1.6 Products and the Hausdorff Axiom

Product and its Existence

Definition

Let $\mathcal{C}$ be a category, $X,Y$ objects in $\mathcal{C}$. An object $Z$ plus two morphisms

is a (fiber) product if it has the following universal mapping property: for all objects $W$ and morphisms

there is a unique morphism $t:W\to Z$ such that $r=p\cdot t,s=q\cdot t$, i.e., such that

commutes.

Lemma

If $X,Y\in \mathcal{C}$ and $X\times Y$ exists, then $\mathrm{st}(X\times Y)\simeq \mathrm{st}(X)\times \mathrm{st}(Y)$, where $\mathrm{st}:\mathcal{C}\to \mathrm{Set},X\mapsto \mathrm{st}(X)$ is a forgetful functor.

\begin{proof} For any $(x_0,y_0)\in \mathrm{st}(X)\times \mathrm{st}(Y)$, define ${\mathbb{A}}^0=\{p\}$ and ${\mathbb{A}}^0\to X,p\mapsto x_0$ and ${\mathbb{A}}^0\to Y,p\mapsto y_0$. By universal property, there exists unique $\theta :{\mathbb{A}}^0\to X\times Y$ and $\theta (p)\leftrightarrow (x_0,y_0)$.

Conversely, for $w_0\in \mathrm{st}(X\times Y)$, we have $p(w_0)=x_0$ and $q(w_0)=y_0$ where $p,q$ are projection map. \end{proof}

in the Category of Affine Varieties

Proposition

Let $X$ and $Y$ be affine varieties, with coordinate rings $R$ and $S$. Then

• there is a product prevariety $X\times Y$.
• $X\times Y$ is affine with coordinate ring $R{\otimes }_{k}S$.
• a basis of the topology is given by the open sets

$$\sum f_i(x)g_i(y)\ne 0,f_i\in R,g_i\in S$$

• ${\underline{o}}_{(x,y)}$ is the localization of ${\underline{o}}_x{\otimes }_k{\underline{o}}_y$ at the maximal ideal $m_x\cdot {\underline{o}}_y+{\underline{o}}_x\cdot m_y$.

Remark. We only need the fiber product is a prevariety. If $W={\cup }_{\text{finite}}{W}_i$ with affine varieties $W_i$, then $W$ is a fiber product iff $\{W_i{\}}_i$ are fiber products.

\begin{proof} It is a sketch of proof generated by Gemini.

1. Construction of $X\times Y$ (Set of Points and Algebraic Structure):

   • If $X\subseteq k^{n_1}$ is defined by ideal $I\subset k[x_1,...,x_{n_1}]$ and $Y\subseteq k^{n_2}$ by ideal $J\subset k[y_1,...,y_{n_2}]$, then the set of points $X\times Y$ can be naturally viewed as the algebraic set $V(I+J)$ in $k^{n_1+n_2}$ defined by the ideal $I+J\subset k[x_1,...,x_{n_1},y_1,...,y_{n_2}]$.

   • Its coordinate ring is $k[x_i,y_j]/(I+J)$. There is a standard isomorphism:

     $$k[x_i,y_j]/(I+J)\cong (k[x_i]/I){\otimes }_k(k[y_j]/J)=R{\otimes }_{k}S$$

   • Key Algebraic Fact: Since $k$ is an algebraically closed field and $X,Y$ are varieties, $R$ and $S$ are integral domains. In this case, $R{\otimes }_{k}S$ is also an integral domain.

   • Corollary: Because $R{\otimes }_{k}S$ is an integral domain, the radical of the ideal generated by $I+J$ in $k[x_i,y_j]$ is a prime ideal. This means the algebraic set $X\times Y$ is irreducible. Therefore, $(X\times Y,{\underline{o}}_{X\times Y})$ is itself an affine variety, whose coordinate ring is precisely $R{\otimes }_{k}S$.

