Products Segre and Veronese Embeddings and Grassmannians

Put $A=k[X]$, $B=k[Y]$, and $C=A{\otimes }_{k}B$. The prime-coordinate-ring criterion makes $A$ and $B$ domains. The algebra $C$ is finitely generated by generators of $A$ tensored with $1$ and elements $1\otimes b$ for generators $b$ of $B$. To prove it is a domain, take nonzero $c={\sum }_i{a}_i\otimes b_i$ and $d={\sum }_j{a}_j^{\mathrm{\prime }}\otimes b_j^{\mathrm{\prime }}$, choosing the $b_i$ linearly independent and, separately, the $b_j^{\mathrm{\prime }}$ linearly independent, with $a_1,a_1^{\mathrm{\prime }}\mathrm{\ne }0$. Since $A$ is a domain, $a_1{a}_1^{\mathrm{\prime }}$ is a nonzero function on $X$, so it is nonzero at some $x\in X$. Evaluation in the first factor sends $c,d$ to nonzero elements of $B$ by linear independence. Their product is nonzero because $B$ is a domain; therefore $cd\mathrm{\ne }0$ in $C$. Also $1\otimes 1\mathrm{\ne }0$ over the field $k$. Hence $C$ is a nonzero reduced affine $k$-algebra.

Now thm-affine-algebraic-sets-coordinate-duality constructs an affine algebraic set $Z$ with coordinate algebra $C$. It is nonempty, since the empty set has zero coordinate ring, whereas $C\mathrm{\ne }0$. Since $C$ is a domain, thm-affine-variety-prime-coordinate-ring makes $Z$ a classical affine variety. Only now apply thm-affine-morphisms-coordinate-ring-anti-equivalence: the two canonical maps $A\to C$, $B\to C$ give morphisms $p:Z\to X$, $q:Z\to Y$.

For a classical affine variety $T$ and morphisms $f:T\to X$, $g:T\to Y$, the tensor coproduct theorem gives a unique $k$-algebra map $C\to k[T]$ extending $f^{\ast }$ and $g^{\ast }$. Since $T$ and $Z$ are now both classical affine varieties, the affine morphism correspondence turns this into the unique morphism $T\to Z$ with projections $f,g$. This is the required universal property, and $k[Z]=C$ proves the ring formula. If a factor is a point, its ring is $k$ and the same argument gives $A{\otimes }_{k}k\cong A$.

By [F1], the affine product has coordinate ring $\mathbb{C}[x,y]$, so the diagonal $D=\{(c,c):c\in \mathbb{C}\}$ is the algebraic set $V(x-y)$ and is Zariski closed.

Each factor has the cofinite Zariski topology by [F2]. For a point $(a,b)$ off the diagonal, every basic product neighbourhood $U\times V$ of it has $U\cap V\mathrm{\ne }\mathrm{\varnothing }$, because both $U$ and $V$ are cofinite in the infinite field $\mathbb{C}$. Choosing $c\in U\cap V$ gives $(c,c)\in (U\times V)\cap D$. Thus no product-topology neighbourhood of $(a,b)$ is contained in the complement of the diagonal.

Hence the diagonal is not product-topology closed although it is Zariski closed. For arbitrary affine varieties $X,Y$, the product property in [F1], applied to a one-point affine variety, identifies points of $X\times Y$ with pairs $(x,y)$; the fibre of the first projection over $x$ is therefore set-theoretically $\{x\}\times Y$. In this example its coordinate ring is $\mathbb{C}[x,y]\mathrm{/}(x-a)\cong \mathbb{C}[y]$. Thus the expected fibres coexist with the topology difference.

Replacing $x,y$ by $\lambda x,\mu y$ replaces every coordinate $x_i{y}_j$ by the single nonzero scalar $\lambda \mu$ times it. Some product $x_{i_0}{y}_{j_0}$ is nonzero, so these coordinates define a well-defined projective point.

Choose $i_0,j_0$ with $x_{i_0}\mathrm{\ne }0$ and $y_{j_0}\mathrm{\ne }0$. From a projective matrix $[z_{ij}]$ in the image, the ratios $z_{i{j}_0}:z_{i_0{j}_0}$ recover $[x_i]$, and $z_{i_{0}j}:z_{i_0{j}_0}$ recover $[y_j]$.

