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1

This follows immediately from Nakai-Moishezon's criterion: a Cartier divisor $D$ on a proper scheme $X$ is ample if and only if, for every subscheme $Y$ of $X$, one has $D^{\dim(Y)} \cdot Y > 0$. You can find this theorem for surfaces in Hartshorne's Algebraic Geometry or, more general, in Debarre's Higher-Dimensional Algebraic Geometry (in this last one, ...


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 $z$ ...


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

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 ...


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

Recall that a subset $F$ of a topological space $X$ is closed if and only if there exists an open cover $\{ X_{i} \}$ of $X$ such that $F \cap X_{i}$ is closed in $X_{i}$. Cover the Grassmannian variety $G(m-k, n)$ with open sets $U_{i_{1}, \ldots, i_{m-k}}$ consisting of $(m-k)$-dimensional vector subspaces of $\mathbb{K}^{n}$, which we represent by a ...


0

First, if $\mathrm{codim}(X) > n - 1$, then by the Trisecant Lemma $P \cap X = \{ x_{1}, \ldots, x_{n} \}$, and therefore, $P \cap X$ is not even a curve. Assume $\mathrm{codim} = n - 1$. In this case, we can apply the Trisecant Lemma only for $(n - 2)$-planes generated by $n - 1$ of the general points $x_{1}, \ldots, x_{n}$. Let $E$ be an irreducible ...


1

I think the problem lies in the definition of "maximal chain". It means, in my opinion, first "unrefinable", that is you can not insert irreducible sets in between elements of the chain, and secondly "unextendable", that is you can not prepend or append further irreducible sets. Now the fact that for an integral affine $k$-Algebra $A$ all maximal chains ...


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 ...


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 $s$ ...


0

This looks solid. Alternatively you can construct mutual inverse rational maps $\phi:\Bbb P^2\dashrightarrow\Bbb P^1\times\Bbb P^1$ and $\psi:\Bbb P^1\times\Bbb P^1\dashrightarrow\Bbb P^2$ using the homogeneous coordinates on $\Bbb P^2$ and $\Bbb P^1\times\Bbb P^1$. These homogeneous coordinates are described by \begin{align*} (t_0,t_1,t_2)\sim(\lambda\cdot ...


1

Every element of $R$ can be written as $G(x_1,\ldots,x_n)$ for $G\in k[X_1,\ldots,X_n]$. But since $d$ is a derivation of $R$ into $\Omega_k(R)$, we have $$d(G(a_1,\ldots,x_n))=\sum_{i=1}^n G_{X_i}(x_1,\ldots,x_n)dx_i.$$ Now every element of $\Omega_k(R)$ is of the form $\sum_i r_i d(s_i)$ for some $r_i\in R$ and some $s_i\in R$ by definition of the module ...


0

If $V$ is a union of two distinct non-empty closed subvarieties, then $V$ is not irreducible. Ignoring that, consider the variety $y^2=x^2$ made up of the two lines $y=x$ and $y=-x$ in the plane. They intersect at the origin $(0,0)$, and at this point the tangent space is 2-dimensional while the lines (the irreducible components) are 1-dimensional. This ...


0

If $P$ vanishes on a nonempty open subset $U$ of $Z(Q)$, then it vanishes on all of $Z(Q)$. This is simply because $Z(P)$ is closed, and if $U\subseteq Z(P)$, then $Z(Q)=\overline{U}\subseteq Z(P)$. The closure of any nonempty $U$ is equal to $Z(Q)$ because $Z(Q)$ is irreducible. Hence, $P\in \sqrt{\langle Q\rangle}=\langle Q\rangle$ as required. We choose ...


1

Consider $z(x)$ in the local ring of $\mathbb P^1$ at $x=0$. The local ring is $k[x]_{(x)}$, with maximal ideal $(x)$. This is a DVR so every function can be written as $x^e u$, where $u$ is invertible in the local ring. In your case $e=1$. For the local ring at $1-x$, your function can be written as $(1-x)^{-1} x$, and here $x$ is invertible. So the order ...


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 ...


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

If $F\in I(V)$, then $\overline{F} = 0$ in $\Gamma_h(V)$, so $\overline{F}z = 0$ too. Because $J_z\supseteq I(V)$, we have $V(J_z)\subseteq V\bigl(I(V)\bigr)$. By the Nullstellensatz, $V\bigl(I(V)\bigr) = V$.