2. Verifying the Universal Property:

   • We need to show that the constructed affine variety $X\times Y=\mathrm{Spec}(R{\otimes }_{k}S)$ satisfies the categorical definition of a product.
   • The projection maps $p:X\times Y\to X$ and $q:X\times Y\to Y$ are natural (algebraically corresponding to $R\hookrightarrow R{\otimes }_{k}S$ and $S\hookrightarrow R{\otimes }_{k}S$) and are morphisms.
   • Given any prevariety $W$ and two morphisms $r:W\to X$, $s:W\to Y$, we can first uniquely define a map of sets $t:W\to X\times Y$ such that $p\circ t=r,q\circ t=s$, namely $t(w)=(r(w),s(w))$.
   • We need to verify that $t$ is a morphism. This can be done by checking that the pullback $t^{\ast }$ maps regular functions on $X\times Y$ to regular functions on $W$. Since $\Gamma (X\times Y)=R{\otimes }_{k}S$ is generated by elements from $R$ and $S$, and since $r^{\ast }$ and $s^{\ast }$ already pull back elements of $R$ and $S$ to $\Gamma (W,{\underline{o}}_W)$ (because $r,s$ are morphisms), using the universal property of tensor products, one can show that $t^{\ast }$ is a ring homomorphism, and thus $t$ is a morphism.

3. Basis of Topology:

   • The basis for the topology on an affine variety $\mathrm{Spec}(A)$ is given by the principal open sets $D(h)=\{P\in \mathrm{Spec}(A)\mid h\notin P\}$, where $h\in A$.
   • In $X\times Y=\mathrm{Spec}(R{\otimes }_{k}S)$, an element $h$ has the form $h=\sum f_i\otimes g_i$. The point $(x,y)$ corresponds to the maximal ideal $m_{(x,y)}$. The condition $h\notin m_{(x,y)}$ is usually interpreted as $h(x,y)\ne 0$, which means $\sum f_i(x)g_i(y)\ne 0$. Thus, these sets form a basis for the topology.

4. Local Ring:

   • The local ring ${\underline{o}}_{Z,z}$ of an affine variety $Z=\mathrm{Spec}(A)$ at a point $z$ (corresponding to maximal ideal $m_z$) is the localization $A_{m_z}$.
   • Therefore, ${\underline{o}}_{(x,y)}=(R{\otimes }_{k}S{)}_{m_{(x,y)}}$.
   • The equivalent description given in the Red Book is that ${\underline{o}}_{(x,y)}$ is the localization of the ring ${\underline{o}}_x{\otimes }_k{\underline{o}}_y$ at the maximal ideal $m_x\cdot {\underline{o}}_y+{\underline{o}}_x\cdot m_y$.
   • (Here ${\underline{o}}_x=R_{m_x}$ and ${\underline{o}}_y=S_{m_y}$. $m_x\cdot {\underline{o}}_y$ denotes the ideal generated by elements $a\otimes b$ where $a\in m_x\subset {\underline{o}}_x,b\in {\underline{o}}_y$, and ${\underline{o}}_x\cdot m_y$ is similar. This ideal corresponds to the functions in ${\underline{o}}_x{\otimes }_k{\underline{o}}_y$ that vanish at $(x,y)$. Note: The corresponding formula in the original notes was incorrect).

Now we finish the proof. \end{proof}

in the Category of Prevarieties

Theorem

Let $X$ and $Y$ be prevarieties over $k$. Then they have a product.

\begin{proof} Assume $Y$ is affine. Then $X={\cup }_{i=1}^{\ell }{X}_i$ with $X_i$ affine and $X_i\times Y$ exists as an affine variety by ^f0237a. Since $(X_1\cap X_2)\times Y$ is an open set in both $X_1\times Y$ and $X_2\times Y$. Then we can glue $X_1\times Y$ and $X_2\times Y$ along $(X_1\cap X_2)\times Y$. Now we get a topological space, and then we construct a sheaf on it by

$${\underline{o}}_{(X_1\cup X_2)\times Y}(U)={\cap }_{(x_0,y_0)\in (X_1\cup X_2)\times Y}{\underline{o}}_{(x_0,y_0)}\subseteq k(X_1)\otimes k(X_2).$$

