Tag Info

Hot answers tagged

10

I'm going to make this more geometric and less algebraic. But it all translates to the algebro-geometric setting of divisors if you wish. You should think of a line bundle as a twisted product, and tensor product means that you concatenate or superimpose the twists. For example, thinking of the Möbius strip as a real line bundle $\mathscr L \to S^1$, then ...


8

To my mind, "invertible with respect to the tensor product" is already the correct definition of a line bundle, in full generality. If that isn't the definition you're using I assume you're using something like "locally free of rank $1$," so let me say something about this. Intuitively you should think of a line bundle as literally a bundle of lines; that ...


4

One way of showing that $\def\ZZ{\mathbb Z}\def\CC{\mathbb C}\CC[X,Y]$ and $\CC[X]\otimes_\ZZ\CC[Y]$ are not isomorphic as rings is to notice that the first one is a noetherian ring while the second is not. Indeed, there is a surjective ring homomorphism $\CC[X]\otimes_\ZZ\CC[Y]\to\CC\otimes_\ZZ\CC$, so it is enough to show that $\CC\otimes_\ZZ\CC$ is not ...


4

This problem is an instance of the (almost trivial) statement: Let $R$ be a commutative ring and $a \in R$, then $R[Y]/(Y-a) \cong R$. Just set $R = k[X]$ and $a = X^2$.


4

Consider the diagonal morphism $M\cap N\to M\times N$ where $x\mapsto(x,x)$. Since $M$ and $N$ are affine, so is $M\times N$. We see that $M\times N\subseteq\mathbb{P}^n\times\mathbb{P}^n$. Let $\Delta$ be the diagonal; it is closed since $$\Delta=\{([x_0:\cdots:x_n],[y_0:\cdots:y_n])\in\mathbb{P}^n\times\mathbb{P}^n:x_iy_j-x_jy_i=0,i,j=0,\ldots,n\}.$$ ...


3

Let $k$ be a commutative ring. If $X$ is a $k$-scheme and $R$ is a commutative $k$-algebra, then one defines $X(R)$ to be the set of $k$-morphisms $\mathrm{Spec}(R) \to X$. If $k$ is a field, and $R=k$, the set of $k$-morphisms $\mathrm{Spec}(k) \to X$ identifies with the set of points $x \in X$ whose residue field extension $k \to k(x)$ is trivial. Notice ...


3

Yes, as a DVR is a local ring there is only one prime number that is not invertible. (Note that a nontrivial ideal cannot include two distinct prime numbers and each non-unit is contained in some maximal ideal.) In more detail: Every non-zero subring of $\mathbb{Q}$ must contain $\mathbb{Z}$ and since a field is not a DVR the ring must not equal ...


3

Since $\overline{\mathbb{Q}}$ is no $\mathbb{C}$-algebra, it doesn't make sense to talk about the $\overline{\mathbb{Q}}$-valued points of a $\mathbb{C}$-variety. Of course you could look at the underlying variety over $\mathbb{Q}$, say, but then there are again no points, as you have observed correctly. Your $\mathbb{C}$-variety has exactly two ...


3

This looks plausible, but the answer is no. Everything is over an algebraically closed field. Let $Y=\mathbf P^1$, which is not affine. Take a smooth conic curve $C \subset \mathbf P^2$ and a point $p$ on $C$. Projection away from $p$ gives a surjective map $X :=C \setminus \{p\} \rightarrow Y$. But any smooth conic is isomorphic to $\mathbf P^1$, so $X$ ...


3

As long as one of $F_X$ or $F_Y$ is nonzero, we can suppose $F_Y\ne 0$, and if $F_X=F_Y=0$, then either $F$ is a constant, which is not the case, since $F$ is irreducible, or $k$ has characteristic $p$ and $F$ is a polynomial in $X^p$ and $Y^p$. Since $k$ is algebraically closed, the $p^{\text{th}}$ roots of all the coefficients exist, so $$F=\sum ...


