New answers tagged

0

By the Nullstellensatz, if $K$ is algebraically closed, then $$I(V(I)) = \sqrt{I} = \{f\in K[x_1,\dots,x_n]|\;\exists r\in \mathbb N,\;f^r\in I\}.$$ Now, $V\subseteq W \Rightarrow I(W)\subseteq I(V)$ so $V(J)\subseteq V(I) \Rightarrow \sqrt{I}\subseteq \sqrt{J}\Rightarrow I\subseteq \sqrt{J}$ and that's all (except if $J$ is radical, i.e. $\sqrt{J}=J$). In ...


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

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 $n$ ...


0

H. Cartan's book on complex analysis has a chapter on analytic functions of several complex variables, which may be helpful because it is concise and clear, and at the same time has exercises and questions.


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 $\bigcup_{j=1}^{j=n}D(f_{i_j})=Spec(A)$.


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 ...


0

Since you know $a$ and since you know some random point $(x_0,y_0)$ on the hyperbola, you can write $$\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1$$ where $b$ is the unknown. This gives $$b^2 = \frac{y_0^2a^2}{x_0^2-a^2}$$ hence $$b = \frac{|y_0|a}{\sqrt{x_0^2-a^2}}.$$


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 ...


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 $...


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

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

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 ...


0

The function $f-h$ will not necessarily have a minimum on $\mathbb{R}$. For example, take $f(x)=-x$ and $h(x)=x^2$. Then $f(x)-h(x)=-x(1+x)$ which is a maximum at $x=-1/2$ and no minimum. The same is true of $g-h$. There can also be many points where the derivatives of these functions are zero.


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

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.


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 ...


1

After these references it really depends on what type of commutative or homological algebra you intend to work in. However, some of the most widely useful general references are as follows. Almost everything in the book Bruns and Herzog is lingua franca and can't be skipped. Most of the chapters of Weibel's Homological algebra. Here its I guess okay to ...


0

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

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-$r$ ...


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 ...


0

I was previously confused about what it means to write a regular function as a single expression, but I think I have a clearer picture now. I still think the issue is that the closed subscheme $p = (1,-1) \in \text{Spec}\,\mathbb{k}[x,y]/(y^2 - x^3)$ is not "cut out by one equation." However, I think the correct way to translate the (vague) quoted phrase ...


5

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 ...


6

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 ...


0

I cannot answer the question but I am trying to gather potentially relevant material to answer this question, work in progress until trying to target the question. I could find the threads and material sometimes relevant for positive polynomials but also focus on nonnegative polynomials Nonnegativity condition for a polynomial in two variables Lecture ...


0

http://www.math.uiuc.edu/K-theory/ is one site which contains some preprints, some survey articles. I thought this may be useful for other users, which is why i made it as an answer.


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, ...


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.


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, ...


0

This is a very long comment, and not an answer! I shall write only about the Grothendieck cohomology of sheaves of Abelian groups over a topological space! Let $X$ be a topological space; one can define \begin{equation*} \forall U\subseteq X\,\text{open},\,\mathcal{C}(U)=\{f:U\to\mathbb{R}\,\text{is continuous}\} \end{equation*} where over $\mathbb{R}$ ...


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 $\lambda A +B$ ...


0

This is more or less a copy of part of my answer here, which I copied over rather than just linked to since that answer contains a lot of material irrelevant to this question. I also don't think that either question is a duplicate of the other, this question is interesting even without applying it to the linked one. Since $\pi: X \rightarrow \operatorname{...


0

You can think of the section as a map $S\rightarrow X$ of $S$-schemes (where $S$ is an $S$-scheme with structure map $id: S\rightarrow S$). Clearly this is a map of finite etale $S$-schemes, and hence this section is also finite etale. The fiber of this map clearly has degree 1, so it's finite etale of degree, 1, and hence is an isomorphism onto its image, ...


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, ...


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 ...


