Questions tagged [zariski-topology]

For questions about the topology of schemes and (classical) algebraic varieties.

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How exactly does Zariski-Topology relate to Standard topology on $\Bbb{C}$?

I'm currently self-studying algebraic geometry and I tried to construct some proofs / counterexamples for relations between open / closed sets in Zariski and Standard Topology on $\mathbb{C}$ but I'm ...
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Let $G$ be a linear algebraic group and $C_G(x)$ be the centralizer of $x$. Show that $C_G(x)$ is a closed subgroup for all $x\in G$.

Let $G$ be a linear algebraic group and $C_G(x)$ be the centralizer of $x$. Show that $C_G(x)$ is a closed subgroup for all $x\in G$. I want to enstablish a homomorphism from $G$ to $G$, and the ...
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$X\times Y$ and the product topology [duplicate]

I'm studing Andreas Gathmann's notes on algebraic geometry (pdf here: https://agag-gathmann.math.rptu.de/de/alggeom.php). In chapter 4 (about Morphisms) he was using the universal property of products ...
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Can $\operatorname{Spec}(R)$ be not homeomorphic to $\operatorname{Spec}(S)$ but be isomorphic as a poset?

Looking at the definition of the spectrum of a ring made me wonder when/whether is its topology determined by the poset of prime ideals of the ring. Let $R$ and $S$ be commutative unital rings. Can ...
Cork Screwdriver's user avatar
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If the product of closed sets is closed in the product topology shouldn't {(0,0)} be closed in $A^1 \times A^1$ zariski topology.

I am not sure what is incorrect about the statement. If the product of closed sets is closed in the product topology shouldn't {(0,0)} be closed in $A^1 \times A^1$ zariski topology, i.e singletons ...
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Is "Zariski connected" equivalent to "ring connected"?

Let A be a ring, in class we defined the following two definitions of connected: A is Zariski connected if Spec A is connected with the Zariski topology. A is ring connected if $$a+b=1, ab=0\...
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Commutative algebra: Irreducible components of $\operatorname{Spec}(A)$

Let $k$ be a field of characteristic other than $2$. Consider the ring $$A:= k[X,Y]/(X(Y+1),X(Y+X^2)).$$ Describe all the irreducible components of $\operatorname{Spec}(A)$. It does not look that ...
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Curves homeomorphic under Zariski topology

Prove any two curves over some field $k$ are homeomorphic, where $k$ might not be algebraically closed. Curves are defined to be varieties (integral separated scheme of finite type) of dimension $1$. ...
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When the Zariski topology is $T_1$ (depending on $k \ge 1)$.

Please explain when the Zariski topology is $T_1$ (depending on $k \ge 1$). Let $X$ be a topological space and let $x$ and $y$ be points in $X$. We say that $x$ and $y$ are separated if each lies in a ...
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The Lie algebra of zariski closure of an algebraic group.

I think this question should be pretty straightforward but through a combination of being self-taught in group theory and being awful at geometry, something is escaping me. I am trying to understand ...
Yushi MuGiwara's user avatar
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The Zariski topology on $\mathbb{K}^n $ is Noetherian.

A topological space is Noetherian if there are no infinite descending chains of closed sets. Show that the Zariski topology on $\mathbb{K}^n $ is Noetherian. I know and understand what is topological ...
Artur111's user avatar
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Let $f:A→B$ be a ring morphism and $φ : \operatorname{Spec}B→\operatorname{Spec}A$ its induced continuous map, then prove that $\rm im\ φ ⊆ V(\ker f)$

Today I was asked this question by my teacher: Let $A$,$\ B$ be two conmutative rings with unity lef $f :A \rightarrow B$ be a ring morphism and $φ : \operatorname{Spec}(B) \rightarrow \operatorname{...
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Dimension of Zariski closure of an infinite discrete group

