Hot answers tagged

8

Reformulate the primitive element theorem as follows: Let $\mathfrak m \subset k[X_1, \dotsc, X_n]$ be a maximal ideal, such that the obtained field extension is separable. For general $(a_1, \dotsc, a_n) \in k^n$, the map $$k[T] \to k[X_1, \dotsc, X_n]/\mathfrak m, T \mapsto a_1X_1 + \dotsb + a_nX_n$$ is surjective. Geometrically, any closed point ...


7

In general, the first cohomology group over $\mathbb{Q}$ is given by$$H^1(X, \mathbb{Q}) = \text{Hom}(\pi_1(X), \mathbb{Q}),$$where $\pi_1(X)$ is the fundamental group. The toric variety $X$ of a complete fan has trivial fundamental group, so$$\pi_1(X) = H^1(X, \mathbb{Q}) = 0.$$ As I follow-up, I think that the higher odd cohomology can be nonzero when the ...


7

Here’s a geometric argument that has nothing to do with algebraic geometry. You may find it insufficiently rigorous, but the idea is certainly sound. Consider a separable extension $K\supset k$. One consequence of separability is that there are only finitely many intermediate fields $E$, $K\supset E\supset k$. Consider the finitely many proper subfields, ...


7

So to simplify the situation, I would suggest throwing out the point $\text{Spec}\, \mathbb{C}[x]/(x-1)$, because it is common to both of the schemes under consideration. You're left with $\text{Spec}\, \mathbb{C}[x]/(x) = \text{Spec}\, \mathbb{C}$ and $\text{Spec}\, \mathbb{C}[x]/(x^2)$. Both of these schemes are points, topologically speaking, as you ...


7

To show that $\text{Spec}\,\mathbb{Q}(t) \otimes_\mathbb{Q} \mathbb{C}$ has closed points in natural correspondence with the transcendental complex numbers, we use the formula for how tensor product behaves with respect to localization. We have $\mathbb{Q}(t) = \mathbb{Q}[t]$ localized at $S$, where $S = $ the elements of $\mathbb{Q}(t)$ of positive degree, ...


6

But $G(3,5)$ has dimension $3(5-3)=6$, which is far less than $9$. The key idea you're missing is that the image of the Plücker map consists of all (projectivized) decomposable $k$-vectors, which is in general a very thin subset of $\Bbb P(\Lambda^k V)$.


6

Suppose we have a complex manifold $M$ described by the equation$$F(z_1, z_2, \ldots, z_n) = 0.$$Now, without imposing this equation, the holomorphic form locally is just$$dz_1 \wedge dz_2 \wedge \ldots \wedge dz_n.$$The simplest way to impose the constraint is to just multiply this by a delta function and integrate over one dimension:$$\Omega = \int \delta(...


5

Using complex number as coordinates, we have \begin{align*} z &= \frac{(a^2-b^2)(a\cos \theta+bi\sin \theta)}{|a\cos \theta+bi\sin \theta|^2}\\ |z| &= \frac{|a^2-b^2||a\cos \theta+bi\sin \theta|}{|a\cos \theta+bi\sin \theta|^2} \\ |z| &= \frac{|a^2-b^2|}{|a\cos \theta+bi\sin \theta|} \\ \arg z &= \tan^{-1} \left( \frac{b}{a} \tan ...


4

The theorem of Lang-Nishimura can fail if either of the aforementioned changes is made. The assumption that $Y$ is proper is dropped. Let $k$ be a finite field. Let $X = \mathbb{A}_k^1$, which is irreducible. Let $Y$ be the open subvariety $X - X(k)$ of $X$. The identity $Y \to Y$ represents a rational map $X \,-\!\!\rightarrow Y$. Moreover, $X$ has ...


3

A variety $X$ over a base variety $S$ is just a variety $X$ together with any morphism $f : X \to S$. A very simple example would be that you can consider varieties over $\operatorname{Spec} \mathbb{C}$ or $\operatorname{Spec} \overline{\mathbb{F}_p}$, which are of course just single points. This is important if you want to form the Cartesian product of ...


3

This is an excellent question -- although it appears to have been answered in various forms elsewhere on this site. I would however like to remark that it speaks to the following interesting open problem: if $k$ is a field, how do we describe the length-$m$ closed subschemes of $\mathbb{A}_k^2$ supported at the origin? By ``describe,'' I mean not only give ...


3

As in here or here, we only need to classify $\mathbb{R}$-local algebras $A$ of lengths $2$ and $3$. Let $m$ be the maximal ideal. Then we consider the filtration associated to $m$. If $m/m^2$ is one dimensional, the same reasoning as in what I linked to above implies that$$A \simeq \mathbb{R}[x]/\langle x^l\rangle.$$Otherwise, the length must be three. The ...


3

Here is an example of the type you require: If $Y$ is a connected smooth projective curve of genus $g\geq2$ its tangent bundle $T_Y$ has degree $2-2g$ and has thus $0\in \Gamma(Y,T_Y)$ as only regular section. On the other hand an elliptic curve $E$ has trivial tangent bundle $T_E=\mathcal O_E$. Thus the product $X:=Y\times E$ has as tangent bundle: $$T_X=...


3

No. Consider the simple case of $V\subset \mathbb{A}^1_k$ being two points, say given by the ideal $I(V)=(x^2-x)$. Then $k[x]/I(V)$ admits non-trivial idempotents, $x+I(V)$ and $1-x+I(V)$. Algebraically closed field does not resolve anything here... In general a ring $R$ admits non-trivial idempotents if an only if $R\simeq R_1\times R_2$ is a product of ...