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 $a = X^2$.


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 ...


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 ...


0

By definition, there exists an open subscheme $U$ of $X$ containing all closed points of $X$ such that $U \to Y$ is smooth. If $U \not= X$, then $X \backslash U$ is a closed subscheme and the morphism from $X$ to $Y$ is singular at each point on $X \backslash U$. But $X \backslash U$ is quasi-compact, hence contains closed points. This contradicts the first ...


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

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 ...


0

A morphism is called of finite type if it is quasi-compact and locally of finite type. If $f_i : X_i \to Y$ are morphisms of finite type for $i \in I$, then clearly the induced morphism $f : \coprod_{i \in I} X_i \to Y$ is locally of finite type, but it is usually not of finite type when $I$ is infinite. So, for example, if $A$ is any commutative ring $ \neq ...


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 $X$ ...


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 ...


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 ...


0

Most of this post is a quilt of examples and comments others have offered me when asked about formal schemes, which I've found very helpful. Hopefully, they'll help you. This is just following up on Sebastian's answer to explain why your intuition is right: the formal completion of E along the "0"-section is like the Taylor series expansion of E about the ...


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.


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 $$\Delta=\{([x_0:\cdots:x_n],[y_0:\cdots:y_n])\in\mathbb{P}^n\times\mathbb{P}^n:x_iy_j-x_jy_i=0,i,j=0,\ldots,n\}.$$ ...


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 ...


0

The theory is about the same in the complex analytic case, and in fact that setting is the original motivation, so it's helpful to study it there. I like (with some reservations) Rick Miranda's book "Algebraic Curves and Riemann Surfaces" for this.


0

You're just supposed to observe that there's a linear transformation taking any subspace to any other of the same dimension. One way to see this is to note that the matrix $$A = \left(\begin{array}{c|c|c|c} & & & \\ v_1 & v_2 & \cdots & v_n \\ & & & \end{array}\right)$$ gives $A e_i = v_i$, so we also have $A^{-1} ...


0

Yes, there is a "categorial" proof. We formulate the following Universal Property. Let $\mathscr{F} \times \mathscr{G}$ be the presheaf of sets $U \mapsto \mathscr{F}(U) \times \mathscr{G}(U)$. Let $f\colon \mathscr{F} \times \mathscr{G} \to \mathscr{H}$ be a map of preheaves that is $\mathcal{O}_X$-bilinear, that is, for $s \in \mathscr{F}(U)$, the map ...


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 ...


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.


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 ...


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 ...


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 ...


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.


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 ...


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 ...


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 ...


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 ...


1

The morphism in question comes from the surjection $B/I \to B/J$. If $f \in B$ is homogeneous of positive degree then we get $(B/I)_f \to (B/J)_f$ and then $((B/I)_f)_0 \to ((B/J)_f)_0$, both surjective. So the only question is injectivity. Say $b/f^n$ maps to $0$ where $b \in B_{n \deg f}$ (everything should be mod $I$, really). Then for $m \gg 0$ we have ...


0

The trick to this question is that we need to choose the point $p$ wisely. For the moment, I will choose $p$ to be a non-basepoint of your line bundle $\mathcal{L}$. Consider the short exact sequence of sheaves $$0 \to \mathcal{O}_C(-p) \to \mathcal{O}_C \to k(p) \to 0$$ where $k(p)$ is the skyscraper sheaf at $p$. Tensoring this with $\mathcal{L}$, we get ...


2

Ok, I have a somewhat ad-hoc construction. The key algebraic fact about smooth curves (to use instead of the fact that rational morphisms from smooth curves to proper varieties extend to genuine morphisms) is that a coherent sheaf on $C$ is locally free if and only if it is torsionfree. This is because $C$ is covered by affine charts which are the spectra of ...


1

I think what is going on is the following. The nonzero section $s$ determines a morphism $\varphi_s : \mathcal{O} \to \mathcal{V}$, while the subbundle $\mathcal{L} \subset \mathcal{V}$ is the image of $\varphi_s$ (see new edit). The point is that scaling $s$ by $\mathcal{O}^*$ changes the particular morphism $\varphi_s$ but leaves the image of $\varphi_s$ ...



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