Let $ \Gamma $ be a subgroup of $ G:=GL(n,\mathbb{C}) $ which is discrete with respect to the manifold topology on $ G $. Let $ \overline{\Gamma} $ be the Zariski closure of $ \Gamma $. Suppose that $ ...
Ian Gershon Teixeira's user avatar
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${\rm Spec}(A/\mathfrak{a})={\rm Spec}(A)$ if and only if $\mathfrak{a}$ is generated by nilpotent elements

I have been introduced to the Zariski topology and I cant solve this problem: Let $A$ be a commutative ring with unity and $\mathfrak{a}$ an ideal of $A$, we define ${\rm Spec}(A) $ as the set of ...
Pablo Borrego's user avatar
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Showing the linear algebraic subgroup $\Bbb U_n$ of $\Bbb{GL}_n(k)$ is closed.

This is part of Exercise 2.1.5(2) of Springer's, "Linear Algebraic Groups (Second Edition)". It is likely to be a simple question. Fix an algebraically closed field $k$. Let $n\in \Bbb N$. ...
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Zariski topology and morphisms that coincide on a dense set

In Miles Reid Undergraduate Algebraic Geometry book it is stated informally, about the Zariski topology: (1) two morphisms which coincide on a dense open set coincide everywhere I am suprised that ...
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Geometric Picture of Open Subsets of Irreducible spaces in Zariski topology

I understand why nonempty open subsets of irreducible topological spaces are dense. I am having trouble comprehending how this translates to Zariski's topology. We know that a finite affine variety is ...
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Determining whether or not a set is algebraic/ irreducible? [duplicate]

I have an understanding of Affine Varieties, Irreducibility/Connectedness, and Dimension in terms of set notation via Zariski Topological Spaces, but I struggle to apply it to actual examples. Does ...
Carson Newman's user avatar
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Serre's condition $(S_n)$ locus is stable under generalization?

Let $R$ be a Noetherian ring and $n\ge 1$ an integer. Consider the set $S_n^R:=\left\{\mathfrak p \in \text{Spec}(R) \text{ }| \text{ } \text{depth } R_{\mathfrak p} \ge \inf \{n, \dim R_{\mathfrak p}\...
strat's user avatar
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Propetries of ideals and zero sets

Why we obtain $I(X_1 \cup X_2) = I(X_1) \cap I(X_2)$ and $Z(J_1 \cup J_2)= Z(J_1) \cap Z(J_2)$? for a subset $X⊆\mathbb{A}^n,$ $I(X)$ is the ideal of $f∈k[x_1,⋯,x_n]$ with $f|X=0$ and for a subset $J⊆...
Artur111's user avatar
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Is a product of infinite subsets of $\Bbb C_p$ Zariski dense in $\Bbb C_p^2$?

Consider the $p$-adic number field $\mathbb Q_p$. Let $\mathbb C_p:=\widehat{\bar{\mathbb Q}_p}$ with maximal ideal $\mathfrak{m}$. Then any infinite subset $S \subset \mathfrak m$ is Zariski-dense in ...
ANG's user avatar
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Krull dimension vs. Dimension in Zariski topology for a field that is not algebraically closed

I'm new to algebraic geometry and struggle with the notion of the dimension of an affine variety. In Robin Hartshorne's book Algebraic Geometry the dimension is introduced using the Zariski topology: ...
joap's user avatar
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Motivating Zariski topology

I always find it difficult to motivate Zariski topology on $\text{Spec}(k[X_1,...,X_n])$, both the underlying set and the topology. As far as I understand, Zariski topology isn't really that essential ...
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Zariski closure of $\{ (z, \bar z) : z\in\Bbb C \}$

The question I am intersted in is: If a polynomial $p\in\Bbb C[x,y]$ has the property $\forall z\in\Bbb C: p(z,\bar z)=0$, is $p$ automatically the zero polynomial? Equivalent formulations: Is every ...
Grinsekotze's user avatar
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Is it Zariski closed: The set of matrices for which the first column is in the span of the other columns

Is the set of matrices $M \in \mathbb{C}^{n \times m}$ for which the first column of $M$ is contained in the span of the other columns of $M$ Zariski closed?
Ben's user avatar
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How is the Zariski topology (?) being used in the proof of the KAN decomposition of a semisimple Lie group?