Assume $Y$ is affine. Then $X={\cup }_{i=1}^{\ell }{X}_i$ with $X_i$ affine and $X_i\times Y$ exists as an affine variety by ^f0237a. Since $(X_1\cap X_2)\times Y$ is an open set in both $X_1\times Y$ and $X_2\times Y$. Then we can glue $X_1\times Y$ and $X_2\times Y$ along $(X_1\cap X_2)\times Y$ using the identity map. Repeating this for all pairs $i,j$ yields a topological space $Z$. We construct a sheaf ${\underline{o}}_Z$ on $Z$ by defining its sections on open sets $U\subseteq Z$:

$${\underline{o}}_Z(U)=\left\{f:U\to k:\forall i,f{|}_{U\cap (X_i\times Y)}\in {\underline{o}}_{X_i\times Y}(U\cap (X_i\times Y))\right\}.$$

Now we verify $(Z,{\underline{o}}_Z)$ is a prevariety. Notice that $\{X_i\times Y{\}}_{i=1}^{\ell }$ is a finite cover of $Z$, where $X_i\times Y$ are irreducible affine varieties. Also note that $X_i\times Y$ and $X$ are connected, then $Z$ is also connected, because a space covered by finitely many connected open sets with overlapping intersections is connected. Therefore, $(Z,{\underline{o}}_Z)$ is a prevariety. Define $Z=X\times Y$.

It remains to check the universal property. For any prevariety $W$ and $r:W\to X,s:W\to Y$, define $t:W\to X\times Y,w\mapsto (r(w),s(w))$. Then it remains to check $t$ is a morphism. Take a finite open cover $\{W_k\}$ of $W$ such that $r(W_k)\subseteq X_i$ for some $i$. Then $t{|}_{W_k}:W_k\to X_i\times Y$ is a morphism by ^f0237a.

By the definition of morphism, it suffices to show for any open set $V\subseteq X\times Y$ and $g\in \Gamma (V,{\underline{o}}_{X\times Y})$, $g\circ t\in \Gamma (t^{-1}(V),{\underline{o}}_W)$. For any $w\in t^{-1}(V)$, there exists $W_0\in \{W_k\}$ containing $w$ such that $t{|}_{W_0}$ is a morphism. Define $W_1=W_0\cap t^{-1}(V)$, then $t{|}_{W_1}$ is also a morphism. It deduces that

$$(g\circ t){|}_{W_1}\in \Gamma (t_{W_1}^{-1}(V),{\underline{o}}_{W_1})=\Gamma (W_1,{\underline{o}}_{W_1}).$$

That is, $g\circ t$ is regular at $w\in W_1$ for any $w\in t^{-1}(V)$. Thus $g\circ t\in \Gamma (t^{-1}(V),{\underline{o}}_W)$ and so $t$ is a morphism. The uniqueness is easy to check.

In general, prevariety $Y$ can be written as $Y={\cup }_j{Y}_j$ with $Y_j$ affine. As $X\times Y_j$ exists, we can glue $X\times Y_1$ and $X\times Y_2$ along $X\times (Y_1\cap Y_2)$. Similarly, we can get a topological space $Z^{\prime }=X\times Y$, which satisfies the universal property.

Remark. For affine $X$ and $Y$, take open $U\subseteq X$, then $U\times Y$ is open in $X\times Y$.

in the Category of Projective Varieties

Theorem

The product of two projective varieties is a projective variety.

\begin{proof} Assume $X\subseteq {\mathbb{P}}^n$ and $Y\subseteq {\mathbb{P}}^m$, then $X\times Y\subseteq X\times {\mathbb{P}}^m\subseteq {\mathbb{P}}^n\times {\mathbb{P}}^m\to {\mathbb{P}}^N$ where $N=(n+1)(m+1)-1$. Define the Segre embedding

$$\mathrm{I}:{\mathbb{P}}^n\times {\mathbb{P}}^m\to {\mathbb{P}}^N,[x_0,\cdots ,x_n],[y_0,\cdots ,y_m]\to [\cdots ,x_i{y}_j,\cdots ]$$

and $\mathrm{I}$ is injective.