Thus equal Segre images have equal two projective factors. The same calculation works when one factor has one homogeneous coordinate, proving the endpoint cases.

If $z_{ij}=x_i{y}_j$, then $z_{ij}{z}_{k\mathrm{\ell }}=x_i{x}_k{y}_j{y}_{\mathrm{\ell }}=z_{i\mathrm{\ell }}{z}_{kj}$, so every Segre point satisfies every displayed minor.

Conversely choose a nonzero entry $z_{i_0{j}_0}$. The minor equations give $z_{ij}{z}_{i_0{j}_0}=z_{i{j}_0}{z}_{i_{0}j}$, hence $z_{ij}=x_i{y}_j$ for $x_i=z_{i{j}_0}$ and $y_j=z_{i_{0}j}\mathrm{/}{z}_{i_0{j}_0}$. Thus the matrix is a Segre image.

The minors are homogeneous quadrics, so their common zero locus is projectively closed, and step 2.1 identifies it with the image. On $z_{i_0{j}_0}\mathrm{\ne }0$, the ratios $\frac{z_{i{j}_0}}{z_{i_0{j}_0}},\frac{z_{i_{0}j}}{z_{i_0{j}_0}}$ are regular and recover the standard affine coordinates of both factors. Thus the asserted chartwise inverse is proved directly; injectivity alone is not used as an embedding criterion.

Inside the rank-one locus, the opens $z_{i_0{j}_0}\mathrm{\ne }0$ cover. On each such open, the regular inverse coordinates from the Segre theorem identify the desired subset with the product of the corresponding affine pieces of $X$ and $Y$, hence with a closed subset of that chart. Closedness is local on this finite open cover, so the restricted Segre image is a closed projective algebraic set.

The inverse coordinate recovery in the Segre theorem identifies this set with pairs $(x,y)\in X{\times }_{\mathrm{s}\mathrm{e}\mathrm{t}}Y$; the displayed regular ratios show on every chart that both coordinate projections are morphisms. Thus the Segre point map and its inverse are morphisms for this constructed structure.

A pair of morphisms into $X,Y$ has on each inverse-image product chart the Segre coordinate formula $[f_i{g}_j]$. These local formulas are regular and agree on overlaps, so they give a morphism into the closed model. The coordinate projections recover the given maps, and the point-pair identification makes the factorization unique. Hence this model has the product universal property.

The nonempty standard affine opens of $X$ and $Y$ are affine varieties. Their pairwise products are affine varieties by the affine-product theorem, and they cover the Segre model. Any two such product opens meet: their factor opens meet by irreducibility of $X$ and $Y$, and choosing one point from each of those two nonempty intersections gives a point of both product opens. Thus the covering affine opens all meet, so their union is irreducible. The model is therefore a nonempty projective variety.

Scaling $x$ by $\lambda$ scales every degree-$d$ monomial by ${\lambda }^d$; not all monomials vanish because some $x_i\mathrm{\ne }0$. Thus the homogeneous-coordinate criterion makes ${\nu }_{n,d}$ a morphism.

On the open where $x_i^d\mathrm{\ne }0$, the ratios $x_j{x}_i^{d-1}\mathrm{/}{x}_i^d=x_j\mathrm{/}{x}_i$ recover the usual affine coordinates. These chart inverses show injectivity and regular local inverse maps.

The relations $Z_{\alpha }{Z}_{\beta }-Z_{\gamma }{Z}_{\delta }$ whenever $\alpha +\beta =\gamma +\delta$ vanish on monomial coordinates. Repeatedly exchanging one variable between two degree-$d$ multi-indices gives $Z_{\alpha }^d={\prod }_j{Z}_{d{e}_j}^{{\alpha }_j}$. Thus every point satisfying the relations has some $Z_{d{e}_i}\mathrm{\ne }0$: otherwise every $Z_{\alpha }$ would vanish. On this chart the same exchanges give $Z_{\alpha }\mathrm{/}{Z}_{d{e}_i}={\prod }_j{\left(Z_{(d-1)e_i+e_j}\mathrm{/}{Z}_{d{e}_i}\right)}^{{\alpha }_j}$. The chart is therefore precisely the image of the affine chart $x_i\mathrm{\ne }0$, with inverse coordinates $x_j\mathrm{/}{x}_i=Z_{(d-1)e_i+e_j}\mathrm{/}{Z}_{d{e}_i}$. The common zero locus of the homogeneous relations is exactly the image, so the image is closed and these chart inverses make the map a closed immersion.

