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1

Let $G$ be the Galois group of $K$ over $k$. The image is the kernel of a map $Br(K)^G \to H^3(G, K^*)$. Look in books on Galois cohomology, preferably one written in French...


2

Here are some oversimplified blurbs about what each one does. (Abstract) algebra deals with operations on sets, especially binary operations. Geometry deals with sets which have groups acting on them. Part of this involves shape. Algebraic geometry applies commutative algebra to sets described by algebraic equations. It gives information about the shape ...


0

I read only 3. $S_f=\{\begin{pmatrix}a&b\\c&3-a\end{pmatrix}|a(3-a)-bc=2\}$. Then $S_f\subset k^3$ is defined by the relation $a(3-a)-bc=2$. Thus $S_f$ is Zariski-closed. The concept of annulator polynomial is different: Let $T_f=\{A|f(A)=(A-I)(A-2I)=0\}$. Then $T_f=S_f\cup\{ I\}\cup \{2I\}$, that is $S_f$ $\cup$ two isolated points.


0

An example is $G = \mathbf{G}_{a, S}$ over $S = \text{Spec}(A)$ and $g$ is given by $a \in A$. Then $\phi_g(x) = a$, i.e., $\phi_g$ is given by the $A$-algebra map $A[x] \to A[x]$ which sends $x$ to $a$. Then $\text{d}x$ maps to zero (in relative differentials) hence it is true in this case. It is true in general too, because $\phi_g$ factors as $G \to S ...


0

This is something simple about commutative compact real lie groups of dimension $g$. I think these are all of the form $G = (S^1)^g \times A$ where $A$ is a finite abelian group (didn't carefully check this, but I'm not worried). Hence $G/2G$ is isomorphic to $A/2A$. On the other hand $G[2] = (\mathbf{Z}/2\mathbf{Z})^g \times A[2]$. And for a finite abelian ...


0

Since no three of the five points are aligned, any conic through the five points is non-singular.


1

A point $(x,y)$ in $L\cap C$ satisfies $y=ax+b$ and $F((x,y))=0$. So every such point is a zero of the polynomial $F(X,aX+b)$. Inductively it is easy to show that $F(X,aX+b)$ defines a polynomial of degree $\leq n$ in $K[X]$. So there are at most $n$ points in $L\cap C$. EDIT: obviously, I assume $C\neq L$.


2

I'm assuming that the assumption that $H$ is closed under formal sums means that if you take a sequence $h_0, h_1, h_2, \dots \in H$ such that $\text{ord}(h_0) < \text{ord}(h_1) < \dots$, then $\sum_l h_l$, which is a well-defined element of $k[[t]]$, is in fact an element of $H$. Nevertheless, as stated, the claim is false. Consider $H = {\mathbb ...


3

Suppose $Y$ is noetherian for simplicity. For a blowing-up morphism $f: X\to Y$, one has $f_*O_X=O_Y$ if $Y$ is integral and normal. Indeed, $f$ is birational and proper, hence $O_Y\to f_*O_X$ is finite by the direct image theorem, and $f_*O_X$ is contained in the function field of $Y$. So the normality of $Y$ implies that $O_Y=f_*O_X$. If $Y$ is ...


3

No. Even Chern numbers are not topologically invariant — an example was given by Borel and Hirzebruch in 1959 (Characteristic classes and homogeneous spaces, §24.11).


1

Easiest way is to show that the corresponding ideal, which is $(Y - X^2)$, is prime. This in turn can be seen easiest by showing that its quotient ring $k[X,Y]/(Y-X^2)$ is an integral domain (or even by directly arguing that $Y - X^2$ is irreducible). Now, what is this quotient ring isomorphic to? But, to actually answer your question: yes, from the fact ...


0

It is exactly what Hoot said. If you know the stalks then you just take the sheaf of sections, meaning locally compatible maps $s:U\rightarrow \coprod_{x\in U}\mathscr{F}_x$. In this case $\mathscr{F}_x=f^{-1}\mathscr{G}_x=\mathscr{G}_{f(x)}$. Another way of seeing this is by the Etalé Space of a presheaf. In general if $\mathscr{F}$ is a presheaf on $Y$ ...


2

As cant_log notes, this theorem is often stated in the form "the hyper elliptic involution is unique (when it exists)". The terminology arises as follows: a complex algebraic curve (equivalently, compact Riemann surface) which can be written as a 2-fold branched cover of $\mathbb P^1$ is called hyperelliptic. If $\phi: X \to \mathbb P^1$ is a double ...