On page 13 of this PDF the existence of KAN decompositions of semisimple Lie groups is proven. The proof uses the fact that the matrix multiplication map is regular to conclude that the image of $(K,A,...
wlad's user avatar
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spec(-) is a contravariant functor

Claim: Let $\varphi: A\rightarrow B$ be a ring homomorphism. Then there exists a continuous map $$f:SpecB\rightarrow SpecA: P\mapsto \varphi^{-1}(P)$$ My thoughts: It suffices to check the preimage of ...
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Fleshing out a proof that $\phi$ is dominant iff $\phi^*$ is injective.

This is a solution-verification/proof-explanation question, since I have a proof that I need to flesh out; I am unsure of a couple of steps. The Details: Most of the terminology here is given in this ...
Shaun's user avatar
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Computing a local parameter: don't understand a step

In trying to compute a local parameter of some projective curve, this paper I am reading uses the fact that if ${C_Z}$ is an affine variety, then the inclusion (which is a morphism) ${i : C_Z\...
CoffeeBean's user avatar
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Existence of morphism from a locally ringed space $X \to Spec(\mathbb{F}_p)$

Let $X$ be a topological space, such that $(X, \mathcal{O}_X)$ is locally ringed. Let $A$ be a ring. We showed in the lecture that there is a natural bijection of $Hom((X, \mathcal{O}_X),(Spec(A), \...
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Ideal-vanishing locus correspondence in affine schemes as natural transformations

$\DeclareMathOperator{\Spec}{Spec}$$\newcommand{\pf}{\mathfrak{p}}$Let $A$ be a ring (commutative, with $1$), then its spectrum $\Spec A$ is the set of prime ideals of $A$. It can be turned into a ...
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Quasicompactness in terms of morphisms and $\operatorname{Spec} \mathbb{Z}$

In The Rising Sea, Vakil states the following right before exercise 8.3.B (p. 231): Following Grothendieck’s philosophy of thinking that the important notions are properties of morphisms, not of ...
Lorenzo Andreaus's user avatar
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A morphism of affine varieties $\phi: X\to Y$ is an isomorphism iff the algebra homomorphism $\phi^*$ is an isomorphism.

This is Exercise 1.4.8(3) of Springer's book, "Linear Algebraic Groups (Second Edition)". The Question: A morphism of affine varieties $\phi: X\to Y$ is an isomorphism if and only if the ...
Shaun's user avatar
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How to determine whether the following subsets are open, closed, or dense in the Zariski topology?

Let $k=\mathbb{C}$ and decide for each of the following subsets $S \subset \mathbb{A}_{\mathbb{C}}^2$ whether they are closed, open or dense in the Zariski topology: (a) $S=\{(t, s t) \mid s, t \in \...
Bill's user avatar
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Zariski open set in $\mathbb{\bar Q}^2$ and a question on shifted-sparse interpolation

In [1 (linked PDF here)], there is mention of a Zariski open set in Corollary 3 and they mention that "generic" means $(b_1, b_2)$ belongs to a Zariski open set in $\mathscr{\bar F}^2$. ...
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How is a morphism between projective curves the union of morphisms of affine curves?

I am reading a paper, and it uses the following definition: A morphism ${f : C \to C'}$ of affine curves is the restriction to C of a map of the following form, where ${f_1}$ and ${f_2}$ are elements ...
CoffeeBean's user avatar
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Atiyah MacDonald Exercise 1.21 iv) Proof of continuity

Let $\Phi: A \rightarrow B$ be a surjective ring homomorphism. Let $X = \text{Spec}(A)$ and $ Y = \text{Spec}(B)$, then $\Phi^*=\Phi^{-1}: Y \rightarrow V(Ker(\Phi))$. I already showed that $\Phi^*$ ...
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Proving a set is algebraic.