In the ordered Veronese coordinates $Z_{\alpha }$, define the linear form $L={\sum }_{\alpha }{c}_{\alpha }{Z}_{\alpha }$ and its hyperplane $H=V_{+}(L)$.

Substitution in the definition of ${\nu }_{n,d}$ gives $L({\nu }_{n,d}([x]))=\sum c_{\alpha }{x}^{\alpha }=F(x)$.

Therefore a point belongs to the pullback hyperplane exactly when it belongs to $V_{+}(F)$, proving the equality.

A change of basis multiplies $v_1\wedge \cdots \wedge v_r$ by its nonzero determinant, so its projective line is independent of the chosen basis. The wedge is nonzero because the basis is independent.

Let $w=v_1\wedge \cdots \wedge v_r$. A vector $u$ satisfies $u\wedge w=0$ exactly when $u\in S$: one direction has a repeated vector, and the other follows because $u,v_1,\dots ,v_r$ are independent when $u\mathrm{\notin }S$, so their wedge is nonzero.

This annihilator description recovers $S$ from the projective line $[w]$. Equal Plucker points therefore determine equal subspaces.

For $r=0$ or $r=n$, the Grassmannian and $\mathbf{P}({\mathrm{\Lambda }}^{r}V)$ are both one point, so the assertion is immediate. Suppose henceforth that $0<r<n$. For ordered tuples $(i_1,\dots ,i_{r-1})$ and $(j_1,\dots ,j_{r+1})$, alternating multilinearity gives the signed Plucker relation ${\sum }_{a=1}^{r+1}(-1{)}^a{p}_{i_1\dots i_{r-1}{j}_a}{p}_{j_1\dots \widehat{j_a}\dots j_{r+1}}=0.$ The ordered-coordinate convention supplies every insertion sign and makes repeated indices contribute zero.

On a chart with $p_{1\cdots r}\mathrm{\ne }0$, rescale so that $p_{1\cdots r}=1$ and define the entries of $A$ by the coordinates with exactly one index outside $\{1,\dots ,r\}$, using their alternating signs. The maximal minors of $(I_r\mid A)$ agree with these coordinates when zero or one index is outside the base set. Induct on the number of outside indices in an ordered $r$-tuple $I$. For an outside index $j\in I$, apply the relation of step 1.1 to the $(r-1)$-tuple $I\setminus \{j\}$ and the $(r+1)$-tuple $(1,\dots ,r,j)$. The term omitting $j$ is, up to sign, $p_I{p}_{1\cdots r}=p_I$; every other term is a product of a coordinate with fewer outside indices and a one-outside coordinate. The minors of $(I_r\mid A)$ satisfy the same relation, so the induction identifies $p_I$ with its corresponding minor. Thus the row span has precisely the given projective coordinates.

The analogous charts cover every nonzero coordinate point satisfying the relations. Thus every such point is decomposable, while step 1.1 proved the reverse inclusion; homogeneous quadratic equations make this locus closed, and the injective Plucker map realizes it as the required projective algebraic set.

Reorder the basis so $I=\{1,\dots ,r\}$. The projection $S\to k^r$ is invertible, so $S$ has the unique row-space representative $(I_r\mid A)$ with $A\in {\mathrm{Mat}}_{r,n-r}(k)$.

The $r(n-r)$ entries of $A$ give affine coordinates, and every Plucker coordinate is a minor of $(I_r\mid A)$, hence a polynomial in them. Conversely these entries are ratios of Plucker coordinates by $p_I$.

Thus this locus is affine space. On an overlap, changing the pivot columns replaces $(I_r\mid A)$ by multiplication with the inverse of an invertible minor, so the transition entries are regular ratios on that overlap.

By [F1], each point has an affine-space neighbourhood of dimension $r(n-r)$. This proves smoothness in the stated local-chart sense and the dimension assertion. If $r=0$ or $r=n$, there is just one subspace, so all the assertions hold for a point. Assume $0<r<n$ henceforth.

