specified by other structures and not determined by the topology of the bundle see this question for a general discussion of this. For both the Mobius band and the cylinder, it is possible to ...


2

Let me explain why the statement in the last paragraph of the question is false as written; it is true under the extra assumption that $h$ has connected fibres. A counterexample for the statement in the question is the following: let $h:C \rightarrow \mathbf P^1$ be a $2:1$ map from an elliptic curve. Let $L=O(2)$. Then $h^0(L)=3$, but $h^*(L)$ has ...


1

Since any abelian variety is isogenous to its dual, the answer is yes.


3

In the case of line bundles, your question is related to the behaviour of the Picard variety in families, and counterexamples can be found by looking for examples in which the Neron--Severi rank jumps. E.g. let $E_t$ be a one-parameter family of elliptic curves, chosen so that $E_0$ is CM, but $E_t$ is not CM for $t \neq 0$. More precisely, choose the ...


0

The basic idea is that if the circle $S^1$ acts on a space, then the orbit of any non-fixed point will be a copy of the circle, so the space decomposes as a union of circles, and the fixed points. Since circles have Euler char. $0$, the Euler char. will then be equal to just the Euler char. of the fixed point set. Here the intuition being used is that ...


2

Yes, you essentially understand the situation : here is the classification of the points of a scheme $X$ of finite type over a (completely arbitrary) field $k$. 0) The closed points are the points $x\in X$ whose residue field $\kappa(x)$ is a finite extension of $k$, i.e. $[\kappa(x):k]\lt \infty$ 1) The other points $y\in Y$ correspond bijectively ...


0

Question 1: Let us represent a point of $\mathbb{A}^n$ as the action of an $n$-tuple of regular functions on the point itself. These regular functions are just the coordinate functions on $\mathbb{A}^n$. So we have that $a=(a_1,\dots,a_n) = (y_1,\dots,y_n)(a_1,\dots,a_n)$. Alternatively, the point $a$ is the image of itself under the identity morphism ...


0

See [BB], in particular Corollary 2: Suppose $G=\mathbb{G}_m$ acts on an algebraic scheme $X.$ Then $\chi(X) = \chi (X^G).$ I'm not certain what is precisely meant by an algebraic scheme in that paper, but it probably includes your notion of a variety. Bialynicki-Birula's result is for any algebraically closed field. If you ever yearn for the same ...


-1

Define $g^*$ in parallel to the way you define $f^*$, where you take $g = f^{-1}$ (which exists since $f$ is an isomorphism). Thus their composition is the identity since $f \circ g = g \circ f = id$. As for dominance, we define the open sets $U$ and $V$ so that this is the case - i.e. define $U$ as above, then define $V$ as $f^*(U)\cap \cap_i V_i$ (where ...


0

As you write, the problem is to show that if $X$ is a variety of dimension at least two, then any ample divisor on $X$ is connected. This is usually proved via the Lemma of Enriques--Severi--Zariski, which is in turn is (following Serre) proved via Serre duality. The key trick is that if $\mathcal O(-D)$ is the ideal sheaf of $D$, then its $n$th tensor ...


1

Because $X$ and $Y$ are non-sigular, the conormal bundle to $Y$ in $X$ is a locally free sheaf on $Y$ of rank $n$. Choose an affine n.h. $U = $ Spec $A$ of $x$ in $X$ such that over the intersection $U \cap Y$ the conormal bundle becomes free. If we let $I$ be the ideal in $A$ cutting out $Y$ (so $U \cap Y =$ Spec $A/I$), then the conormal bundle is the ...


1

The case $ax^4-by^2\pm\ldots=0$ with $a,b>0$ is an open curve pretty much for the same reason that $ax^2-by^2\pm\ldots=0$ is an open curve as well, called hyperbola. Then all you have to do is compute $y_{_{1,2}}(x)=\dfrac{-(s+qx)\pm\sqrt{(s+qx)^2-4r(t+px)^4}}{2r}$ , the area being $\displaystyle\int_v^w|y_{_1}(x)-y_{_2}(x)|~dx=$ ...


1

I'd like to make some complement to Mohamed Hashi's answer. Let $X$ be the blow-up of $\mathbb{A}^2$ at $O=(0,0)$,that is $$X=\{((x,y),(u:t))\in \mathbb{A}^2 \times \mathbb{P}^1\mid xu=yt\}.$$ Let $\mathbb{A}^1_u,\mathbb{A}^1_t$ be the affine cover of $\mathbb{P}^1$,then $$\mathbb{A}^2 \times \mathbb{P}^1=(\mathbb{A}^2 \times\mathbb{A}^1_u) ...