I want to investigate about if the set of points of the form $(x,1)$ with $x\neq 0$ (horizontal line without a point) is closed or not in $\mathbb{A}_k^n$ with the zariski topology. My intuition say ...
3435's user avatar
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Dimension of affine varieties preserves for open subsets

My problem: Let $X$ be an affine variety and $U$ an open subset of $X$. Then $dim U = dim X$. My attempt: If I take a chain of irreducible and closed sets $Z_1\subsetneq... \subsetneq Z_n$ in $U$, ...
3435's user avatar
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When is there a one to one correspondence between closed sets in the Zariski topology of the prime spectrum of a ring and radical ideals of that ring?

My current understanding: this is always true Let $A$ be a ring. Let $S$ denote a subset of the ring. Let $V(S)$ be the set of prime ideals of $A$ containing $S$. Define closed sets in $X= \...
Mr Lolo's user avatar
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Understanding $\mathcal V(I)$, $\mathcal I(X)$, and their relationship to each other.

The Details: Since definitions vary: A topological space $(X,\tau)$ is a set $\tau$ of subsets of $X$, called closed subsets, such that $\varnothing, X\in\tau$, The intersection $$\bigcap_{i\in I}...
Shaun's user avatar
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Let $k$ be an algebraically closed field. Show that $\{(x,y)\in k^2\mid xy=0\}$ is closed and connected but not irreducible in $k^2$.

This is Exercise 1.2.8(3) of Springer's, "Linear Algebraic Groups (Second Edition)". According to this search on Approach0, it is new to MSE. The exercise comprises of three parts, whose ...
Shaun's user avatar
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Can I recover the Zariski open subobjects from the Grothendieck topology they generate?

Let $\mathbf{cRing}$ be a category of commutative rings and let $\mathbf{Set}$ be a category of sets relative to which $\mathbf{cRing}$ is small (Grothendieck universes). The opposite $\mathbf{Aff}$ ...
Nico's user avatar
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Dense open subset of a topological space with strictly smaller dimension

An exercise in the first chapter of Hartshorne asks you to give an example of a topological space $X$ and a dense open subset $U$ in $X$ such that $\dim U < \dim X$. I have an example, but it feels ...
stillconfused's user avatar
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1 answer
112 views

What do we gain by considering spectra?

There are two topologies which are known as the "Zariski topology" - one of them a topology on an algebraic set (henceforth a 'variety') $V$ in $A^n(k)$, and the other on $Spec(R)$ where $R$ ...
ducksforever's user avatar
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How to show irreducible closed subsets of Spec are vanishing sets of primes?

I would like to show if $R$ is a comm ring, and $Y \subset\operatorname{Spec}R$ is a closed irreducible subset, then there is a prime ideal $p \subset R$ such that $Y = V(p)$, i.e. $Y = \{ q \in \...
Indraneel Tambe 2's user avatar
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Does the Zariski-closure of an irreducible subgroup $G$, also act irreducible?

Let $G \subset GL(V)$ be a subgroup for a complex vectorspace $V$. Let $H=\overline{G}$ be its Zariski-closure. Then I want to check if the following statement (1) implies statement (2): $G$ acts ...
Lucy's user avatar
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Is a closed subgroup of a linear algebraic group, again a linear algebraic group?

Let $G$ be a linear algebraic group over an algebraically closed field $K$. For a subgroup $H \subset G$, we denote with $\overline{H}$ the closure of $H$ in $G$ with respect to the Zariski topology. ...
user12345's user avatar
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Definition of spectral spaces using closed set

As we know in any spectral space the compact open subsets necessarily form a basis of the topology and are closed under finite intersection. Is there any equivalent statement using closed sets instead ...
Mona's user avatar
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Proof of Zariski lemma in Wikipedia

That wikipedia page has a proof of Zariski's lemma, which states that if $A$ is a Jacobson ring (in particular when $A$ is a field) and $B$ is a finitely-generated algebra of $A$, which is a field, ...
uri gluck's user avatar

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