Let $f\colon X\to Y$ be a finite $r$-to-one morphism between smooth projective varieties.  How does one show that $f_\ast\mathcal{O}_X$ is a locally free sheaf of rank $r$ on $Y$ ?  Any insights for $f_\ast\mathcal{O}_X$ would be very much appreciated. A local section over $U\subset Y$ for $f_\ast\mathcal{O}_X$ is just a regular function over $f^{-1}(U)$, so it seems as if one can write this uniquely as a sum of $r$ "pieces"...?  An aside second question: is there a naturally defined map $f^{\ast}f_\ast\mathcal{O}_X(D)\to\mathcal{O}_X(D)$ where $D$ is a divisor on $X$ ?
122766
Let $f: X \to Y$ be a covering (proper, surjective, finite regular map) of smooth projective varieties of degree $d$. How one can show that in this case $f_* \mathcal{O}_X$ is a locally free sheaf of rank $d$?
111920
In exercise 1.8 of chap I in Hartshorne algebraic geometry,       Let $Y$ be an affine variety of dimension $r$ in $\mathbf A^n$. Let $H$ be a hypersurface in $\mathbf A^n$, and assume that $Y \nsubseteq H$. Then every irreducible component of $Y \cap H$ has dimension $r-1$.     I refered to a . In this solution, why $f$ is not a unit in $B$?
103764
Let $k$ be a local field, and consider the map $\phi: k^+ \hookrightarrow \widehat {k^+}$ given by $\eta \mapsto \eta X(\cdot)=X(\eta \cdot)$ where $X$ is a non-trivial character.  Tate argues in his thesis that $\phi$ has dense image simply because we have the implication $X(ηξ)=1 ∀η⟹ξ=0$. However, I am not sure how this shows that we can multiply $X$ by some $\eta$ to get arbitrarily close of another given $X'$. For that we would have to show that for any finite collection of compact subsets $B_{1,...n}$ in $k^+$ and open sets $U_{1...n}\subset S^1$ such that $X'(B_i)\subset U_i$ we can find $\eta$ such that $X(\eta B_i) \subset U_i$ as well.
114160
I'm trying to do the following exercise from the book An Invitation to Algebraic Geometry:     Show that if $X \to Y$ is a surjective morphism of affine algebraic varieties, then the dimension of $X$ is at least as large as the dimension of $Y$.   Could anyone give me a hint? I'm looking for a proof that avoids commutative algebra since it is not mentionned in the book at this point.
40881
Let $X,Y$ be varieties over $\mathbb{C}$, $k(X), K(Y)$ be function fields of $X, Y$. Suppose $\pi: X \to Y$ is a dominant,  $\textit{injective}\ $ morphism, why the degree of the function field extension $[K(X) : K(Y)] =1$?  If $\phi : X \to  Y$ is a finite morphism, then the fibre is finite, and by semicontinuity theorem, let $n$ be the number of the points in the generic fibre, then I feel one should similarly have $[K(X) : K(Y)] =n$. But I don't know how to show that. Any suggestions or reference on this question?  $\textbf{Edit}$: I really want the morphism $\phi$ to be a morphism between locally finite type and finite morphism. To be precise, for any affine open set $U=\rm{Spec}(B) \subset Y$, there is an affine open over of $\pi^{-1}(U)$, such that each $\rm{Spec}(A_i)$ in this cover has the property $A_i$ is a finitely generated $B-$module. I don't know the corresponding definition of this sort of morphism, or is it just a finite morpism?
59015
Let $X$ be a smooth projective variety over $\mathbb{C}$, then the Neron-Severi group $NS(X)$ of $X$ is defined to be the Picard group of $X$ modulo algebraically equivalent relations.   On the other hand, by the exponential sequence, there is a first Chern class map  $$c_1: {\rm Pic}(X) \to H^2(X, \mathbb{Z}).$$ It is claimed that the image of $c_1$ coincides with $NS(X)$.   I want to know why this is true. Any suggestion or reference is greatly welcome!
61251
Let $n\in\mathbf{N}$ be fixed, and $f$ entire and $|f^{-1}(\left\lbrace w\right\rbrace)|\leq n$ for every $w\in\mathbf{C}$. Then $f$ is a polynomial of degree at most $n$.    I try to prove this statement, and I think one can prove it as follows: consider $f(1/z)$. $0$ cannot be an essential singularity of $f(1/z)$, for the big Picard theorem would imply that on any neighborhood of $0$ $f(1/z)$ takes on all possible complex values (with at most one exception) infinitely often, but this is contrary to $|f^{-1}(\left\lbrace w\right\rbrace)|\leq n$. Then $f(1/z)$ has a pole of order $k$ say, and since $f$ is holomorphic it is a polynomial of degree $k$ (for its principal part vanishes). By the fundamental theorem of algebra, $k\leq n$.      Is it possible here to avoid using Picard's theorem?
