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.

On chart overlaps these maps are coordinate ratios with nonzero denominator, so they are affine-chart isomorphisms.

$f^h=\sum c_{\alpha }{x}_0^{d-\mathrm{\mid }\alpha \mathrm{\mid }}{x}^{\alpha }$, 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 $A$.

$G$ vanishes on this chart exactly when $G(1,x)\in I$; homogenizing this condition is exactly $x_0^{r}G\in I^h$ for some $r$.

Homogeneous equations vanish on a set exactly when they vanish on its closure, hence the ideal is $I^h:x_0^{\mathrm{\infty }}$.

Every homogeneous polynomial vanishing on $X$ vanishes on $C$. Conversely, [given, algebra] 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 $0=P(\lambda a)={\sum }_d{\lambda }^d{P}_d(a)$. Since the algebraically closed field $k$ is infinite, each $P_d(a)$ 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 [given, algebra] $FG\in I_{+}(X)$. 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, so $I_{+}(X)$ is prime.

Conversely, suppose $I_{+}(X)$ is prime and $X=Y\cup Z$ with $Y,Z$ [given, algebra] 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 [step 1.1, algebra] $V_{+}(J)\mathrm{\ne }\mathrm{\varnothing }$. Its affine zero locus is the cone over $V_{+}(J)$. The affine Nullstellensatz 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, because $F^r$ vanishing on $X$ forces $F$ to vanish there. This is the stated radical-ideal correspondence.

$G\mathrm{/}H=G\mathrm{\prime }\mathrm{/}H\mathrm{\prime }$ is equivalent to $GH\mathrm{\prime }=G\mathrm{\prime }H$ after multiplying by the nonzero product $HH\mathrm{\prime }$.

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

Therefore the affine descriptions agree exactly as stated.

Put $A=S(X)$. Irreducibility makes $A$ a graded domain, so $f$ is a [given, algebra] degree-zero element of $\mathrm{Frac}(A)$. If $x_i=0$ in $A$, take $N_i=1$, and then $x_i^{N_i}f=0\in A_{N_i}$. Otherwise $U_i=X\cap D_{+}(x_i)$ is nonempty. Normalizing $x_i=1$ identifies its defining ideal with the dehomogenizations of the homogeneous elements of $I_{+}(X)$; hence its affine coordinate ring is canonically the degree-zero localization $A_{(x_i)}$. The affine global-functions theorem places $f{\mathrm{\mid }}_{U_i}$ in $A_{(x_i)}$, so it has the form $a\mathrm{/}{x}_i^{N_i}$ with $a\in A_{N_i}$. Therefore, for every $i$, there is $N_i\ge 0$ such that $x_i^{N_i}f\in A_{N_i}$.

Choose $N\ge N_i$ for all $i$. If $d>(n+1)(N-1)$, every degree-$d$ monomial is divisible by some $x_i^N$, so multiplication by $f$ sends the finite-dimensional space $A_d$ into itself. This space is nonzero: choose $p\in X$ and a coordinate $x_i$ nonzero at $p$; then $x_i^d$ is nonzero in $A_d$.

Cayley--Hamilton applied to the $k$-linear endomorphism $a\mapsto fa$ of $A_d$ gives a nonzero polynomial $P\in k[T]$ with $P(f)a=0$ for every $a\in A_d$. Taking $0\mathrm{\ne }a\in A_d$ and working in the field $\mathrm{Frac}(A)$ gives $P(f)=0$. Since $k$ is algebraically closed, $P$ splits into linear factors, and the domain property forces $f=c$ for some $c\in k$. Thus every global regular function is constant.

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 projective radical-ideal correspondence gives $I_{+}(V_{+}(J))=J$.

$V_{+}(J)$ is closed by definition, so its inclusion is a closed embedding in the classical coordinate sense.

Substitution in the coordinate-ring definition gives the displayed quotient.

$I_{+}(X)$ is prime.

The cone coordinate ring is $k[x_0,\dots ,x_n]\mathrm{/}{I}_{+}(X)$ and is therefore a domain.

The affine prime-coordinate-ring criterion makes $C(X)$ 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.