3

Let $X = \{ (x, y) \in \mathbb{C}^2 : x^2 - y^3 = 0 \}$, let $Y = \mathbb{C}$ and let $X \to Y$ be the first projection. This is a homeomorphism (!) but it is not even an immersion: indeed, the corresponding ring homomorphism $\mathbb{C} [t] \to \mathbb{C} [x, y] / (x^2 - y^3)$ is the one that sends $t$ to $x$, and it is clear that $y$ is not in the image of ...


3

Dimension of the variety $X$ is the transcendence degree of its function field $k(X)$ which is isomorphic to $k(U)$ for an open set $U \subset X$ so the dimension consequence.


2

The Zariski tangent space of the moduli space of stable sheaves at a point $[F]$ for a stable sheaf $F$ can be canonically identified with $Ext^1(F,F)$. Now if $F$ is locally free, then this space is just $H^1(X,\mathcal{E}nd(F))$. You can read about all that in Huybrechts-Lehn: The Geometry of Moduli Spaces of Sheaves. Where did you read that the moduli ...


2

If I had to give a tl;dr to my incoherent drivel above, it would probably be as follows. $V \otimes V^*$ has a canonical basis if and only if $V$ is $1$-dimensional. We have a circle. We draw the circle the opposite direction.


2

Here's how to do it in the affine case. Suppose $X = \text{Spec } k[x_1, \dots x_n]/(f_1, \dots f_m)$. The condition that a point is singular can be phrased in terms of the rank of the matrix with entries the partial derivatives $\frac{\partial f_i}{\partial x_j}$: generically this rank takes a particular maximum value, and it drops at precisely the singular ...


2

Take your curve to be $\mathbb{P}^1$. And take any rational function $z(x) :=p(x)/q(x)$; where $p,q \in \mathbb{C}[x]$. Then the zeroes of $z(x)$ are exactly the zeroes of the polynomial $p$ and the poles are exactly at the zeroes of $q$. Depending on the degree of $z$, defined as $deg(p) - deg(q)$ the behaviour at $\infty$ changes. If you are looking ...


2

In short, no, I do not think the proof is complete. It looks like you are trying to show $A(\mathbb{A}^1)\cong A(V(y-x^2))$ and then note $A(\mathbb{A}^1)\cong k[t]$ and $A(V(y-x^2))\cong k[x,y]/(y-x^2)$. Although the claim that the algebraic set $\{(x,y)\in \mathbb{A}^2: x=t, y=t^2\}$ is the zero set of $y-x^2$ is true, how do you know it is exactly the ...


2

Lemma. Let $p\in\mathbb R[x,y]$ be a polynomial and let $T\subset \mathbb R$ be an infinite set with an accumulation point $a$. Assume $p(t,\sin t)=0$ for all $t\in T$. Then $p$ is the zero polynomial. Proof. Because $t\mapsto p(t,\sin t)$ is analytic, it follows that $p(t,\sin t)=0$ for all $t\in\mathbb R$. Write $$p(x,y)=\sum_{k=1}^n x^kr_k(y)$$ with ...


2

It is false: take any DVR $V$, $\pi$ a uniformising element and $K$ its field of fractions. Then $K/\pi K=0$, but $K$ can't be a finitely generated $V$-module, since this would mean $K$ is integral over $V$ and hence $K=V$, which is impossible since by definition, a DVR is not a field.


1

Write $P(\alpha(X)) = Q(X)$, with $X=(x_1,x_2,x_3)$, and use Taylor formula : $$(1) \quad Q(X+H) = Q(X) + DQ(X).H + \underbrace{ {}^t H.D^2Q(X).H } + O(H^3).$$ where : $H=(h_1,h_2,h_3)$, $DQ(X).H = \sum_{i=1}^3 \tfrac{\partial Q}{\partial x_i}(X).h_i$, $D^2Q(X) = \mathcal{H}_Q(X)$ and $O(H^3)$ is some polynomial in $X$ and $H$ of total degree $\geq 3$ in ...