0

Yes, geometric properties of objects can be captured in the sheaves of functions on them. For example, consider Castelnuovo's criterion, which gives a condition for when a curve on a smooth algebraic surface to be "contracted" to a point, to produce a new smooth surface. The conditions are purely geometric, but if you look at the proof given in Ch. V of ...


1

Vakil's question is about the fibres at various points, not the stalk. The fibre at points away from the cusp is one-dimensional, while at the cusp, the fibre is two-dimensional. This shows that the module is not free (it has torsion supported at the cusp, as Martin Brandenburg remarks). You are suggesting that you should localize at $(X,Y)$, but to ...


1

By definition of fiber bundle, there exists $\theta_{U,V}:U\cap V\to K$ such that $\phi_V(b,k)=\phi_U(b,k\cdot\theta_{U,V}(b))$ on $p^{-1}(U\cap V)$. Evaluating this equality on $(p(x),h_1)$, we get: $$\phi_U(p(x),h)=x=\phi_V(p(x),h_1)=\phi_U(p(x),h_1\cdot\theta(p(x)))$$ This means that $\theta_{U,V}(p(x))=h_1^{-1}\cdot h$. In particular, ...


2

Torsion means torsion over $\mathcal O_X$. For a coherent sheaf on a curve, it is equivalent to having support a finite number of closed points.


0

$\Omega_{A}$ is the free $A$-module generated by the differentials $dX, dY$ subject to the relation $2Y dY = 3X^2 dX$. Since $\Omega$ commutes with localization, it follows that $\Omega_{A_{(X,Y)}}$ is the free $A_{(X,Y)}$-module generated by $dX, dY$ subject to the relation $2Y dY = 3X^2 dX$. You seem to claim that this is free of rank $2$. This is wrong. ...


3

A morphism of affine varieties is invertible if and only if it the associated map of affine coordinate rings is an isomorphism. In your case, this says that $F: k^n \rightarrow k^n$ is invertible if and only if the map $F^*: k[x_1,\dots,x_n] \rightarrow k[x_1,\dots,x_n]$ given by $g(x_1,\dots,x_n) \mapsto g(F_1,\dots,F_n)$ is a ring isomorphism. In ...


0

I think I have an answer. Please let me know if it isn't right. Suppose $X $ is a variety of positive dimension. Since $X $ can be viewed as a subset of $\mathbb{P}^{n}$, we get $|X|\le |k|$. $X$ contains an open affine subset $U$. $U$ has a surjective regular function (for example: projection on first coordinate). Hence $|X|\ge |k|$. Does this make ...


2

I dont think you have defined an action yet. If we take $g \in K$ then we use the map $\varphi$ to define the action where if $(u,h) \in U\times K$ then $$g(u,h)=(u,gh)$$ now you need to use the fact that $K$ is the structural group to show this this is well defined. The orbit space is $B$ since $K$ acts transitively. If you have a section, set the ...


2

This follows precisely from the fact that $\Omega_{A/k}^1$ is locally free of rank $d$ (by smoothness). So, locally $\wedge^{d+1}\Omega_{A^k}^1$ is locally zero, and so globally zero.


0

If you have two distinct points $P$ and $Q$, then the parametric line $s(t) = (1-t)P + tQ$ passes through them. Pick $n+1$ points in some plane and one point outside the plane. You want to send these to $n+1$ points of "the standard plane" and one more point. That's a system of $n+2$ equations in the unknown entries of the matrix, but the "scale factors" ...


6

The cohomology ring of $G(k,n)$ is isomorphic to $$\frac{\mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$ where $T$ and $Q$ are respectively the tautological and the quotient bundle. The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas: $c_i(T) = ...


1

Given $u \in U$ we have defined $\theta(u)$ as a map that takes $y \in F$ to $\mu \langle u,y \rangle \in F$. Thus $\theta(u): F \to F$ certainly belongs to $\operatorname{Maps}(F,F)$, and $\theta : U \to \operatorname{Maps}(F,F)$. What remains to be seen is that the image $\theta(u)$ is in fact a homeomorphism living in $K \subset \operatorname{Maps}(F,F)$, ...


3

You can interpret Klein's point of view on geometry very, very broadly as being the following: geometry is dictated by the morphisms you choose between geometric objects. In the usual interpretation "morphism" is taken to mean "isomorphism," but from the point of view of category theory there's no reason to make this restriction. In other words, geometry is ...


