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    The Zariski topology

    D-BIVAU

    (Zariski topology)

    The Zariski topology on \(\AA^n\) has as its closed sets the vanishing loci \[ V(J) \da \ts{ p \in \AA^n \st f(p) = 0 \text{ for all } f \in J } \] of ideals \(J \subseteq k[x_1,\ldots,x_n]\). The distinguished open set attached to \(f\) is \(D_f \da \AA^n \sm V(f)\), and these form a basis.

    On \(\AA^1\), and on any irreducible curve, the proper closed sets are the finite sets, so the Zariski topology is the cofinite topology. For \(\AA^1\) over \(k = \bar{k}\): since \(k[x]\) is a PID, every ideal is \((f)\), and factoring \(f\) into linear factors gives \(V(f) = \ts{a_1, \ldots, a_r}\) when \(f \neq 0\). Since \(k\) is infinite, any two nonempty open subsets of \(\AA^1\) meet, because their complements are finite; so \(\AA^1\) is irreducible and not Hausdorff.

    What are the closed subsets of $\AA^n$?
    What is a distinguished open set?
    What is the Zariski topology on $\AA^1$?
    Show that the Zariski topology on $\AA^1$ is not Hausdorff.
    Show that $\AA^1$ is irreducible without using the Nullstellensatz.
    Show that $\AA^n$ is irreducible.
    Show that $\dim \AA^1 = 1$ and $\dim \AA^n = n$.

    Backlinks

    • D-VKR54$\Spec A$ as a locally ringed spaceuses
    • PR-WZGOQHypersurface complements are affineuses
    • FE-ISIPRThe Zariski topology on $\AA^2$ is not the product topologyuses

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