1

I'll assume what Pavel Čoupek indicated in his comment to the question. The desired statement then essentially follows from the following lemma (this is Lemma 3.32 from the book Algebraic Geometry 1 by Görtz and Wedhorn where you will also find a more complete proof on p. 79): Let $B$ be a ring and let $A$ be a $B$-algebra. Assume that there are finitely ...


1

Surely $y-x^2, z-x^3 \in I(Y)$. Let $I= \langle y-x^2, z-x^3 \rangle $, let $f(x,y,z) \in I(Y)$ so we have $f(t,t^2,t^3)=0$ for any $t \in k$ or $g(x) = f(x,x^2,x^3) \equiv 0 $ since $g(t)=0$ for all $t \in k$. We will now show $f(x,y,z) + I = 0+I$ which will do the job. $f(x,y,z) + I = f(x,x^2,x^3)+I = 0 +I$. Hence $I = I (Y)$.


1

This is a very nice example! In fact, provided that $p+q$ is not a canonical divisor, the linear system $K+p+q$ separates "most" points, since $K$ is the only $g^1_2$ on the curve. Thus for any $r$ and $s$ not equal to $p$ and $q$, $\ell(K+p+q-r-s) = 1 = 3-2$, so $r$ and $s$ are separated by the linear system (the dimension drops twice, so you know that $s$ ...


1

Let $A$ be a noetherian ring which is regular in codimension $1$. That means $A_P$ is a discrete valuation ring for all $P$ with $\mathrm{height}\, P = 1$. The ring $A'=k[X,Y]$ fulfills this condition. In fact every prime of height one is principal $P=(f)$ with $f$ irreducible and the valuation of $P$ is given by the exponent of $f$ in the prime factor ...


1

The phrasing of Hartshorne's last sentence is maybe a little unclear. I interpret the phrase $\mathscr{F}(U)$ is equal to the set of all continuous sections of $\mathop{\mathrm{Spe}}(\mathscr{F})$ as meaning the canonical homomorphism $\mathscr{F}(U)\to \{\text{continuous sections }U\to \mathop{\mathrm{Spe}}(\mathscr{F})\}$ sending $s\in ...


1

Yes, one can avoid the use of Nullstellensatz. Let $I=(Y-X^2)$, $\alpha:k[X,Y]\to k[X,Y]/I$ be a natural homomorphism. Then we have two obviuous facts: $\alpha(k[X])=k[X,Y]/I$; $k[X]\cap I=\{0\}$. It follows, that restriction $\alpha'=\alpha|_{k[X]}$ epimorphic and injective, hence $\alpha':k[X]\to k[X,Y]/I$ is an isomorphism. Verification of statement ...


1

Write $z = a/b$, with $a, b \in \Gamma_{h}(V)$ forms of the same degree $d$. Note that $\overline{b}z = \overline{a} \in \Gamma_{h}(V)$, and therefore $b \in J_{z}$. Let $P \in V(J_{z})$. Then $F(P) = 0$, for every polynomial $F \in k[X_{1}, \ldots, X_{n+1}]$ such that $\overline{F}z \in \Gamma_{h}(V)$. Since $b \in J_{z}$, we have $b(P) = 0$, and thus $z$ ...


1

If your curve is a local complete intersection, then one can show in general: Assume $Z$ is a local complete intersection of codimension $m$ in an algebraic variety $Y$. Let $F$ and $G$ be coherent sheaves on $Z$ and assume that $F$ is locally free, then one has: $\mathcal{E}xt^k(i_{*}F,i_{*}G)=i_{*}(\Lambda^k N_{Z/Y}\otimes F^{\vee}\otimes G)$ for $0\leq ...


1

To say that $$[f(t)]^e[g(t)]^f$$ with $e+f\le m$ form a linearly dependent set in $k[t]$ is the same as saying that there are $A_{e, f} \in k$, not all zero, so that $$\sum_{e + f \le m} A_{e, f}[f(t)]^e[g(t)]^f = 0$$ (I suppose you are saying that the set is dependent in $k[t]$, treating $k[t]$ as a $k$-vector space). Thus the curve $(f(t), g(t))$ lie ...



Only top voted, non community-wiki answers of a minimum length are eligible