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    The homogeneous coordinate ring is not an isomorphism invariant

    P-AGH39HOMCOORDNOTINV

    Area
    Algebraic Geometry
    Topics
    Coordinate Rings, Isomorphism, Veronese Embedding
    Status
    draft

    The homogeneous coordinate ring of a projective variety is not invariant under isomorphism. Let \(X = \PP^1\) and let \(Y\) be the \(2\)-uple embedding of \(\PP^1\) in \(\PP^2\). Then \(X \cong Y\), but show that \(S(X) \not\cong S(Y)\).

    Solution

    Choose homogeneous coordinates \([s:t]\) on \(X=\PP^1\) and \([u:v:w]\) on \(\PP^2\) so that the quadratic Veronese map is \[ [s:t]\longmapsto[s^2:st:t^2]. \]

    1. The homogeneous coordinate rings are\[ S(X)=k[s,t], \qquad S(Y)=k[u,v,w]/(uw-v^2). \]

    The projective line has zero homogeneous ideal, so \(S(X)=k[s,t]\). The displayed parametrization satisfies \(uw-v^2=0\). By P-AGH212DUPLE, its image is a projective variety isomorphic to \(\PP^1\) and is exactly the quadratic Veronese conic. The polynomial \(uw-v^2\) is irreducible: viewing it as a degree-one polynomial in \(u\) over \(k[v,w]\), any nonconstant factor independent of \(u\) would have to divide both \(w\) and \(v^2\), whose greatest common divisor is \(1\). Thus \(Z(uw-v^2)\) is an irreducible hypersurface in \(\PP^2\), and its homogeneous ideal is \((uw-v^2)\) by P-AGH28HYPERSURFACE. Hence the second displayed description follows.

    2. The ring \(S(X)\) is a unique factorization domain.

    It is the polynomial ring in two variables over the field \(k\), hence is a unique factorization domain.

    3. The ring\[ R=k[u,v,w]/(uw-v^2) \]is not a unique factorization domain.

    The polynomial \(uw-v^2\) is irreducible in the UFD \(k[u,v,w]\), hence prime, so \(R\) is a domain. Give \(R\) its standard grading, and write \(\bar u,\bar v,\bar w\) for the degree-one residue classes.

    First, every nonzero homogeneous element of degree one in \(R\) is irreducible. Indeed, suppose such an element \(h\) factors as \(h=ab\). Write \(a\) and \(b\) as sums of homogeneous components, and let \(r,s\) be their highest occurring degrees and \(p,q\) their lowest occurring degrees. Since \(R\) is a domain, the highest and lowest nonzero homogeneous components of \(ab\) occur in degrees \(r+s\) and \(p+q\), respectively. Because \(h\) is homogeneous of degree one, \[ r+s=p+q=1. \] The inequalities \(p\le r\) and \(q\le s\) then force \(p=r\) and \(q=s\). Thus both factors are homogeneous, and one has degree zero. Since \(R_0=k\), that factor is a unit.

    Consequently \(\bar u,\bar v,\bar w\) are irreducible. They are pairwise nonassociate: all units are nonzero constants by the same highest-degree argument, and the degree-one component of \(R\) has basis \(\bar u,\bar v,\bar w\) because the defining relation has degree two. But in \(R\), \[ \bar u\,\bar w=\bar v^2. \] This gives two factorizations into irreducibles which cannot be matched by reordering and multiplication by units. Therefore \(R\) is not a unique factorization domain.

    4. Q.E.D.

    By P-AGH34DUPLEISO, the quadratic Veronese map gives \(X\cong Y\). Steps <1>2–<1>3 show that \(S(X)\) is a UFD while \(S(Y)\) is not. Since being a UFD is preserved by ring isomorphism, \[ S(X)\not\cong S(Y). \]

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