It suffices to show $({\mathbb{P}}^n{)}_{x_0}\times ({\mathbb{P}}^m{)}_{y_0}\to {\mathbb{P}}^N$ is a closed embedding, that is, the image of $I$ is a subvariety of ${\mathbb{P}}^N$. Note that

$$\begin{array}{rl}({\mathbb{P}}^n{)}_{x_0}\times ({\mathbb{P}}^m{)}_{y_0}& \to {\mathbb{P}}^N\\ ((s_1,\cdots ,s_n),(t_1,\cdots ,t_n))& \mapsto (\cdots ,R_{ij},\cdots )\end{array}$$

where $s_i=x_i/x_0$, $t_j=y_j/y_0$ and $R_{ij}=s_i{t}_j$ if $ij\ne 0$ and $R_{0j}=t_j$, $R_{i0}=s_i$. Then $\mathrm{im}(\mathrm{I})=I(R_{ij}-R_{i0}\cdot R_{0j}:i,j⩾1)$, that is, the ideal generated by $R_{ij}-R_{i0}{R}_{0j}$.

Check that

$$\begin{array}{r}k[s_1,\cdots ,s_n]\otimes k[t_1,\cdots ,t_m]\stackrel{\simeq }{\leftarrow }k[\cdots ,R_{ij},\cdots ]/(R_{ij}-R_{i0}{R}_{0j}{)}_{i,j}\\ \Gamma (\mathrm{im}(\mathrm{I}))\to \Gamma (({\mathbb{P}}^n{)}_{x_0}\times \Gamma ({\mathbb{P}}^m{)}_{y_0}),R_{ij}\mapsto s_i\otimes t_j\end{array}$$

is a morphism between affine variety. So $f$ is an isomorphism. \end{proof}

Hausdorff Axiom

Hausdorff axiom

Let $X$ be a prevariety. $X$ is a variety if for all prevarieties $Y$ and for all morphisms

$\{y\in Y\mid f(y)=g(y)\}$ is a closed subset of $Y$.

Too similar definitions to differ

• Affine variety: Defined by polynomial equations in affine space.
• Prevariety: A space locally modeled on affine varieties (can be glued from them).
• Variety: A prevariety satisfying the Hausdorff axiom — morphisms into it are distinguishable (equalizer sets are closed).

Example. Let $Y=X\times X,f=p_1$, $g=p_2$. Also let $\Delta :X⟶X\times X$ be a morphism $\left({\mathrm{id}}_X,{\mathrm{id}}_X\right)$. $\Delta$ is called the diagonal morphism, and

$$\Delta (X)=\left\{z\in X\times X\mid p_1(z)=p_2(z)\right\}.$$

Therefore the Hausdorff axiom implies $\Delta (X)$ is closed. In fact, we can prove that for a topological space $X$, $X$ is Hausdorff iff $\Delta (X)$ is closed.

Proposition

Let $X$ be a prevariety. Then $X$ is a variety if and only if $\Delta (X)$ is closed in $X\times X$.

\begin{proof} Suppose $f,g:Y\to X$ are given. Then they induce a morphism $(f,g):Y\to$ $X\times X$. Since

$$\{y\in Y\mid f(y)=g(y)\}={\theta }^{-1}(\Delta (X))$$

$\Delta (X)$ being closed implies the Hausdorff axiom, so $X$ is a variety.

Conversely, if $X$ is a variety, then take $Y=X\times X$ with $f=p_1,g=p_2$ and $\Delta (X)=\{z\in Y:p_1(z)=p_2(z)\}$ is closed. \end{proof}

Example. a line with two points at zero does not satisfy Hausdorff axiom, see here.

Separateness = "limit is unique"

Why is this called the “Hausdorff Axiom”? This term draws an analogy with the Hausdorff property in topology:

• Topology: Hausdorff separation means distinct points $x\ne y$ can be separated by disjoint open sets.
  • This is equivalent to the diagonal $\Delta (X)$ being closed in $X\times X$ (product topology).
• Algebraic Geometry: The separation axiom here distinguishes morphisms. For any $f,g:Y\to X$, the set where they agree, $E=\{y\in Y\mid f(y)=g(y)\}$, must be closed in $Y$ (Zariski topology).
  • Connection to Limits/Closure: Taking the closure of $E$ is similar to taking a limit. As all open sets are dense in $Y$ (if $Y$ is irreducible), a morphism is uniquely determined by its values on an open set.
  • This axiom is also equivalent to the diagonal $\Delta (X)$ being closed in $X\times X$ (product Zariski topology).

