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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 ...


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 ...


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 ...


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 $n + 1$.


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.


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).


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 ...


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.


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

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 ...


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

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.


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 ...


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 ...


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 ...


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 ...


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 ...


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

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 ...


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

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 ...


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 ...


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 ...


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 ...


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 ...


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.


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 ...


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 ...


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 $C\neq L$.



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