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## Hot answers tagged algebraic-geometry

4

A morphism $\phi: \mathbb{P}^m \rightarrow \mathbb{P}^n$ can be given by $\phi =(F_0: \dotsc : F_n)$, where the $F_i$ are homogeneous polynomials of same degree in $m$ variables (See Remark 3.2 on the same chapter of Silverman). The only problem is that the $F_i$ might have common zeros in $\mathbb{P}^m$. Say $P$ is one such point, then $\phi(P) = (0: \... 3 Ok Slup, here goes. Let$R$be any commutative ring and let$A$be a polynomial ring over$R$. Let$P$be any projective module over$R$. Then Quillen (and Suslin a bit later in this generality) proved that if for every maximal ideal$\mathfrak{m}$of$R$,$P_{\mathfrak{m}}$is of the form$Q\otimes_{R_{\mathfrak{m}}} A_{\mathfrak{m}}$for some projective$...

3

A handful for $n$ odd 13 a: 230153 b: 12792 c: 283945 d: 284233 u: 507 v: 164 15 a: 32625 b: 3472 c: 61455 d: 61553 u: 56 v: 31 23 a: 523367 b: 57072 c: 1413145 d: 1414297 u: 984 v: 667 27 a: 206703 b: 3848 c: 231345 d: 231377 u: 156 v: 37 89 a: 11534489 b: 700920 c: 63439289 d: 63443161 u: 649 v: 540 105 a:...

3

Note that $\dim k[x,y]=2$. A prime ideal $I\subset k[x,y]$ is of height 2 if $I$ is a maximal ideal. Now the ideal $I=(y^4+x^4-x^2)$ is clearly not maximal since for example $I\subset (x,y)$. This shows that $\mathrm{ht}\: I\neq 2$. On the other hand $\mathrm{ht}\: I\neq 0$ because $k[x_1, x_2]$ is an integral domain. This shows that $\mathrm{ht}\: I = 1$, ...

3

Consider the natural morphism $f:G\times C\to X$ and let $E=f^{-1}D$. Then $E$ is a closed subset of $G\times C$. Then consider the proper map, $E\to G$. Transitivity says that $E$ is codimension one and thus the morphism $E\to G$ has fibers either finite or one dimensional. Not all fibers are one dimensional by $\dim E\leq \dim G$. Then the set of points ...

3

Consider the map $f:k[u,v]\rightarrow k[x,y]$ sending $u\mapsto x^2$, $v\mapsto xy$. Note that $f(u)$ and $f(v)$ are homogeneous polynomials of degree $2$. Also $f$ maps the monomials $u^av^b$ to distinct monomials in $k[x,y]$, so $f$ is injective. Thus $I=0$ and every point is a common zero of $I$. However there are no $x,y\in k$ for which $(x^2,xy)=(0,1)$.

3

If we take the definition of a (global) lefschetz number as in $7.1$ of these notes, then the proof goes as follows: If $f$ is homotopic to $g$, the inclusion maps $id×f$ and $id×g$ of graph $f$ and graph $g$ are homotopic. Thus the global Lefschetz number of a map is homotopy invariant.

2

Theorem: If Spec$(R)$ is Noetherian, then so is Spec$(R[X])$. [Theorem 2.5 in Rings with Noetherian spectrum'' by Ohm and Pendleton] So for the example, take $R[X]$, where $R$ is any non-Noetherian ring with Noetherian spectrum.

2

Let us say that a ring $A$ satisfies condition $(S)$ if every finite type projective $A$-module is free. Clearly a necessary condition for $(S)$ to hold is $K_0(A)=\mathbf{Z}$. Example. (i) If $A$ is local Noetherian then $(S)$ holds. (ii) If $A$ is a Dedekind domain then $K_0(A)=\mathbf{Z}\oplus\mathrm{Pic}(A)$, and thus a necessary condition for $(S)$ is $... 2 First of all you must assume more. Namely that a section corresponds to a monomorphism: $$\mathcal{O}_X\rightarrow \mathcal{O}_X(D)$$ You may clearly get rid of this by assuming that$X$is integral. Then there is a bijection between set of such sections up to a scaling by$\Gamma(X,\mathcal{O}^*_X)$and effective divisors linearly equivalent with$D$. My ... 2 An Attempt You have$a^2+\left(n^2+1\right)b^2=d^2$. All integral solutions$(a,b,d)$to this equation satisfies (1)$|a|=|d|$and$b=0$, or (2)$d\neq 0$and$\left(\frac{a}{d},\frac{b}{d}\right)=\left(\frac{\left(n^2+1\right)r^2-1}{\left(n^2+1\right)r^2+1},\frac{2r}{\left(n^2+1\right)r^2+1}\right)$for some$r\in\mathbb{Q}$. Now, if$b^2+c^2=d^2$, then (... 2 Following Mike Miller's suggestion, consider the cylinder$X =S^1 \times \mathbb{R}$(as a Riemann surface, you may view it as either$\mathbb{C} \setminus \{0\}$or$\mathbb{C}/\mathbb{Z}$). As this deformation retracts onto the base circle and homology is a homotopy invariant, we know that$H_2(X;\mathbb{C}) \cong H_2(S^1;\mathbb{C}) =0$. As$\mathbb{C}$... 2 Use Eisenstein to deduce that$y^3-x^3(x+1)$is irreducible. As an alternative, note that the polynomial (in the variable$y$) is primitive, of degree$3$and has no roots. Then invoke the Gauß lemma. 2 No, it's not right. To do relative deRham cohomology, in your relative cochain complex$C^k(X,A)$consists of$k$-forms on$X$whose pullback to$A$vanishes. This is a bit of a subtle game. But, in particular, when$A$is a point, any$k$-form with$k\ge 1$will pull back to be$0$. It's hard to find references for this, but I know of a book (in French) by ... 1 It should be true for any PID. See the book by Lam, Serre's Conjecture 1 Let$U_i = \{ z_i \neq 0 \} = \{ z_i=1 \}$. By defintion we have that$\Gamma(U_i, \mathcal O_{\mathbb P^n}(1))$is the degree$1$-part of the localization$k[z_0, \dotsc, z_n]_{z_i}$, i.e. the sections are of the form $$\frac{f(z_0, \dotsc, z_n)}{z_i^{\deg f -1}}.$$ After letting$z_i=1$, this is nothing else but an arbitrary polynomial in the$z_j$'s with ... 1 It suffices to show that if$\phi_U$is dominant, then$\phi_{U\cap V}$will also be dominant for$U\cap V \neq \emptyset$. Note that $$\phi_{U\cap V}(U\cap V) = \phi_U(U\cap V).$$ Now, suppose for a contradiction that there is some nonempty open set$W\subset Y$such that$W\cap \phi_{U\cap V}(U\cap V) = \emptyset$. By the above equality,$W\cap \phi_U(...