0

For the notation $S[1/p]$ to make sense, you have to suppose $\mathcal O$ has characteristic $0$. If $S$ is irreducible, then $S[1/p]$, being an open subset of $S$, is also irreducible. But $S_p$ ($S$ mod $p$) can be essentially anything. If $S[1/p]$ is irreducible and $S$ is flat over $\mathcal O$ (meaning that $R$ has no torsion over $\mathcal O$: ...


3

Another way to make a curve with many branches through a point is to start with $\mathbb A^1$ over $\mathbb C$ (say) and then glue together the points $t = 0, 1, \ldots, n$ (for some the value of $n$). That is, we let $A$ be the $\mathbb C$-subalgebra of $\mathbb C[t]$ consisting of polynomials $f$ such that $f(0) = f(1) = \ldots = f(n).$ This a finitely ...


4

Consider the curve singularity at the origin of the image $C$ of the map $z \mapsto (z^e, z^{e + 1} \ldots, z^{e + n})$ where $e$ is a large integer (say larger than $n + 1$). Any polynomial equation for $C$ must have vanishing constant and linear terms. Hence the Zariski tangent space of $C$ at $0$ has dimension $n + 1$.


3

Yes: $\mathrm{GL}(n,\mathbf{Q}_p)$ acts transitively on the variety of complete flags, which is compact, and the stabilizer of a point is the set of upper triangular matrices, which is solvable and cocompact. The same holds with $\mathbf{Q}_p$ replaced by any non-discrete locally compact field.


2

Adding to Asal Beag Dubh's answer, use a little bit of topology (for example, $\mathbb{C}$ is the universal covering space of both curves) to show that your holomorphic map lifts to a linear map on $\mathbb{C}$, translated by a constant. In other words, if $f$ sends 0 to 0 (I'm assuming you know that a genus 1 curve is a group), it is a group homomorphism.


9

They're isomorphic. Consider the map $K[a,b,c,d] \to K[x^4,x^3y,xy^3,y^4]$ given by $a \mapsto x^4$, $b \mapsto x^3 y$, $c \mapsto xy^3$, $d \mapsto y^4$. What we need to see is that the kernel of this map is the ideal $(ad - bc, a^2c - b^3, bd^2 - c^3, ac^2 - b^2d)$. One way to do this is look at the ideal $I := (a - x^4, b - x^3y, c - xy^3, d - y^4)$ of ...


2

The Riemann–Hurwitz formula says that if $f: C \rightarrow D$ is a finite holomorphic map of degree $d$ between any two algebraic curves, then $$\chi(C) = d \cdot \chi(D) - \sum_{P \in C} (e_P-1)$$ where $e_P$ is the so-called ramification index of $f$ at the point $P \in C$. (This is a natural number with the property that $e_P=1$ if and only if ...


2

In geometric terms: A complex variety $V$ of dimension $n$ is rational if there exists a birational map $\mathbb P^n --\to V $ or, equivalently, if its function field is purely transcendental i.e. there exists a field isomorphism $Rat (V)\cong \mathbb C(t_1,\cdots, t_n)$ . More generally $V$ is called unirational if there exists a rational surjective ...


5

The symmetric matrix corresponding to your quadratic form is $$A = \begin{pmatrix} 1 & 1 & 0 & -1 \\ 1 & 2 & -1 & 1 \\ 0 & -1 & m+1 & -2 \\ -1 & 1 & -2 & m^2 + 4 \end{pmatrix}.$$ A straightforward calculation reveals that $\det(A) = m^3 - m$. Thus, for $m \neq -1, 0, 1$, the surface is a smooth quadric ...


1

Since your map is polynomial, it cannot be bijective from $\mathbb{R}^2$ to $\mathbb{R}^3$ as it goes from dimension $2$ to dimension $3$. In fact the image needs to be in a variety of dimension $\le 2$, which corresponds to say that the points of the image satisfy some polynomial equation. In your case, you can easily check that $(0,0,1)$ is not in the ...


1

Strictly speaking, $\mathcal{O}(1)$ is not a line bundle but a sheaf. You can construct a rank 1 vector bundle from it: locally, on $D_+(T_i)$, $\mathcal{O}(1)$ is isomorphic to $\mathcal{O}_{\mathbb{P}^n}$ via multiplication with $T_i^{-1}$ and these local isomorphisms and the cover $D_+(T_i)$ define a line bundle.


