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    A projective variety has codimension one exactly when it is a hypersurface

    P-AGH28HYPERSURFACE

    Area
    Algebraic Geometry
    Topics
    Dimension, Height Of Ideals, Hypersurfaces
    Status
    draft

    Let \(k\) be algebraically closed and let \(n\ge0\). A projective variety \(Y \subseteq \PP^n\) has dimension \(n-1\) if and only if it is the zero set of a single irreducible homogeneous polynomial \(f\) of positive degree. In this case \(Y\) is called a hypersurface in \(\PP^n\).

    Solution

    Put \(S=k[x_0,\ldots,x_n]\) and \(I=I(Y)\), so \(S(Y)=S/I\) is the homogeneous coordinate ring. The projective ideal correspondence makes \(I\) a homogeneous prime ideal and gives \(Z(I)=Y\). Use the cone dimension formula \(\dim S(Y)=\dim Y+1\) [@Har10a, Exercise I.2.6] and the dimension formula \(\operatorname{height}\mathfrak p+\dim(S/\mathfrak p)=n+1\) for primes in the polynomial ring [@Har10a, Chapter I, §1].

    1. If \(\dim Y=n-1\), then \(I\) is generated by one irreducible homogeneous polynomial of positive degree.

    The dimension formulas give \(\dim(S/I)=n\) and \(\operatorname{height}I=1\). The ring \(S\) is a unique factorization domain. Choose a nonzero element of \(I\) and an irreducible factor \(f\) which belongs to \(I\); primality ensures that some factor does. The ideal \((f)\) is a nonzero prime contained in \(I\). If it were strictly smaller than \(I\), the chain \((0)\subsetneq(f)\subsetneq I\) would contradict \(\operatorname{height}I=1\). Thus \(I=(f)\).

    The generator \(f\) is homogeneous. To prove this, choose a nonzero homogeneous component \(f_d\) of \(f\). Since \(I\) is homogeneous, \(f_d\in I=(f)\), so \(f_d=fg\) for some nonzero \(g\in S\). In a nonnegatively graded domain, the least and greatest nonzero degrees of a product are the respective sums of those degrees in its factors. The product \(fg\) has just one degree, so both factors have just one degree. Hence \(f\) is homogeneous, and its degree is positive because \(I\) is proper. Finally \(Y=Z(I)=Z(f)\).

    2. If \(Y=Z(f)\) for an irreducible homogeneous \(f\) of positive degree, then \(\dim Y=n-1\).

    The irreducible element \(f\) generates a prime ideal in the UFD \(S\). Since \(Y\) is nonempty, P-AGH23ZIPROPS, part (d), gives \[ I(Y)=I(Z(f))=\sqrt{(f)}=(f). \] The principal ideal theorem and \(f\ne0\) give \(\operatorname{height}(f)=1\). The polynomial-ring dimension formula yields \(\dim(S/(f))=n\). Therefore \[ \dim Y=\dim S(Y)-1=\dim(S/(f))-1=n-1. \] This uses the homogeneous coordinate ring, whose dimension exceeds that of the projective variety by one.

    3. Q.E.D.

    Steps <1>1–<1>2 prove both implications. When \(n=0\), the only projective variety in \(\PP^0\) is its single point, of dimension zero, while every nonzero positive-degree homogeneous polynomial in \(k[x_0]\) has empty projective zero set. Thus neither condition holds in that case, consistent with the equivalence.

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    Dependencies

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

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