81501
In Dummit and Foote, Exercise 15.18 (p. 727 3rd edition) one is asked to prove that for a commutative ring with unity $R$, and $f$ a non-nilpotent element of $R$, that the localization $R_f$ is isomorphic to $R[x]/(xf - 1)$.  I have done this, using the strategy outlined in the problem, using the universal property of $R_f$ proved in Theorem 36. The problem I have is this:  If $R_f \cong R[x]/(xf - 1)$ and $\phi:R[x] \to R_f$ is a surjective homomorphism, it stands to reason that $\text{ker}(\phi)$ should be $(xf - 1)$. I cannot show this directly. The surjective homomorphism I have in mind is, for:  $p(x) = a_0 + a_1x + \cdots + a_mx^m$  $\phi(p) = \dfrac{a_0}{1} + \dfrac{a_1}{f} + \cdots + \dfrac{a_m}{f^m}$  It is clear to me that $\phi$ annihilates $(xf - 1)$, so that ideal is contained in the kernel. What I cannot show is the other containment, or equivalently:  If $\phi(g) = 0_{R_f}$ then $g(x) \in (xf - 1)$. The trouble I am running into is that $R$ might be a "bad" ring, containing a lot of zero-divisors, $f$ might even be a zero-divisor (being non-nilpotent is a considerably weaker property), so the usual polynomial tricks of the division algorithm or using the irreducibility of $xf - 1$ aren't much use, here.  I feel I am missing something obvious....what is it?
27782
I can't understand the proof of theorem 8.17 from Hartshorne's "Algebraic Geometry". Namely, he says that we have an exact sequence  $$ 0 \to \mathcal J'/\mathcal J'^2 \to \Omega_{X/k} \otimes \mathcal O_{Y'} \to \Omega_{Y'/k} \to 0  $$ and there are $x_1, \dots,x_r$ such that $dx_1, \dots, dx_r$ generate a free subsheaf of rank $r$ in a neighborhood of $y \in Y$. It follows that $\Omega_{Y'/k}$ is locally free. Why?
14371
This question is similar to      and    except that I am concerned with smooth manifolds. If $M$ is a smooth manifold and $V$ and $W$ are (smooth, real) vector bundles on $M$, are there reasonable conditions that ensure that the map $\Gamma(M,V)\otimes_{\mathscr{O}_M(M)}\Gamma(M,W)\rightarrow\Gamma(M,V\otimes W)$ is an isomorphism? Equivalently, are there any reasonable conditions under which taking global sections of finite locally free $\mathscr{O}_M$-modules commutes with tensor product?  I don't really know much smooth manifold theory, and my intuition is mostly derived from algebraic geometry, where the analogous question, as is indicated in the questions I linked to, is not a simple one, even for invertible sheaves. I'd always assumed the situation for smooth manifolds would be the same, but I was just looking at the Wikipedia page on vector-valued differential forms, and it seems (unless I'm misunderstanding which is certainly possible) to claim there that for $V$ an arbitrary vector bundle and $W$ the $p$-th exterior power of the cotangent bundle, $p\geq 1$, the answer to my question is "yes," with the justification being that "$\Gamma(M,-)$ is a monoidal functor." Looking at the definition of a monoidal functor, it seems to me that all this implies is that there is a map from the tensor product of the global sections to the global sections of the tensor product (which is clear from the definitions) and not that this map is an isomorphism.
36387
Let $S$ be a scheme and let $f:X\to Y$ be a morphism of separated $S$-schemes.  shows that, when $X$ is reduced, the obvious morphism $g_f:\Gamma_f \to X\times_S Y\to X$ is an isomorphism, where $\Gamma_f$ is the scheme-theoretic image of $\Delta_f:X\to X\times_s Y$.     Can we show that $g_f$ without assuming that $X$ is reduced?   I see how the reduced induced structure is being used there, but I do not see if $X$ being reduced affects the statament, $X$ cannot have double/triple/... points if $\Gamma_f$ does not! So, I wonder if there is a probably more elaborate proof that shows $g_f$ to be an isomorphism.   The only assumption I have is that $S$ is Noetherian.   A hint or a reference would be appreciated.
8621
Let $k$ be a field, n a positive integer.  Vakil's notes, 17.4.B: Show that all the automorphisms of the projective scheme $P_k^n$ correspond to $(n+1)\times(n+1)$ invertible matrices over k, modulo scalars.   His hint is to show that $f^\star \mathcal{O}(1) \cong \mathcal{O}(1).$ (f is the automorphism. I don't if $\mathcal{O}(1)$ is the conventional notation; if it's unclear, it's an invertible sheaf over $P_k^n$) I can show what he wants assuming this, but can someone help me find a clean way to show this?