3

The exactness of $1\to H\to G\to K$ means $H\to G$ induces an isomorphism from $H$ to the kernel of $G\to K$ which is the fibere product $G\times_K \{1_K\}$. The exactness of $G\to K\to 1$ means $G\to K$ is faithfully flat. This is equivalent to $G\to K$ is surjective when $G, K$ are smooth over $k$. For any field extension $F/k$, $$ 1\to H(F)\to G(F) \to ...


3

Yes to your first question. Of course, one of the key things is to show that it doesn't matter which variable you use. To see this the easiest thing is to choose resolutions for both and consider the double complex and do an argument with zig-zags. This argument is given for the case of tor groups of modules in Section Tag 00LY. Exactly the same works for ...


1

I can't comment. Here are a few remarks: I guess $V=U$ in the last paragraph. Consider a morphism $g: W\to U$ which is bijective and flat but not an isomorphism. Let $f: X=\mathbb P^1_W\to W$ and let $\pi=g\circ f$. Then $\pi$ maps any fiber $X_s$ to one point of $W$ (namely the point $g^{-1}(s)$), but there is no $U\to W$. A version of rigidity lemma ...


2

If $L/K$ is a Galois extension, then $L\otimes_K L =L[G]$ where $G$ is the Galois group of $L/K$. As $L[G]\simeq L^n$ as $L$-algebra, we have $T\times_K L=T\times_L (L\otimes_K L) \simeq T\times_L (L^n)$ is $n$ disjoint copies of $T$. So $\mathrm{Hom}_L(T\times_K L, X)$ is the product of $n$ copies of $\mathrm{Hom}_L(T, X)$. In general, the set of ...


2

The d-uple embedding can be described in a coordinate-free way as the map $\mathbb{P}(V^*)$ to $\mathbb{P}(Sym^d(V^*))$ given by sending a quotient ($V \rightarrow L$) (considered as a point of $\mathbb{P}(V^*)$) to the quotient ($Sym^dV \rightarrow Sym^dL$) (considered as a point of $\mathbb{P}(Sym^dV^*)$). If $W$ is any quotient of $V$, with say $\phi:V ...


0

The question has been answered in MO.${}$${}$${}$${}$${}$${}$${}$${}$${}$${}$${}$${}$${}$


4

You can show that, for $n>>0$, $L^n$ has a lot of global sections. ($\dim_k(H^0(X,L^n))\geq \dim X+1$). If $L^{-1}$ has a nonzero global section, then so does $L^{-n}$. Tensor these with those of $L^n$ to get lots of global sections of $\mathcal{O}_X$. But we know there is only one global section (up to scalar of course).


2

This is not an answer, but I am not eligible to comment yet. It seems that there is problem in the proposed argument in the affine case. Let {$e_1,...,e_n$} be a basis of $L$ over $K$. Then, an element in $Hom_L(L[X_1,...,X_d]/(f_1,...,f_r),A\otimes _K L)$ should be given by elements $$ \Sigma_{i} (x_{i1}e_i),...,\Sigma_{i} (x_{id}e_d)$$ where $x_{ij}\in ...


4

Assuming by variety, you mean that the associated topological space is irreducible, then yes, $f = g \circ \text{pr}_Y$ on all of $X \times Y$. (Also, I'm assuming that you implicitly have $W$ is a proper closed subset of $Y$, i.e. $W \neq Y$.) This follows from the more general fact that if $f, g : X \rightarrow Y$ are two morphisms of varieties, then $Z ...


1

$K(X)$ is a finitely generated field extension of $k$, so any field between $k$ and $K(X)$ is also such an extension. This is believable but harder to prove than one might expect. See, for example, this MO answer.


4

If you mean $V=V((x+y)^s) \subseteq \mathbb C^2$, then we alyways have $V(I)=V(\sqrt I)$ (since $f(x)^s=0 \Rightarrow f(x)=0$), so that actually $V=V(x+y)$. So as a variety, $V$ is in fact irreducible. However, as a scheme, $V$ is still irreducible, but it is "fat". For a scheme theorist, $V$ is like the line $y=-x$ $s$ times.


2

To answer your question in full would take quite a bit of work (to get all details straight -- if the thing is true) which I am not willing to do. I suggest looking back at where you found this statement and look for a proof there or a reference. But here is a baby case. Suppose we have a ring map $A \to B$ and a finitely generated ideal $I$ of $B$. Then we ...