Fix $I_0=\{1,\dots ,r\}$. In its chart write a plane as the row space of $(I_r\mid A)$. For any $r$-subset $J$, retain the identity columns indexed by $J\cap I_0$ and assign the columns indexed by $J\setminus I_0$ bijectively to the remaining standard basis vectors of $k^r$. Complete the other columns of $A$ arbitrarily. The minor indexed by $J$ is then $1$ or $-1$. Thus $U_J\cap U_{I_0}$ is a nonempty open subset of both irreducible affine charts.

Each intersection in step 2.1 is dense in $U_J$, since $U_J$ is irreducible. Hence the closure of the irreducible set $U_{I_0}$ contains every $U_J$, and so is the whole Grassmannian. A closure of an irreducible set is irreducible. The chart isomorphisms in [F1] are for the Plucker-image topology itself (their maps are polynomial minors with regular inverses), so this proves irreducibility in that topology. Together with [F2], it makes the image a projective subvariety.

A point $[v]$ belongs to an $r$-plane with decomposable wedge $w$ exactly when $v\wedge w=0$. Expanding this wedge gives homogeneous bilinear equations in the point and Plucker coordinates.

These equations vanish precisely on ${\mathcal{I}}_{1,r}(V)$ by the annihilator characterization used for the Plucker map. The ambient product is closedly modeled by Segre and Plucker coordinates, so their common zero locus there is closed.

For the containment locus, cover both Grassmannians by their finitely many standard affine charts. On a product of two such charts, the planes $A$ and $B$ have canonical row-frame matrices. The condition $A\subseteq B$ is equivalent to the stacked $(a+b)\times \dim V$ matrix having rank at most $b$, which is cut out by all of its $(b+1)\times (b+1)$ minors. Thus the containment locus is closed on every member of this finite open cover. Closedness is local on an open cover, so it is closed globally. Together with steps 1.1--2.1 this proves both assertions without choosing global basis vectors of the varying plane $A$.

The diagonal ${\mathrm{\Delta }}_Y\subseteq Y\times Y$ is closed: embed $Y$ projectively and use the Segre/projective coordinate model, where equality of two projective points is cut out on the standard charts by coordinate differences.

Let $p_X:P\to X$ and $p_Y:P\to Y$ be the structure maps. The morphism $F=⟨f\circ p_X,p_Y⟩:P⟶Y\times Y$ is defined by the two product universal properties, without choosing a pair-set realization of $P$. Its inverse image of ${\mathrm{\Delta }}_Y$ is exactly the image of $⟨{\mathrm{id}}_X,f⟩$.

Since inverse images of closed algebraic sets under a morphism are closed, $F^{-1}({\mathrm{\Delta }}_Y)={\mathrm{\Gamma }}_f$ is closed. This is precisely Milne's projective-target graph criterion under the stated construction hypothesis.

The multiplication map $m:k[X]\otimes k[X]\to k[X]$, $a\otimes b\mapsto ab$, is the pullback of the diagonal map. It kills every $x_i\otimes 1-1\otimes x_i$.

Quotienting by these differences identifies the two copies of every coordinate class, hence identifies the quotient with $k[X]$. Therefore their ideal is $\ker m$.

The closed set of this kernel consists exactly of pairs whose coordinate values agree, namely the diagonal.

Points of a classical affine chart over $k$ are closed by the point/maximal-ideal correspondence in Classical affine algebraic sets correspond to radical ideals, and irreducible sets to prime ideals, so $\{s\}$ is closed in $S$. Regular maps are continuous by the chart convention and A classical morphism pulls Zariski closed sets back to closed sets. Thus $F=f^{-1}(s)$ is closed in $X$. Every point $x\in F$ has affine charts $W\subseteq X$ and $V\subseteq S$ around $x$ and $s$ with $f(W)\subseteq V$: shrink a source chart within the open preimage of $V$ to a principal affine chart.

On such charts put $J_W=k[W]f^{\ast }({\mathfrak{m}}_s)$ and $F_W=V_W(J_W)$. Its points are exactly $W\cap F$, because the coordinate functions of $V$ distinguish $s$ from every other point. The relative Nullstellensatz Classical affine algebraic sets correspond to radical ideals, and irreducible sets to prime ideals gives $I_W(F_W)=\sqrt{J_W}$, hence $k[F_W]=k[W]\mathrm{/}\sqrt{J_W}$. These reduced affine algebraic-set structures agree under restriction on overlaps. On a common principal-open refinement, localization commutes with radicals: $\sqrt{J_W}k[W{]}_h=\sqrt{J_{W}k[W{]}_h}$; both chart presentations therefore give the same reduced quotient and the same sheaf of locally regular functions on the closed locus. They glue to a classical prevariety structure on $F$, including the empty case. It is quasi-compact because $F$ is closed in the quasi-compact $X$. The inclusion $F\to X$ is regular and its composite with $f$ is constant at $s$.