Proposition

$X$ is variety, and $Z\subseteq X$ is a closed sub-prevariety. Then $Z$ is a variety.

\begin{proof}

Since $X$ is variety, $\{y\in Y:(j\circ f)(y)=(j\circ g)(y)\}$ is closed and so $\{y\in Y:f(y)=g(y)\}$ is closed. Hence $Z$ is a variety. \end{proof}

Proposition

$X,Y$ are varieties, then $X\times Y$ is variety.

\begin{proof} Let $W$ be an arbitrary prevariety, and let $f,g:W\to X\times Y$ be two arbitrary morphisms. By definition, it suffices to show

$$E=\{w\in W\mid f(w)=g(w)\}$$

is a closed subset of $W$.

Let $p_X:X\times Y\to X$ and $p_Y:X\times Y\to Y$ be the projection morphisms. We can decompose the morphisms $f$ and $g$ into their components:

• $f_1=p_X\circ f:W\to X$
• $f_2=p_Y\circ f:W\to Y$
• $g_1=p_X\circ g:W\to X$
• $g_2=p_Y\circ g:W\to Y$

Since compositions of morphisms are morphisms, $f_1,f_2,g_1,g_2$ are all morphisms. Note that $w\in E$ iff both $f_1(w)=g_1(w)$ and $f_2(w)=g_2(w)$ hold. Let $E_1=\{w\in W\mid f_1(w)=g_1(w)\}$ and $E_2=\{w\in W\mid f_2(w)=g_2(w)\}$, then one know $E=E_1\cap E_2$. Since $X$ and $Y$ are varieties, $E_1$ and $E_2$ are closed and so for $E$. \end{proof}

Proposition

Affine variety is a variety.

\begin{proof} Let $X$ be an affine variety, and let $R=\Gamma (X,{\underline{o}}_X)$ be its coordinate ring. To show $X$ is an variety, it suffices to show for any prevariety $Y$ and any morphisms $f,g:Y\to X$, the equalizer $E=\{y\in Y\mid f(y)=g(y)\}$ is closed in $Y$.

For any $s\in R$, the functions $s\circ f$ and $s\circ g$ are global regular functions on $Y$ since $f,g$ are morphisms. Then their difference $(s\circ f)-(s\circ g)$ is also a global regular function on $Y$. Note that $y\in E$ iff $s(f(y))=s(g(y))$ for all $s\in R$. So $E$ can be written that

$$E=V(\{(s\circ f)-(s\circ g)\mid s\in R\}).$$

By definition of a prevariety, $E$ is a closed subset of $Y$. Now we finish the proof.

Alternating proof. Let $X$ be an affine variety, embedded as a closed subset in some affine space ${\mathbb{A}}^n$. Then the product $X\times X$ is a closed subset of ${\mathbb{A}}^n\times {\mathbb{A}}^n\cong {\mathbb{A}}^{2n}$. Note that

$$\Delta ({\mathbb{A}}^n)=\{(a,a)\mid a\in {\mathbb{A}}^n\}\subseteq {\mathbb{A}}^n\times {\mathbb{A}}^n$$

is defined by the equations $x_1-y_1=0,\dots ,x_n-y_n=0$, where $(x_1,\dots ,x_n,y_1,\dots ,y_n)$ are coordinates on ${\mathbb{A}}^n\times {\mathbb{A}}^n$. This is clearly a closed set.

It follows that $\Delta (X)=\Delta ({\mathbb{A}}^n)\cap (X\times X)$ is still closed in ${\mathbb{A}}^{2n}$ and is also a closed subset of $X\times X$. By ^t398yf, $X$ is a variety. \end{proof}

Proposition

$\Delta (X)$ is locally closed. Remark that we say $Z\subseteq X$ is locally closed if for any $z_0\in Z$, there exists open $U\subseteq X$ containing $z_0$ such that $Z\cap U$ is closed in $U$.