0

It seems to me -- as a warning, I haven't written anything down -- that you could define $f^{-1}\mathscr{G}$ as follows. Over an open set $U \subset X$ a section is an element $(s_p)_{p \in U}$ of $\prod_{p \in U} \mathscr{G}_{f(p)}$ such that for each $p \in U$ there is a neighborhood $U' \subset U$ of $p$, an open set $V$ of $Y$ containing $f(U')$, and a ...


1

To homogenize an ideal, you must homogenize a Gröbner basis for $I$. For more on what can happen, see for example this MO answer. However, in the case of two generators, it is usually very easy to find a Gröbner basis. For example, if it the initial terms (in some order) are prime, then then the two polynomials are already a Gröbner basis, so you can just ...


5

$v_P(f+g)\geq \operatorname {min}(v_P(f),v_P(g))\quad \quad v_P(\lambda f)=v_P(f) _\;(\lambda \neq0)$


2

Am I right in saying that any two lifts of $A$ to an endomorphism of $F_2$ differ by some inner automorphism of $F_2$? No, using a free basis $F_2 = \langle a,b \rangle$, the map given by $f(a)=a$, $f(b) = a b^3 a^3 b^{-1} a^{-4} b^{-1}$ is not an inner automorphism but it induces the identity on $\mathbb{Z}^2$, as does the identity automorphism ...


0

There're numerous online resources for preliminary materials such as MIT open courseware. There's a growing list in this MO topic Video Lectures in Algebraic Geometry containing rather more advanced materials. Including more math areas, there's the following list Video lectures of mathematics courses available online for free as well.


2

In order to learn algebraic geometry you will be best off with a good grounding in abstract algebra (groups, rings etc.) and then some commutative algebra (though I guess it's possible to learn some of this as you go along). Searching MIT opencourseware and similar sites for these topics might work out well, and ...


2

Some quick answers: -Yes, the $\geq$ version of your definition is correct. (In the definition of moving part you have a typo; replace $D$ by $F$.) -In general $|M|$ is not basepoint-free. Certainly that is true for curves, because there "fixed part" and "base locus" coincide. For surfaces $|M|$ might not be basepoint-free, but it is a theorem of Zariski ...


2

Well, the definition is quite natural: To give a presheaf, you tell what its section are. So let $U$ be an open subset of $X$. Then $$ f_{pre}^{-1}\mathcal G (U) := \lim_{V \supseteq f(U)} \mathcal G(V)$$ where $V$ ranges over the open subsets of $Y$ containing $f(U)$. The reason we have to get fancy and use limits, is that $f$ is not necessarily an open ...


1

Sym$^2(V)$ decomposes into one dimensional eigenspaces with respect to the given cartan subalgebra and has eigenvalues $$E = \{2L_2,2L_1,2L_3,L_1+L_2,L_2+L_3,L_1+L_3\}$$ The eigenvalues of Sym$^n(V)$ are $E_n = \{a_1+\cdots+a_n\;|\;a_1,\cdots,a_n\in E\}$. The number of times $\mathbb{C}$ appears is the difference between multiplicity of eigenvalue 0 and ...


8

Khan Academy is one of the best resources to start with. Just guess what level you think you are at, if it doesn't make sense, just go to the previous level until you can follow it. Then work your way up.


0

This question wasn't stated clearly enough so I asked another more precise question whose answer answers this question. See What does it mean geometrically for a variety to be locally a complete intersection? The answer is yes, we can use the property of being a complete intersection to write a variety as locally the level set of a submersion.


1

Two automated approaches. With Gröbner basis: see the first entry of this result. With resultants: first compute this and then this.


3

Thanks to some outside help I figured out the answer to the question. Actually everything needed was here http://stacks.math.columbia.edu/download/varieties.pdf. Since the identity is a $k$-rational point contained in $G^0$ and this one is connected, we can apply Lemma 5.14. Then $G^0$ is geometrically connected, this allows us to conclude using Lemma 5.4.


2

You are already almost there! Remember that the maps need not be defined everywhere, and $[u:v] = [u/v:1]\in\mathbb P^1$. Let $U\subseteq X$ be the open set where $z\neq 0$, and $V\subseteq\mathbb P^1$ be the open set where $v\neq 0$. The map $\varphi:U\to V, [x:y:z]\to[xy/z^2:1] = [xy:z^2]$ is well-defined and regular on $U$. We can show that $\varphi$ is ...



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