Search Results

Let $S$ be some (fixed) subset of $\mathbb{Z} [X_1, \dots , X_n]$ which contains only homogeneous polynomials, and if $F$ is a field, let $X(F)$ be the set of $ x \in P^{n-1}(F)$ such that $f (x) = 0$ …

Let $X$ be a smooth projective scheme over $\mathbb{Z}_p$, with special fiber $X_s$ over $\mathbb{F}_p$, generic fiber $X_{\eta}$ over $\mathbb{Q}_p$, and geometric generic fiber $\bar{X_{\eta}}$ over …

I assume you mean $H^0(X, K_X)^{\otimes m}$ rather than $\oplus_{i=1}^m H^0(X, K_X)$. If $X$ is a smooth projective connected complex curve of genus $g \geq 2$, then the map $$H^0(X, K_X)^{\otimes m} …

Huneke only states an inclusion. But you are right, one gets a slightly stronger statement, with weaker assumptions. Namely, for any ring $R$, and any sequence $x_1,\dots,x_n$ a of elements of $R$, co …

An $n$-tuple $F=(F_1,\dots,F_n)$ is "unimodular" iff $\exists x \in k^n, F(x) \neq 0$ in $k$. Thus an equivalent question is: if $F =(F_1,\dots,F_n) \in k[X_1,\dots,X_n]^n$ satisfies $J(F) = 1$ and …

No (in general). Take $S = \mathrm{Spec}(A)$ and $X_k = \mathrm{Spec}(A[T]/(T^2))$, with affine transition maps given by $T \mapsto f T$ for some $f \in A$. The limit $X$ is the spectrum of $A \oplus …

Many expositions of Popescu's desingularization theorem indicate that an other proof of this theorem can be found in "Cinq exposés sur la désingularisation" by M. Andre, Ecole Polytechnique Feder …

In scheme theory applied to complex geometry one usually does not encounter coherent rings which are not noetherian as well. However if $X$ is (for example) a Stein manifold then the ring $R = \mathca …

The field $K((X))$ has infinite transcendence degree over $K$ (if $K$ is at most countable, this just follows from a cardinality argument). Thus we can find a countable family $(t_i)_{i \geq 1}$ of el …

Let $X_0$ be a smooth, geometrically connected scheme over $\mathbb{F}_q$. As usual, let $\tau : \bar{\mathbb{Q}}_{\ell} \simeq \mathbb{C}$ be a fixed isomorphism. Let $\mathcal{C}$ be the category of …

As $X$ is proper, the Swan conductor of $\mathcal{F}$ vanishes. Hence the identity $\chi(X,\mathcal F)=\operatorname{rk}\mathcal F\cdot\chi(X,\underline{\mathbb F_\ell}_X)$ follows from Theorem $4.2.9 …

No (in general). Let $A$ be a non-zero ring and let $S = \mathrm{Spec}(A[T]/(T^2))$. Let $M$ be a free $A$-module of infinite rank, viewed as an $A[T]/(T^2)$-module via the section $A[T]/(T^2) \righta …

Sure, by $(1)$ and $(4)$. Any integral morphism is affine by definition. If $f$ is an affine morphism with $f_* \mathcal{O}_X = \mathcal{O}_Y$, then $f$ is clearly an isomorphism.

The following is a particular case of (SGA 4.3, XVII Th. 5.5.21) : Let $X$ be a quasi-projective scheme over an algebraically closed field $k$. Then for any $n \geq 0$ and any $r \geq 1$ we have $$ R …

This is false. The noetherian local ring $R = \mathbb{F}_3[[X,Y]]/(Y^2 - X^3)$ has dimension one, and if $x,y$ are the images of $X,Y$ in $R$ then the sequence $$ R \supseteq (x,y) \supseteq (x^2,y) …