Projective Algebraic Sets Projective Morphisms and Cones

Each monomial of $F$ scales by ${\lambda }^d$, so $F(b)={\lambda }^{d}F(a)$.

Since ${\lambda }^d\mathrm{\ne }0$, $F(b)=0$ if and only if $F(a)=0$.

The empty and whole sets are $V_{+}((1))$ and $V_{+}((0))$.

Direct evaluation gives $V_{+}(IJ)=V_{+}(I)\cup V_{+}(J)$.

Direct evaluation gives $V_{+}({\sum }_{\alpha }{J}_{\alpha })={\bigcap }_{\alpha }{V}_{+}(J_{\alpha })$.

These identities prove the closed-set axioms, and the complement of $V_{+}((x_i))$ is $D_{+}(x_i)$.

Multiplication by $a_i^{-1}$ gives the unique representative with $i$th coordinate $1$.

Retaining the remaining coordinates and reinserting $1$ give mutually inverse maps.

A homogeneous polynomial restricted to the representative with $i$th coordinate 1 becomes an ordinary polynomial in the other coordinates. Conversely, for any polynomial $h$ in those coordinates, multiply $h(x_j\mathrm{/}{x}_i{)}_{j\mathrm{\ne }i}$ by $x_i^{\mathrm{deg}h}$ to obtain a homogeneous polynomial with the same zero set on the chart. Thus relative projective closed sets correspond exactly to affine polynomial zero sets, proving that the bijection is a homeomorphism. The affine closed-set axioms also follow directly: intersections combine equations and finite unions use pairwise products of equations.

On the overlap with chart $j$, write $u_l=a_l\mathrm{/}{a}_i$ and $u_i=1$. The new coordinates are $v_l=u_l\mathrm{/}{u}_j$, with $u_j\mathrm{\ne }0$; the inverse has the same form with $i,j$ exchanged. Substituting these ratios into a local polynomial quotient and clearing powers of $u_j$ again gives a polynomial quotient with nonvanishing denominator. Hence transition maps preserve regular functions in both directions. The local quotient condition is preserved under restriction and gluing of functions, so the transported structures agree on overlaps and define the claimed chart isomorphisms, including the one-point case $n=0$.

$f^h=\sum c_{\alpha }{x}_0^{d-\mathrm{\mid }\alpha \mathrm{\mid }}{x}^{\alpha }$, so each monomial has degree $d$ and $x_0=1$ recovers $f$.

Factoring $x_0^d$ from every homogeneous monomial of $F$ proves the displayed reverse formula.

Dehomogenization on $D_{+}(x_0)$ identifies $V_{+}(I^h)$ there with $A$.

Let G be homogeneous of degree d. It vanishes on the embedded A exactly when $g=G(1,x)\in I(A)=I$. If g is nonzero of degree e, homogenization gives $G=x_0^{d-e}{g}^h\in I^h$. If g=0, the same homogenization identity gives G=0. Conversely, if $x_0^{r}G\in I^h$, dehomogenizing an ideal expression gives $g\in I$, because every generator $f^h$ dehomogenizes to f in I. Thus the homogeneous elements vanishing on A are exactly the homogeneous elements of $I^h:x_0^{\mathrm{\infty }}$.

A homogeneous equation vanishes on A exactly when its projective closed zero locus contains the projective closure of A. Both the homogeneous vanishing ideal of that closure and $I^h:x_0^{\mathrm{\infty }}$ are homogeneous ideals; agreement in every homogeneous degree therefore proves equality for all polynomials. If A is empty, I is the unit ideal, as are both ideals in the formula. No Nullstellensatz or choice principle is needed because $I=I(A)$ is given.

