# Tag Info

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 ...

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