1

Is it true for compact simply-connected oriented smooth 4-manifolds that a real 4-dimensional vector bundle is determined by its Stiefel-Whitney and Pontrjagin classes? No. For example, the tangent bundle of $S^4$ is stably trivial (because the outward pointing normal gives a trivialization of its normal bundle in $\mathbb{R}^5$; true more generally of ...


4

Yes. By the the explicit construction of the fiber product of locally ringed spaces (in particular, of schemes), it follows that the continuous map $|X \times_Z Y|\to |X| \times_{|Z|} |Y|$ is surjective and that the fiber over some point $(x,y,z)$ in $|X| \times_{|Z|} |Y|$ is $\mathrm{Spec}(\kappa(x) \otimes_{\kappa(z)} \kappa(y))$. A basis of the ...


0

Yes to both questions, but I'm not sure what you mean by wanting a global proof. Reducedness is a local property! The proof will consist of picking a point $x \in X$ and an affine chart $U = \text{Spec}(R)$ containing $x$, then checking reducedness on that chart. In particular, in the affine case, if $R$ is reduced, then all its localizations are reduced. ...


3

The ideal $f(I)A$ is the ideal of the scheme-theoretic fiber $\text{Spec}(A) \times_f \text{Spec}(B/I)$, that is, the restriction of the morphism $\text{Spec}(A) \to \text{Spec}(B)$ to the subscheme $\text{Spec}(B/I)$. So algebraically, $f(I)A$ is the kernel of the map of $B$-algebras $$A \to A \otimes_B B/I \cong A/f(I)A.$$ So it is the (ideal of the) ...


5

In short: "varieties are nice schemes over fields". In particular, the study of varieties is a subscase of the study of schemes. A variety (depending on who you ask) over a field $k$ is a finite type separated (usually reduced, sometimes geometrically integral!) scheme over $\text{spec}(k)$. The reason that the two fields look so different, is purely in ...


2

Here is a reference: on page 136 of the book of Görtz-Wedhorn, Proposition 5.51 (ii) says that a $k$-scheme $X$ is geometrically integral over $k$ if and only if for every integral $k$-scheme $Y$, $X\times_kY$ is integral. This doesn't quite give you what you want immediately, in that there is still an argument to be made, but this is the key ingredient. The ...


3

Well, since ramification is a purely local phenomenon (since you compute it in the stalks of the respective schemes), it suffices to consider the affine analogue of your question. Namely, let $f(x,y)$ be the affinization of your curve $C$ (which, without loss of generality, I assume is monic in $y$) and replace $\mathbb{P}^1$ with $\mathbb{A}^1$. You then ...


5

Yes. It doesn't matter which definition you choose, smoothness of $\emptyset \to X$ is satisfied for trivial reasons. More generally, any open immersion is smooth.


1

Then he defines the betti number $\beta_{i,j}$ of the free module $F_i$ as the minimal generator of degree j that $F_i$ requires. No, he defines the betti number to be the number of minimal generators of degree $j$ required to generate $F_i$. The module $F_1$ is just a direct sum of three copies of $S$, so it is generated by $(1,0,0), (0,1,0), ...


3

The morphism $\mathrm{Spec}(k(x))\to X$ need not be locally of finite type, so it in general has no chance to be smooth. For example, take $X=\mathbf{A}_k^1$ for a field $k$ and $x$ the generic point. The morphism corresponds to the inclusion $k[T]\hookrightarrow k(T)$, and this is not of finite type, hence not smooth. At least if $X$ is a $k$-scheme for ...


3

What is a strict transform? Let $X\subset \mathbb{A}^2$ be any affine variety containing the origin. Then $\varphi^{-1}(X)$ consists of two irreducible components. Namely $\tilde{X}$, the closure of $\varphi^{-1}(X-O)$, and $\varphi^{-1}(O)\cong \mathbb{P}^{1}$. The first is called the strict transform and the second the exceptional curve. So the exceptional ...


1

$u$ and $t$ represent the slope of the tangent. In the first case $u=\frac{y}{x}$ assuming $x\neq 0$. so dividing the original equation by $x^2$ we get $u^2=x+1$. This should answer question $1$. At the origin there are two tangents, given by $x+y=0$ and $x-y=0$. The intersection of $\tilde{Y}$ at $(0,0,\pm 1)$ gives $x=y=0$ so $u^2=1$ and $u=\pm 1$ the ...



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