Let $C$ consist of zero and all nonzero representatives of points of $X$. Every homogeneous polynomial vanishing on $X$ vanishes on $C$. Conversely, write a polynomial vanishing on $C$ as $P={\sum }_d{P}_d$ with each $P_d$ homogeneous. For a representative $a$ of a point of $X$ and every $\lambda \in k$, one has $0=P(\lambda a)={\sum }_d{\lambda }^d{P}_d(a)$. The field $k$ is infinite: if it were finite, the nonconstant polynomial ${\prod }_{b\in k}(T-b)+1$ would take the value $1$ at every element of $k$, contradicting algebraic closedness. The polynomial ${\sum }_d{T}^d{P}_d(a)$ therefore has infinitely many roots; by [F2] it is zero, so each coefficient $P_d(a)$ is zero. Thus all homogeneous components of $P$ lie in $I_{+}(X)$, and $I(C)=I_{+}(X)$.

Suppose $X$ is irreducible and homogeneous $F,G$ satisfy $FG\in I_{+}(X)$. Then $X\subseteq V_{+}(F)\cup V_{+}(G)$, so irreducibility gives $F\in I_{+}(X)$ or $G\in I_{+}(X)$. A homogeneous ideal is prime exactly when this test holds for homogeneous elements: in the graded quotient, the product of two nonzero elements has nonzero component at the sum of their lowest nonzero degrees. Here the ideal is proper since $X$ is nonempty. Thus $I_{+}(X)$ is prime.

Conversely, suppose $I_{+}(X)$ is prime and $X=Y\cup Z$ with $Y,Z$ projective algebraic subsets of $X$. If both are proper, choose $p\in X\setminus Y$ and $q\in X\setminus Z$. Homogeneous defining equations give $F\in I_{+}(Y)$ with $F(p)\mathrm{\ne }0$ and $G\in I_{+}(Z)$ with $G(q)\mathrm{\ne }0$. Then $FG$ vanishes on $X$, contrary to primality of $I_{+}(X)$. Hence $X=Y$ or $X=Z$, so $X$ is irreducible.

Now assume the Axiom of Choice and let $J$ be homogeneous radical with $V_{+}(J)\mathrm{\ne }\mathrm{\varnothing }$. Its affine zero locus is the cone over $V_{+}(J)$. The affine strong Nullstellensatz F1, the exact inherited use of AC in this proof, gives $I(V(J))=\sqrt{J}=J$, while step 1.1 gives $I(V(J))=I_{+}(V_{+}(J))$. Conversely, every $I_{+}(X)$ is homogeneous and radical by its equality with the affine vanishing ideal $I(C)$ in step 1.1: $P^r$ vanishing on $C$ forces $P$ to vanish there. Finally $V_{+}(I_{+}(X))=X$, since homogeneous defining equations for $X$ belong to $I_{+}(X)$. This is the stated radical-ideal correspondence.

The intersection of the two charts with $D_{+}(HH\mathrm{\prime })\cap X$ contains the given point, so it is a nonempty open subset of the irreducible variety $X$. If the two evaluations agree on that overlap, the homogeneous cross difference $GH\mathrm{\prime }-G\mathrm{\prime }H$ vanishes there. Its zero locus is closed, and every nonempty open subset of an irreducible space is dense; hence the cross difference vanishes on all of $X$ and is zero in $S(X)$. Conversely, if $GH\mathrm{\prime }=G\mathrm{\prime }H$ in $S(X)$, evaluation and division by the nonzero product $HH\mathrm{\prime }$ show that the fractions agree throughout their common domain. Equivalently, this is equality of the two fractions in $\mathrm{Frac}(S(X))$.

Dehomogenization in either chart turns this into the same equality of ordinary affine fractions.

Therefore the affine descriptions agree exactly as stated.

In A let D be the homogeneous ideal generated by all homogeneous denominators h for which $fh\in A_{\mathrm{deg}h}$. Global regularity says that at each point of X at least one such h is nonzero. This uses the set of all admissible denominators, not a simultaneous choice of local representatives. Let $\tilde{D}$ be the inverse image of D in $k[x_0,\dots ,x_n]$. Its affine zero set is contained in the cone over X, and has no nonzero point: any such point represents a point of X where all admissible denominators would vanish.

