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Questions tagged [zariski-topology]

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

The Zariski topology allows using tools of topology for the study of algebraic varieties, even when the underlying field is not a topological field.

This is one of the basic ideas of scheme theory, which allows one to build general algebraic varieties by gluing together affine varieties in a way similar to that in manifold theory, where manifolds are built by gluing together charts, which are open subsets of real affine spaces.

The Zariski topology of an algebraic variety is the topology whose closed sets are the algebraic subsets of the variety. In the case of an algebraic variety over the complex numbers, the Zariski topology is thus coarser than the usual topology, as every algebraic set is closed for the usual topology.

See also: Zariski topology at Wikipedia.

417 questions
86
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Why Zariski topology?

Why in algebraic geometry we usually consider the Zariski topology on $\mathbb A^n_k$? Ultimately it seems a not very interesting topology, infact the open sets are very large and it doesn't satisfy the Hausdorff separation axiom. Ok the basis is…
Dubious
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25
votes
1 answer

Principal $G$-bundles in Zariski vs étale topology

Let $G$ be an (affine) algebraic group over say $\mathbb{C}$. A principal $G$-bundle is a scheme $P$ with a $G$-action and a $G$-invariant morphism of schemes $\pi:P\to X$ that is étale locally on $X$ isomorphic to the trivial $G$-bundle $U\times G…
alex
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Are there Zariski-continuous maps between algebraic sets that are not polynomial maps?

Suppose that $S_1$ and $S_2$ are the vanishing sets of a system of polynomial equations in n variables over a field $\mathbb{k}$ (ideal in $\mathbb{k}[X_1,\dots,X_n]$) and a system of polynomial equations in m variables over a field $\mathbb{k}$…
Vladimir Sotirov
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14
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1 answer

Is the construction of Zariski topology from polynomial rings functorial?

Given a polynomial ring $k[X_1,...,X_n]$ over a field $k$, we can consider the space $k^n$ equipped with Zariski topology whose closed sets are exactly the algebraic sets. Is this construction functorial ? I.e. given two polynomial rings over…
user456828
13
votes
2 answers

How does one find the Zariski closure of a set?

I've started to learn algebraic geometry this week (so I do not have much knowledge in the subjet) and, after reading the definition of the Zariski closure $V(I(S))$ of a set $S$, I've tried to do the following exercises without much success: Find…
u1571372
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13
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1 answer

General linear group is an affine variety

I am trying to prove that the general linear group $GL(n)$ is an $\underline{\text{affine}}$ variety. Unfortunately, I am having trouble with showing that $GL(n)$ is indeed affine. Before I show my progress I present the definitions as given to me…
Marc
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13
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2 answers

Why Zariski topology is not Hausdorff

I am reading the book about Algebraic geometry. I am confused about the following two things the book mentioned: Zariski topology is 1. different from the topology studied in real and complex analysis. 2. not Hausdorff. Well, I roughly now about…
sleeve chen
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12
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2 answers

Is there an intrinsic or natural topology on a field, motivating the definition of the Zariski topology

Let $k$ be a field. The usual motivation for the Zariski topology on affine space $\mathbb{A}^n(k)$ is that it is the coarsest topology for which the algebraic sets, the zero loci of polynomials, are closed. This can be phrased in more topological…
ziggurism
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1 answer

Is the axiom of choice necessary to prove that closed points in the Zariski topology are maximal ideals?

I would like to solve the beginner's standard exercise which claims that a point of $\mathrm{Spec} \ R$ is closed iff it is a maximal ideal. The reverse implication is easy. The direct one seems much subtler. If $P = V(I)$ then, since there exist a…
Alex M.
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1 answer

zariski continuous function with regular restrictions that is not regular in any exponent

I've been stuck on this for a few days. I'm supposed to find an example of a continuous function $f$ (with values in the field) defined on an affine variety $V=V_1\cup V_2$ with two irreducible components, such that the restrictions to each…
Feelix
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11
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2 answers

The role of the Zariski topology in algebraic geometry

I am having trouble understading the relevance of the Zariski topology being a topology. Every time I see the proof that sets of the form $V(I)=\{p\in\mathbb{A}^n\mid f(p)=0 \ \forall f\in I\}$ consitute a topology on the affine space…
Marco Flores
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0 answers

Finding irreducible components of Spec$(R/I^n)$

Let $R= k[x,y,z]/(xy,yz,zx)$. Let $I=(x)$. What are the irreducible components of $\mathrm{Spec}(R/I^n)$ where $n \geq 2$ and $k$ is a field? For solving this problem I'm trying to use following exercise from Atiyah and Macdonald book: Let $A$ be…
Arpit Kansal
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10
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1 answer

Why are 1 and -1 eigenvalues of this matrix?

This is a subject I've been working on for a very long time now, but still did not manage to fully understand the interesting properties of this matrix. I have already asked a (viewed but unanswered) question about the same matrices (cf. here), but…
anderstood
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9
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2 answers

Zariski topology, non-empty intersection of open sets

Let $k$ be a field and $X=\mathbb k^n$. A subset $Y \subset X$ is closed if there are $f_1,\ldots,f_m \in k[x_1,.\ldots,x_n]$ such that $Y=\{a \in k^n :f_i(a)=0 \space \forall \space i\}$, we write $Y=V(f_1,\ldots,f_m)$. The Zariski topology is…
user156441
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1 answer

Why do we take the closure of the support?

In topology and analysis we define the support of a continuous real function $f:X\rightarrow \mathbb R$ to be $ \left\{ x\in X:f(x)\neq 0\right\}$. This is the complement of the fiber $f^{-1} \left\{0 \right\}$. So it looks like the support is…
user153312
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