A projective variety \(Y \subseteq \PP^n\) is projectively normal, with respect to the given embedding, if its homogeneous coordinate ring \(S(Y)\) is integrally closed.
If \(Y\) is projectively normal, show that \(Y\) is normal.
There are normal varieties in projective space which are not projectively normal. Let \(Y\) be the twisted quartic curve in \(\PP^3\) given parametrically by \((x,y,z,w) = (t^4, t^3 u, t u^3, u^4)\). Show that \(Y\) is normal but not projectively normal. See (III, Ex. 5.6) for more examples.
Show that the twisted quartic curve \(Y\) above is isomorphic to \(\PP^1\), which is projectively normal. Thus projective normality depends on the embedding.
Solution
1. If a positively graded domain \(S\) is integrally closed, then for every homogeneous element \(f\in S\) of positive degree, the degree-zero subring\[ (S_f)_0 \]of the graded localization is integrally closed.
The localization \(S_f\) is integrally closed by the localization result in P-AGH317NORMAL. Let \[ \alpha\in\operatorname{Frac}((S_f)_0) \] be integral over \((S_f)_0\). The same monic equation shows that \(\alpha\) is integral over \(S_f\), so integrally closedness gives \[ \alpha\in S_f. \]
Every element of \((S_f)_0\) is homogeneous of degree zero in the \(\ZZ\)-grading on \(S_f\), so every quotient of two nonzero such elements also has degree zero in the fraction field of \(S_f\). Hence \(\alpha\) is homogeneous of degree zero. Since it already lies in \(S_f\), it belongs to \((S_f)_0\). Thus \((S_f)_0\) is integrally closed.
2. A projectively normal variety \(Y\subseteq\PP^n\) is normal, proving (a).
Let \(S=S(Y)\) be the homogeneous coordinate ring and assume that it is integrally closed. For each standard projective chart \(D_+(x_i)\), the affine open subset \[ Y_i=Y\cap D_+(x_i) \] has coordinate ring \[ A(Y_i)=(S_{x_i})_0 \] [@Har10a, Theorem I.3.4(b)]. By step 1 this ring is integrally closed. The affine normality criterion from P-AGH317NORMAL therefore makes every \(Y_i\) normal. These standard opens cover \(Y\), so every local ring of \(Y\) is integrally closed and \(Y\) is normal.
3. The parametrization\[ \nu:\PP^1\longrightarrow\PP^3, \qquad [t:u]\longmapsto[t^4:t^3u:tu^3:u^4] \]is an isomorphism onto the twisted quartic \(Y\).
The four homogeneous coordinate polynomials have the same degree and no common zero on \(\PP^1\), so they define a morphism. By definition their image is \(Y\).
The opens \[ Y\cap D_+(x), \qquad Y\cap D_+(w) \] cover \(Y\): for a parameter point, \(x=t^4\) and \(w=u^4\), which cannot vanish simultaneously. On these opens define \[ \psi_x([x:y:z:w])=[x:y], \qquad \psi_w([x:y:z:w])=[z:w]. \] They are morphisms because the displayed coordinate pairs do not vanish on their respective domains. For a parameter point with \(t\ne0\), \[ [x:y]=[t^4:t^3u]=[t:u], \] and for one with \(u\ne0\), \[ [z:w]=[tu^3:u^4]=[t:u]. \] Thus both local maps are the set-theoretic inverse of \(\nu\) on their domains. They agree on the overlap and glue to a morphism \[ \psi:Y\to\PP^1 \] inverse to \(\nu\). Hence \(Y\cong\PP^1\).
4. The twisted quartic \(Y\) is normal.
The projective line is normal: its two standard affine charts have coordinate ring \(k[s]\), a UFD and hence an integrally closed domain, so P-AGH317NORMAL applies. By step 3, \(Y\cong\PP^1\). Normality is invariant under isomorphism, so \(Y\) is normal.
5. The homogeneous coordinate ring of \(Y\) is\[ S(Y)\cong k[t^4,t^3u,tu^3,u^4]\subseteq k[t,u]. \]
The parametrization induces a homomorphism \[ k[x,y,z,w]\longrightarrow k[t,u], \] \[ x\longmapsto t^4, \qquad y\longmapsto t^3u, \qquad z\longmapsto tu^3, \qquad w\longmapsto u^4. \] A homogeneous source polynomial of degree \(d\) maps to a homogeneous polynomial of degree \(4d\), so the kernel is homogeneous. For a homogeneous polynomial \(F\), membership in the kernel is equivalent to vanishing on every parameter point \([t:u]\): the substituted homogeneous polynomial vanishes on all nonzero pairs in \(k^2\), hence is zero. Thus the kernel is exactly \(I(Y)\), and the first isomorphism theorem identifies the homogeneous coordinate ring with the displayed image subring.
6. The ring \(S(Y)\) is not integrally closed, so \(Y\) is not projectively normal.
Put \[ q=t^2u^2. \] The element \(q\) lies in the fraction field of \(S(Y)\) because \[ q=\frac{yw}{z} =\frac{(t^3u)(u^4)}{tu^3}. \] It is integral over \(S(Y)\), since \[ q^2=t^4u^4=xw \] and hence it satisfies the monic equation \[ T^2-xw=0. \]
However, \(q\notin S(Y)\). Give the image subring its grading by total degree in \(t,u\). Every generator has degree four, so the degree-four part of \(S(Y)\) is exactly \[ \operatorname{span}_k\{t^4,t^3u,tu^3,u^4\}. \] The distinct monomial \(t^2u^2\) is not in this span. Terms of higher polynomial degree in the four generators have total degree at least eight, so they cannot contribute to a homogeneous element of total degree four. Thus \(q\notin S(Y)\).
This integral element of the fraction field proves that \(S(Y)\) is not integrally closed. Together with step 4, this proves (b).
7. The standard projective line is projectively normal.
Its homogeneous coordinate ring is \[ S(\PP^1)=k[t,u], \] a UFD and hence an integrally closed domain. Therefore \(\PP^1\) is projectively normal. Combined with the isomorphism of step 3 and the failure in step 6, this proves (c): projective normality depends on the chosen embedding.
8. Q.E.D.
Step 2 proves (a), steps <1>3–<1>6 prove (b), and steps 3 and 7 prove (c).
Source Collections
Dependencies
P-AGH317NORMALNormal varieties and the normalization of an affine varietyuses