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    $\dim \PP^n = n$, and a quasi-projective variety has the dimension of its closure

    P-AGH27DIMPN

    Area
    Algebraic Geometry
    Topics
    Dimension, Projective Varieties, Quasi-Projective Varieties
    Status
    draft

    Let \(k\) be algebraically closed and let \(n\ge0\).

    1. Show that \(\dim \PP_k^n = n\).

    2. If \(Y \subseteq \PP_k^n\) is a quasi-projective variety, show that \(\dim Y = \dim \bar{Y}\).

    Hint

    For (b), use Exercise I.2.6 to reduce to Proposition I.1.10.

    Solution

    Use dimension defined by lengths of strict chains of nonempty irreducible closed subsets. Let \(U_i=D_+(x_i)\subseteq\PP_k^n\) be the standard affine opens.

    1. If \((V_i)\) is an open cover of a topological space \(T\), then \(\dim T=\sup_i\dim V_i\).

    A strict chain of irreducible closed subsets of \(V_i\) gives such a chain in \(T\) by taking closures. The closures remain distinct, since their intersections with \(V_i\) recover the original subsets. This proves \(\dim V_i\le\dim T\).

    Conversely, take a chain \(Z_0\subsetneq\cdots\subsetneq Z_d\) of nonempty irreducible closed subsets of \(T\). Choose a member \(V_i\) of the cover meeting \(Z_0\). It then meets every \(Z_j\). Each intersection \(Z_j\cap V_i\) is nonempty, irreducible and closed in \(V_i\). It is dense in \(Z_j\), because it is a nonempty open subset of an irreducible space. Thus these intersections are distinct: equality of two would give equality of their closures in \(T\). They form a chain of the same length in \(V_i\). Taking the supremum over chains proves the reverse inequality.

    2. One has \(\boxed{\dim\PP_k^n=n}\).

    The ratios \(x_j/x_i\), with \(j\ne i\), identify \(U_i\) with \(\AA_k^n\). The dimension of affine space is \(n\), since its coordinate ring is a polynomial ring in \(n\) algebraically independent variables [@Har10a, Proposition I.1.9]. The \(U_i\) cover \(\PP^n\), so step 1 gives the claimed dimension, including \(n=0\).

    3. For a quasi-projective variety \(Y\), one has \(\boxed{\dim Y=\dim\overline Y}\).

    Put \(W=\overline Y\subseteq\PP^n\). By the definition of a quasi-projective variety, \(Y\) is a nonempty open subset of its irreducible projective closure \(W\). For every \(i\) with \(W_i=W\cap U_i\ne\varnothing\), the set \(W_i\) is an affine variety in \(U_i\cong\AA^n\). The intersection \(Y_i=Y\cap U_i\) is nonempty and dense in \(W_i\), since two nonempty open subsets of the irreducible space \(W\) meet. It is a quasi-affine variety whose closure in \(\AA^n\) is precisely \(W_i\). The dimension result for quasi-affine varieties therefore gives \[ \dim Y_i=\dim W_i \] [@Har10a, Proposition I.1.10]. This applies the affine closure theorem only to an affine ambient space.

    The \(Y_i\) and \(W_i\) are open covers of \(Y\) and \(W\), respectively. Step 1 consequently yields \[ \dim Y=\sup_i\dim Y_i=\sup_i\dim W_i=\dim W. \] Equivalently, every nonempty \(W_i\) has the function field \(k(W)\) and dimension \(\operatorname{trdeg}_k k(W)\) by the affine dimension theorem; the same is then true of its dense open \(Y_i\).

    4. Q.E.D.

    Step 2 proves (a), and step 3 proves (b).

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    Dependencies

    • P-AGH26HOMDIMThe homogeneous coordinate ring satisfies $\dim S(Y) = \dim Y + 1$uses