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 $f-g$?


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 $\Bbb P(\Lambda^k V)$.


1

$C$ is made from the two matrices $A$ and $B$ each defining a conic. Let $\alpha=0$ and you get the one, let $\alpha=\infty$ and you get the other. The whole pencil is described by varying $\alpha$. (If you don't feel comfortable with one conic appearing only in the limit $\alpha\to\infty$, consider $\lambda A+(1-\lambda) B$ instead.) Assuming the conics ...


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

I used a software to produce the hypocycloid and then I tried this with an Astroid. Here is what happened: Hypocycloid been drawn As you can see, the circumference of the circle doesn't go over the shape of the hypocycloid. It doesn't work like that unfortunately.


1

Let $f(x)\in \mathbb{Z}[x]$ be a cubic monic polynomial which is $p$-Eisenstein for some prime $p\neq 3$. Assume to that the splitting field of $f$ is degree $3$ over $\mathbb{Q}$ (which is equivalent to the discriminant being a square). Let $K$ be that splitting field. Let $\mathcal{O}_K$ be its ring of integers and $\mathfrak{p}$ be a prime above $p$. ...


0

I think the current answer doesn't solve the problem, since $a)$ I don't follow the argument for why it is sufficient to consider the case of closed points, and $b)$, it applies a previous theorem about projective space over a field to projective space over a general ring, where I'm not sure it holds. Here's a complete solution. The problem with the ...


0

I've now understood what was confusing me, and will post this answer for future readers who may have the same problem: A regular scheme is, by definition, locally Noetherian! The confusion arose from the fact that we only define regular points for locally Noetherian schemes, and so $C'$ is in fact locally Noetherian, which I didn't realise at the time. ...


1

Since $f\in\mathfrak{a}$, $Z\subseteq H$. So $(\mathbb{A}^n-H)\cap Y=(\mathbb{A}^n-H)\cap \overline{Y}$, which is closed in $\mathbb{A}^n-H$.


2

Answer to part 1: If $f$ is an isomorphism of algebraic varieties, it must be a regular map (which is therefore analytic, as polynomials and their appropriately-taken quotients are analytic). Similarly, it's two-sided inverse must also be regular and thus analytic, and so we have that $f$ is also an isomorphism of analytic varieties. Therefore it must be a ...


1

By definition, if $X = \mathbb{P}^1$, so that $N_m = q^m + 1$, then \[ Z(X,s) = \exp\left(\sum_{m = 1}^{\infty} \frac{q^m + 1}{m} q^{-ms}\right) = \exp\left(\sum_{m = 1}^{\infty} \frac{q^m}{m} q^{-ms} + \sum_{m = 1}^{\infty} \frac{1}{m} q^{-ms}\right),\] which is \[\exp\left(\sum_{m = 1}^{\infty} \frac{q^{-m(s - 1)}}{m}\right) \exp\left(\sum_{m = 1}^{\infty} ...


0

For question 1, if we write it out in terms of rings, we have $t\in A$ that maps to a zero divisor in $B$. That zero divisor must be contained in some associated prime. Take the inverse image of that associated prime in $A$. It contains $t$, but the ideal generated by $t$ is maximal (if we restrict to an open subset disjoint from the zeros of $t$ away from [...


2

For (1), note that $\mathcal{O}_D$ is in fact shorthand for $i_\ast \mathcal{O}_D$, the pushforward of the structure sheaf $\mathcal{O}_D$ by the inclusion $D \hookrightarrow X$. In particular, which $\mathcal{O}_D$ is a (trivial) line bundle on $D$, it's not the case that $i_\ast \mathcal{O}_D$ is a line bundle on $X$. For instance, if $x \in X$ is any ...


2

The intended interpretation is that $K^*=K\setminus\{0\}$ (the group of multiplicative units of $K$), so an element of $K^*$ is just a nonzero element of $K$. Another common notation for this set is $K^\times$.



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