\begin{proof} Take $Y=X\times X$ with $f=p_1,g=p_2$. Then $Z=\Delta (X)$ is closed in $Y$. For any $z\in Z$, let $V$ be an affine open neighborhood of $f(z)$ in $X$. Remark that $f(z)=g(z)$. Then

$$Z\cap \left[f^{-1}(V)\cap g^{-1}(V)\right]=\left\{\begin{array}{ll}z\in f^{-1}(V)\cap g^{-1}(V)& \begin{array}{l}f(z)=g(z)\\ \text{ in the affine }\\ \text{ variety }\end{array}\end{array}\right\}$$

where $U:=f^{-1}(V)\cap g^{-1}(V)$ is an open set containing $z$. Define $f^{\prime }=f{|}_U$ and $g^{\prime }=g{|}_U$, then $f^{\prime },g^{\prime }$ are morphisms from $U$ to $V$. Since $V$ is a variety, $\{z^{\prime }\in U:f^{\prime }(z)=g^{\prime }(z)\}$ is closed in $U$, where $\{z^{\prime }\in U:f^{\prime }(z)=g^{\prime }(z)\}=U\cap Z$. Now we finish the proof. \end{proof}

Proposition

Let $X$ be a prevariety. If for any $x,y\in X$, there exists an open affine subvariety $U\subseteq X$ containing $x,y$. Then $X$ is a variety.

\begin{proof} Define $Z=\{y\in Y:f(y)=g(y)\}$. Take $z_0\in \overline{Z}\subseteq Y$. Assume $f(z_0)=x$ and $g(z_0)=y$, and we aim to show $x=y$. By assumption, there exists an affine open set $U$ such that $x,y\in U$. Then $z_0\in V:=f^{-1}(U)\cap g^{-1}(U)$ is an open set in $Y$. Since $U$ is affine, $Z\cap V=\{v\in V:f{|}_V(v)=g{|}_V(v)\}$ is closed in $V$ by ^w3d7va. It deduces that ${\overline{Z\cap V}}^V=Z\cap V$.

Note that $z_0\in {\overline{Z\cap V}}^V=Z\cap V$. Thus $z_0\in Z$ and so $Z$ is closed. \end{proof}

Corollary

${\mathbb{P}}^n$ is a variety.

\begin{proof} For any $x,y\in {\mathbb{P}}^n$, take $H=\{\sum a_i{x}_i=0\}$ such that $x,y\notin H$. Then ${\mathbb{A}}^n\simeq {\mathbb{P}}^n\setminus H$ contains $x,y$. By ^c0400a, ${\mathbb{P}}^n$ is a variety. \end{proof}

Proposition

Let $X$ be a variety, and let $U,V$ be affine open subvarieties. Then $U\cap V$ is affine.

Counterexample. ${\mathbb{A}}^2\setminus \{0\}$ is not a variety.

\begin{proof} Since $X$ is a variety, the set $\Delta (X)$ is closed by ^t398yf. Since $U,V$ are affine open subvarieties, we know $U\times V$ is affine by ^f0237a. Define

$$\tau :U\cap V\to X\times X,x\mapsto (x,x),$$

and let ${\pi }_1,{\pi }_2$ be the projections to the first and second factors, respectively. Since ${\pi }_1\circ \tau$ and ${\pi }_2\circ \tau$ are morphism, by universal property $\tau$ is also a morphism. Set $Z^{\prime }=\Delta (X)\cap (U\times V)$, and define $\theta ={\pi }_1{|}_{Z^{\prime }}:Z^{\prime }\to U\cap V$. It is easy to check $\theta \circ \tau ={\mathrm{id}}_{U\cap V}$ and $\tau \circ \theta ={\mathrm{id}}_{Z^{\prime }}$. Note that $Z^{\prime }$ is closed in $U\times V$ and $U\times V$ is an affine variety, then $Z^{\prime }$ is also an affine variety.

Define $\Gamma (U)=R$ and $\Gamma (V)=S$, then $\Gamma (U\times V)=R{\otimes }_{k}S$ by ^f0237a. Since $Z^{\prime }$ is closed in $U\times V$, we can set $\Gamma (Z^{\prime })=R{\otimes }_{k}S/I(Z^{\prime })$. Then $\tau$ induces the map ${\tau }^{\ast }:\Gamma (Z^{\prime })\to \Gamma (U\cap V)$, and $\theta$ induces the map ${\theta }^{\ast }:\Gamma (U\cap V)\to \Gamma (Z^{\prime })$. Note that ${\tau }^{\ast }$ and ${\theta }^{\ast }$ are inverse, and we have $U\cap V\stackrel{\tau }{\simeq }{Z}^{\prime }$. Therefore, $U\cap V$ is affine. \end{proof}