2

One source (which I have not read the whole article) might be the survey article "History of Homological Algebra" written by C. Weibel, http://www.math.uiuc.edu/K-theory/0245/survey.pdf.


2

To add a slightly more concrete perspective to Ted Shifrin's answer, why don't you try the simpler example of $G(2,4)$ instead? There you have 6 minors, giving an embedding into $\mathbf P^5$. But you can check that the 6 minors of a $2 \times 4$ matrix always satisfy a certain degree-2 equation. (I won't write the equation here, but it is easy to look up, ...


2

The proof that is sketched here does not quite work. In the first bullet point, $U$ is chosen to be an open set such that $X \setminus U$ has codimension at least 2. In the last bullet point, it is claimed that there should be an open subset $V \subset Y$ isomorphic to $U$. But there is no reason that both of these properties should be attainable at the ...


2

Pick another point on the curve of degree $p$. Find a linear combination of the two curves of degree $n$ that vanishes at this new point. It intersects a curve of degree $p$ at $np + 1$ points, hence by the inverse of Bézout's theorem, they share an irreducible component. Assuming the curve of degree $p$ is irreducible, we then remove it from the degree $n$ ...


1

If $V$ is defined using transition functions $f_{ij} : U_i \cap U_j \to GL_n(\mathbb{C})$, thought of as acting on $\mathbb{C}^n$, and $W$ is defined using transition functions $g_{ij} : U_i \cap U_j \to GL_m(\mathbb{C})$, thought of as acting on $\mathbb{C}^m$ then the hom bundle $[V, W]$ is defined using transition functions $$T \mapsto g_{ij} T f_{ij}^{-...


1

If $\mathfrak{a}_i \subseteq A$ is an ideal for each $i \in Y$, $Y$ a set of any size, then $\sum_i \mathfrak{a}_i$ is (by definition) the set of all finite sums of elements of the $\mathfrak{a}_i$. So if $1 \in \sum_i \mathfrak{a}_i$, then there exist $i_1,\ldots,i_n$ and elements $f_{i_k} \in \mathfrak{a}_{i_k}$ such that $$f_{i_1}+\cdots+f_{i_n}=1.$$


1

$1\in\sum(f_i)$ implies that $1=\sum_{j=1}^{j=n}a_{i_j}f_{i_j}$, thus $\bigcup_{j=1}^{j=n}D(f_{i_j})=Spec(A)$.


1

I am not familiar with algebraic groups or varieties, but purely on the level of abstract groups the following should take place. Let $B<G$ be a subgroup with $N_G(B)=B$. If $\mathcal{B}$ denotes the set of conjugacy classes of subgroups that $B$ is a member of, then by the orbit-stabilizer theorem we know that $\mathcal{B}\cong G/B$ as $G$-sets, where $...


1

To get back to affine space to apply your lemma, take the cone of everything. Hartshorne page 12, Ex. 2.10. Then you can apply Hartshorne page 8, Ex. 1.8 (which is basically the lemma above).


1

Here is a very general point about how codimension is defined. Given a closed embedding of schemes $i : X \hookrightarrow Y$, we say that the embedding is codimension-$r$ at $p \in X$ if in the local ring $\mathcal{O}_{Y,i(p)}$, the ideal cutting out $X$ is generated by a length-$r$ regular sequence. The embedding is codimension-$r$ if it is codimension-$r$ ...


1

An additive basis is a basis as a vector space (or more often free abelian group) as opposed to perhaps a generating set as an algebra. The latter is not usually called a basis, and "additive" is really used for emphasis here.


1

a) $f_*(\mathcal{O}_X)$ will in general, not split as direct sum of line bundles. For an example, take an abelian surface $X$ for some suitable $d$. If the direct image splits as sum of line bundles, it will force $H^1(\mathcal{O}_X)=0$, which is not true for an abelian surface. b) $d=2$ case is the only one which can be understood well. One has the exact ...


1

To show that any point of $X$ has a point in $X_K$ lying over it, just recall that surjectivity is stable under base change (and $\operatorname{Spec}(K)\rightarrow \operatorname{Spec}(k)$ is surjective). To see why this finishes the argument (for future askers of this question), notice that we have a map $\Gamma(X,\mathcal{L})\rightarrow \Gamma(X_K,\mathcal{...


1

Let me first recap the setup and known parts. $A = k[x,y,z,w]/(xy-zw)$ and $p = (y,z)A$. In order to show that $Cl(A)\cong \mathbb{Z}$, it suffices to show that for any $n\ge 1$, $p^{(n)}$ is not principal. Suppose that for some $n \ge 1$, $p^{(n)}$ is principal. Write $S = k[x,y,z,w]$, and let $f \in S$ whose image in $A$ generates $p^{(n)}$. Since $p^n \...


1

Here's a hint: Observe that the open set $D(f-g)$ contains no closed points. It is therefore an open set disjoint from a dense set. What does this imply about $D(f-g)$? What does that implication tell you about $f-g$?


1

OK, I looked at the paper. I think the passage you are asking about is very confusingly written. What they appear to be saying is that for this specific value, $\alpha A +B$ is the (unique) degenerate conic in the pencil. So the plural "conics" in the first line and the pronoun "they" later on are incorrect. So to sum up: 1) the pencil is $\lambda A +B$ ...



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