Each coordinate x_i therefore vanishes on $V(\tilde{D})$. F2 implies $x_i^{N_i}\in \tilde{D}$ for some $N_i\ge 1$, so its class lies in D. This is the precise inherited AC use in the argument. Expressing that homogeneous class as a finite sum of homogeneous generators of D and taking the component of degree N_i shows $f{x}_i^{N_i}\in A_{N_i}$: multiply each generator by f in its prescribed degree. This also covers coordinates zero in A.

Choose $N\ge 1$ at least every N_i. Then $f{x}_i^N\in A_N$ for all i. Choose an integer $d>(n+1)(N-1)$. Every monomial of degree d contains some x_i to exponent at least N, so multiplying by f carries its class into A_d. The finitely many monomials of degree d span A_d, hence multiplication by f defines a k-linear endomorphism of this finite-dimensional space. It is nonzero as a space: at any point of X some coordinate x_i is nonzero, and then $x_i^d$ is nonzero in A_d.

F3 supplies a monic polynomial $P\in k[T]$ with $P(f)a=0$ for every $a\in A_d$. Choose nonzero $a\in A_d$; cancellation in $\mathrm{Frac}(A)$ gives $P(f)=0$. As k is algebraically closed, P splits into linear factors, and the field has no zero divisors. Thus $f=c$ in the fraction field for some $c\in k$. Its local evaluations are consequently that same constant at every point of X.

Rescaling $a$ by $\lambda$ rescales every $F_i(a)$ by ${\lambda }^d$, so the target class is representative-independent.

On $D_{+}(y_j)$ the coordinate functions are $F_i\mathrm{/}{F}_j$, regular equal-degree fractions wherever $F_j\mathrm{\ne }0$.

The target standard opens cover the image and the ratios agree on overlaps, so the local maps glue to a morphism.

The base field is algebraically closed by the standing projective-space convention in [F1]. Since $J$ is homogeneous and radical and $X=V_{+}(J)$ is nonempty, all hypotheses of [F2] hold; under [A1] it gives $I_{+}(X)=J$.

By [F1], $X=V_{+}(J)$ is projectively closed. Its inclusion identifies $X$ with this closed algebraic subset and has the identity as inverse onto its image, so it is a closed projective embedding in the classical coordinate sense used here.

Substituting $I_{+}(X)=J$ from step 1.1 into [F3] gives $S(X)=k[x_0,\dots ,x_n]\mathrm{/}J$. The quotient carries the induced grading because $J$ is homogeneous. This proves the stated coordinate-ring formula together with the closed embedding in step 2.1.

By the irreducibility clause of projective irreducibility homogeneous prime, $I_{+}(X)$ is prime. This clause and its cone-ideal argument require no AC. Homogeneous defining equations show that $C(X)$ consists exactly of zero and the nonzero representatives of points of $X$. Decomposing a polynomial vanishing on this cone into homogeneous components and evaluating on all scalar multiples shows, over the infinite field $k$, that each component vanishes on $X$. Hence $I(C(X))=I_{+}(X)$.

Suppose $C(X)=A\cup B$ for two proper relatively closed subsets. Choose a point of $C(X)\setminus A$ and a polynomial $f$ from defining equations of $A$ that is nonzero there. Likewise choose a polynomial $g$ vanishing on $B$ but not on all of $C(X)$. Then $fg$ vanishes on the cone, so $fg\in I(C(X))=I_{+}(X)$, although neither factor belongs to that prime ideal. This is impossible.

The cone is nonempty since it contains zero, and step 2.1 excludes a union of two proper closed subsets. Thus it is irreducible.

By definition it is the intersection of all projective closed sets containing the image of $A$.

Hence every closed subset of the closure containing the image is the closure itself, which is precisely density.

Normalize a point of $D_{+}(x_i)$ by $x_i=1$; $F$ vanishes exactly when the dehomogenized equation vanishes.

The chart normalization therefore identifies $X\cap D_{+}(x_i)$ with that affine hypersurface.

On overlaps the two normalizations differ by division by a nonzero coordinate, giving the stated transitions.