1

First, we show that any such curve $X$ must lie on a quadratic surface. Consider the sequence$$0 \to \mathcal{I}_X(2) \to \mathcal{O}_{\mathbb{P}^3}(2) \to \mathcal{O}_X(2) \to 0.$$We have $\text{dim}\,H^0(\mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(2)) = 10$, so if $\text{dim}\,H^0(X, \mathcal{O}_X(2)) < 10$, then $\text{dim}\,H^0(\mathbb{P}^3, \mathcal{I}... 1 I have this weird thing where I stare at something for two hours, rage quit, then I get on stackexchange and type up the question. Then right after I post the question I realize how to do it. The functions$f, g, h$are just the compositions of$k^{\ast} \rightarrow k \times k^{\ast} \times k$with the projection maps. 1 If the number is rational, using multiplication formulas for$\cos$and$\sin$you can express the original parametrization in terms of$\cos$and$\sin$of multiples of a single argument. Then you can use a rational parametrization of the circle to replace those$\cos$and$\sin$by rational functions of the parameters, and then you can eliminate the ... 1 Consider an ideal$I=(f_1,\dotsc, f_s)$in a noetherian factorial ring and let$f$be the greatest common divisor of the$f_i$. Then$I$is principal if and only if$f \in (f_1, \dotsc, f_s)$holds. In practice, working in the polynomial ring, figuring out whether this holds, you need Groebner bases. Thus, in practice, you can also directly use the ... 1 'Diophantine Geometry: An Introduction', by Silverman and Hindry, contains a proof in section A.4.3 (Page 74). A few minor steps are skipped, but it should be fairly easy to follow. 1 We want to show that stability of an individual map$(C,\left\{p_i\right\},\mu:C \to \mathbb P_k^r)$in the family is equivalent to ampleness of $$\omega_C(p_1 + \cdots + p_n)\otimes \mu^*(\mathcal O_{\mathbb P_k^r}(3))$$ on the curve$C$. I'll do one direction. If you know a bit about the basic theory of stable curves, you know that a pointed curve is ... 1 The category of affine schemes is the opposite of the category of commutative rings, and taking opposites switches limits and colimits, so in general the categorical quotient$\left( \text{Spec } R \right) / G$is the spectrum$\text{Spec } R^G$of the fixed point subring. Hence you want to compute the subring of$R = k[x, y]$($k$a field containing$\omega$... 1 Your proof uses the following claim (paraphrasing slightly): If$r$is a linear combination of monomials with coef in$\mathbb C$, where no monomial is divisible by$uv$, then$r$must be a polynomial in$w$alone. This is not true. For a counterexample, consider$r(u,v,w)=2u+v^3+vw$. This is not a polynomial in$w$alone, but none of its monomials are ... 1 First of all, in order that the subset$M\subset \mathbb R^ndefined by the system \begin{align*} f_1 = 0 \\ \cdots \\ f_k = 0 \end{align*}be a submanifold you have to assume that the Jacobian matrixJ(f)(x)=(\frac{\partial f_i}{\partial x_j}(x))$has rank$k$at every$x\in M$. If this is the case the tangent space to$M$at$x$consists of the$n-...

1

Since $\mathbb{P}^n$ is a rational homogeneous variety this follows from the fact that an exceptional sheaf is already a homogeneous bundle, see for example Proposition 2.1.4 here.

1

Assume there are polynomials $f$ with $\phi(\overline f)=0$ and $f\notin (y-x^2,z-x^3)$. Among these. let $f_0$ have minimal degree in $z$. If $f_0$ has positive degree $k$ in $z$, say $f(x,y,z)=p_0(x,y)+p_1(x,y)z+\ldots +p_k(x,y)z^k$, then $f_0(x,y,z)-p(x,y)z^{k-1}x^3$ is equivalent to $f_0$ but of lower degree in $z$. We conclude that $f_0$ has degree $0$ ...

1

The product in a category which has a terminal object is the fiber product over the terminal object.

1

I don't know how to answer for algebraic groups in general, only affine algebraic groups over an algebraically closed field $k$. Remember that an affine algebraic group is up to isomorphism the same thing as a closed subgroup of $\textrm{GL}_n$ for some $n$. However, an isomorphism of such a group into $\textrm{GL}_n$ is not canonical. Question 1: For any ...

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