Let $T$ be a classical prevariety with regular map $g:T\to X$ such that $f\circ g$ is constant at $s$. On affine charts of $T$ mapping into a chart $W$ of step 2.1, the pullback $g^{\ast }:k[W]\to k[T_0]$ kills $J_W$. The target coordinate ring is reduced, so it also kills $\sqrt{J_W}$. The affine morphism/coordinate-ring correspondence Classical affine morphisms are contravariantly equivalent to coordinate-ring homomorphisms factors $g$ uniquely through $F_W$. These local factorizations agree on overlaps because they all have the same underlying map into the closed subspace $F$ and the same pullback of locally regular functions. They glue to a unique regular map $T\to F$. Thus $F$ has the pullback property of Base change of classical varieties when the pullback exists, so any previously constructed pullback is canonically isomorphic to it. If $X$ and $S$ are separated, the closed subspace $F\subseteq X$ is separated, proving the last clause.

Replacing $(x,y)$ by $(\lambda x,\mu y)$ multiplies every $F_i$ by ${\lambda }^a{\mu }^b$, a common nonzero scalar. Thus the projective point is well defined.

Use the closed Segre model of the product from cor-projective-variety-product-exists, with coordinates $z_{ij}=x_i{y}_j$. On its chart $z_{pq}\mathrm{\ne }0$, both $x_p$ and $y_q$ are nonzero. Put $d=\max (a,b)$ and form $H_i=x_p^{d-a}{y}_q^{d-b}{F}_i$ for every $i$. These have common bidegree $(d,d)$. Each of their monomials has $d$ factors among the $x$-variables and $d$ among the $y$-variables; pair those factors to write it as a product of $d$ coordinates $z_{ij}$. Thus each $H_i$ is the restriction of a homogeneous degree-$d$ polynomial $G_i$ in the Segre ambient coordinates. On this chart the common multiplier $x_p^{d-a}{y}_q^{d-b}$ is nonzero, so the $G_i$ have no common zero and $[G_0:\cdots :G_N]=[F_0:\cdots :F_N]$.

The charts $z_{pq}\mathrm{\ne }0$ cover the constructed projective variety. On each chart step 2.1 supplies an actual tuple of homogeneous polynomials in its ambient projective coordinates, of one common degree and with no common zero there. On overlaps the tuples give the same point by step 1.1, so they satisfy the cross-multiplication compatibility of def-morphism-to-projective-space-homogeneous-coordinates. That definition now applies directly and proves the morphism assertion. It includes $a=0$ or $b=0$; when both are zero, the nonzero constant tuple gives a constant morphism, and factors ${\mathbf{P}}^0$ cause no change.

Apply ${\nu }_{m,a}$ and ${\nu }_{n,b}$ to the two factors. Their target coordinates are respectively all degree-$a$ and degree-$b$ monomials.

Applying Segre to those two images produces exactly the products $x^{\alpha }{y}^{\beta }$, in the fixed product ordering. It is therefore the fixed-bidegree map of the statement.

Let $X$ and $Y$ be the two Veronese images. The Veronese lemma makes them closed projective subvarieties with regular inverse maps. The construction in cor-projective-variety-product-exists, applied to $X,Y$, realizes their Segre image as a closed subset of the target projective space, with regular projections recovering its two factors. Compose these projections with the Veronese inverses. By the product universal property they give a regular map from this closed image to ${\mathbf{P}}^m\times {\mathbf{P}}^n$, inverse to the map in step 2.1. That forward map is a morphism by the multihomogeneous theorem, since a nonzero coordinate $x_i$ and a nonzero coordinate $y_j$ give the nonzero monomial $x_i^a{y}_j^b$. Thus the displayed map is an isomorphism onto a closed subvariety, as asserted. The argument includes a degree equal to one and a factor ${\mathbf{P}}^0$.