20416
Let $X$ be a topological space containing two closed points $x,y$ and let $i : \{x,y\} \to X$ denote the inclusion map. Notice that $\{x,y\}$ carries the discrete topology. Let $F$ be a sheaf on $X$. Then $i^{-1} F$ is a presheaf on $\{x,y\}$ which is given by $(i^{-1} F)(\emptyset)=1$ (the terminal set), $(i^{-1} F)(\{x\}) = F_x$ (the stalk at $x$), $(i^{-1} F)(\{y\})=F_y$ (the stalk at $y$) and $$(i^{-1} F)(\{x,y\}) = \varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U).$$ Why is, $i^{-1} F$ a sheaf if and only if the canonical map $$\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y.$$ is an isomorphism ?.   Thanks a lot to all of you.
100013
I have a problem when I am proving the following question:  Let $f:A \to A'$ be a surjective homomorphism of rings, and assume that $A$ is local, $A' \neq 0$, show that $A'$ is local.  I have found an answer in . However in my proof, suppose that $\mathfrak{m}$ is the unique maximal ideal in $A$, and I have proved that $f(\mathfrak{m})$ is an ideal and contains any other ideals. To finish this proof, I should prove $f(\mathfrak{m})\neq A'$. But I can't prove $f(1)\notin   f(\mathfrak{m})$.  For example, let  $g:\mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z}$ be the natural projection onto its first coordinate. We know $\mathbb{Z}\oplus 2\mathbb{Z}$ is a maximal ideal in $A$, but $g((1,1)) \in g(\mathbb{Z}\oplus 2\mathbb{Z})$.So it's not true for a general ring.    Is there any direct way to prove $f(1)\notin   f(\mathfrak{m})$ for a local ring?
14960
Apologies if the title is garbage. I couldn't think up anything more pertinent, as it's really just a notational question.  In Hartshorne, p.141, Proposition 6.11, it is shown that Weil divisors are Cartier (under some assumptions).  My question is what on earth does it mean for $\mbox{Spec } \mathcal{O}_x$ to be a local scheme; moreover, how does one induce a Weil divisor $D_x$ on it?  In my head this is supposed to be restricting $D$ to a given point; the construction then gives a local equation for $D$ at $x$.  However, I am struggling to see how the scheme $\mbox{Spec } \mathcal{O}_x$ corresponds at all to $x$: it's not even a point unless the stalk is a field.  Thanks for any help offered.
22132
The qcqs lemma (in Ravi Vakil's notes) says that if $X$ is a quasi-compact (qc) and quasi-separated (qs) scheme, for any global section $f$, the natural map from $\Gamma(X, O_X)_f \to \Gamma(X_f, O_X)$ is an isomorphism.  (We are defining $X_f$ to be the complement of the set where $f(p) = 0$, i.e. where $f = 0$ in $O_{X,x} / m_x$. So the natural map is the one guaranteed by the universal property of localization as $f$ becomes a unit over $X_f$.)  The proof of this lemma uses in a crucial way these finiteness assumptions in order to take advantage of the willingness of localization to commute with finite products. (One builds the exact sequence describing the global sections via the sheaf axioms, plugs a finite affine cover with a finite affine cover of the overlaps into it. Then localizing this sequence expresses $\Gamma(X, O_X)_s$ as the kernel of the exact sequence describing $\Gamma(X_s, O_X)$, bootstrapping off of the case of this theorem in the affine case.)  I want counter examples in the situation when this hypothesis has been dropped.  Here is what I have been thinking so far:  I tried the example of the only qc but not qs scheme that I know: the infinite affine space with a doubled origin. The functions on this space and on any open subscheme are the same as functions on the corresponding open $\mathbb{A}^{\infty}$, since the gluing identifies the variables on a dense open set. So I don't think this will provide an example (but I am not sure - now after thinking about this I am more convinced that this example would provide a counter example, but I don't see how). Here one would (obviously) need to use that infinitely many affines are required to cover the intersection of the two $\mathbb{A}^n$ in order to produce something...   I think I found an example of a quasi-seperated non-quasi compact scheme that provides a counter example. Essentially the point is that localization doesn't commute with infinite products in general, so one can take the infinite disjoint union of $Spec Z / (2^i)$, for all $i \geq 0$, and try to invert the element $2$. $X_2 = \emptyset$, but $(\Pi Z / 2^i)[2^{-1}] \not = 0$, since for instance $(1,1,1,1, \ldots)$ is not 2-torsion.
74276
For an affine variety $X\subset \mathbb{A}^n$, we can associate it with $\overline{X}$, which is the closure of $X$ in $\mathbb{P}^n$.   Does $\overline{X}$ depend on the choice of embedding $X\subset\mathbb{A}^n $? Suppose given $f\colon X\subset\mathbb{A}^n\to Y\subset\mathbb{A}^n$，does it induce morphism $\overline{f}\colon \overline{X}\to \overline{Y}$?