1. Ringed space and morphism.
In the previous note about sheaf theory, we can see that if $X$ is an affine variety and $\mathscr{O}_X$ is the set of all regular function on $X$, then $\mathscr{O}_X$ has ring structure and it is the sheaf on $X$. The pair $(X,\mathscr{O}_X)$ is called a ringed space, and $\mathscr{O}_X$ is called the structure sheaf of $X$. In general, if $X$ is a topological space, and $\mathscr{F}$ is a sheaf on $X$, then the pair $(X,\mathscr{F})$ is called ringed space. The sheaf $\mathscr{F}$ is called structure sheaf of $X$. We come to the very important definition.
Definition 1.1. Let $(X,\mathscr{O}_X)$ and $(Y,\mathscr{O}_Y)$ be two ringed space. A continuous map $f:X\rightarrow Y$ is called a map from ringed space $(X,\mathscr{O}_X)$ to $(Y,\mathscr{O}_Y)$ if for all $\varphi\in(Y,\mathscr{O}_Y)$, $\varphi\circ f\in \mathscr{O}_X$. The map $f$ is called a morphism if for all open subset $U\subset Y$, and all $\varphi\in\mathscr{O}_Y(U)$, $\varphi\circ f\in\mathscr{O}_X(f^{-1}(U))$. We call $f$ an isomorphism, if $f$ is bijective and both $f$ and $f^{-1}$ are morphisms.
In our case, when $X,Y$ is an affine variety, and $U$ is an open subset of $Y$, then $\mathscr{O}_Y(U)$ is an algebra over $k$. Hence, if $f$ is a morphism, then $f^*: \mathscr{O}_Y(U)\rightarrow \mathscr{O}_X(f^{-1}(U))$ is actually a $k$-algebra homomorphism. We can see later that there is a bijective map between the set $\{\text{morphism from }X \text{ to } Y\}$ and the set $\{k\text{-algebra from }\mathscr{O}_Y\text{ to } \mathscr{O}_X\}$. First, it can be easily checked that these properties of ringed space
(i) Composition of morphisms is a morphism: If $f: X\rightarrow Y$, and $g: Y\rightarrow Z$ are morphisms, then $g\circ f$ is also a morphism.
(ii) Restriction of a morphism is a morphism: For any open subset $U\subset X$ and $V\subset Y$, such that $f(U)\subset V$, then $f_{|U}$ is the morphism from the ringed space $(U,\mathscr{O}_X(U))$ to $(V,\mathscr{O}_Y(V))$.
We are now ready to check that
Proposition 1.2. Let $X,Y$ are two affine varieties, then $f: X\rightarrow Y$ is a polynomial mapping, then $f$ is a morphism.
Proof. Left as an exercise.
The simplest morphism is mentioned in the following
Proposition 1.3. Let $X$ be an affine variety and an open subset $U\subset X$, then the set of all morphisms from $U\to \mathbb{A}^1$ is identical with $\mathscr{O}_X(U)$.
Proof. The fact that if $f\in\mathscr{O}_Y(U)$ then $f$ is a morphism is left as an exercise. Conversely, let $f$ be a morphism from $U$ to $\mathbb{A}^1$, then it will induce the map $f^*$ from $k[\mathbb{A}^1]$ to $\mathscr{O}_Y(U)$, that sends $\phi$ to $\phi\circ f$, for all $\phi\in k[\mathbb{A}^1]$. It can be seen that $k[\mathbb{A}^1]=k[x]$, and we can choose $\phi$ such that $\phi(x)=x$. Then clearly $\phi\circ f(u)= f(u)$ for all $u\in U$. And hence, $f\in \mathscr{O}_X(U)$. (Q.E.D).
The simplest morphism is mentioned in the following
Proposition 1.3. Let $X$ be an affine variety and an open subset $U\subset X$, then the set of all morphisms from $U\to \mathbb{A}^1$ is identical with $\mathscr{O}_X(U)$.
Proof. The fact that if $f\in\mathscr{O}_Y(U)$ then $f$ is a morphism is left as an exercise. Conversely, let $f$ be a morphism from $U$ to $\mathbb{A}^1$, then it will induce the map $f^*$ from $k[\mathbb{A}^1]$ to $\mathscr{O}_Y(U)$, that sends $\phi$ to $\phi\circ f$, for all $\phi\in k[\mathbb{A}^1]$. It can be seen that $k[\mathbb{A}^1]=k[x]$, and we can choose $\phi$ such that $\phi(x)=x$. Then clearly $\phi\circ f(u)= f(u)$ for all $u\in U$. And hence, $f\in \mathscr{O}_X(U)$. (Q.E.D).
In a more general point of view, we can prove
Theorem 1.3. Let $U$ be an open subset of an affine variety $X$, and $f: U\rightarrow Y$ a morphism, where $Y$ is an affine variety in $\mathbb{A}^n$ then $f(u) = (f_1(u),...,f_n(u))$ for all $u\in U$, then $f_i\in\mathscr{O}_X(U)$. In particular, morphisms from $U$ to $\mathbb{A}^1$ are regular functions, and morphisms from $X$ to $Y$ are polynomial mappings.
Proof. First, if $f_i$ is a regular functions from $U$ to $\mathbb{A}^1$, then we can check that
(i) $f$ is a continuous map from $U$ to $Y$, for $F$ is a closed subset of $Y$, then $F$ is the set of zeros of $g_j$ in the coordinate ring $k[Y]$. It follows $f^{-1}(F)=\{u\in U|g_j\circ f (u)=0\}$. It can be seen that $g_j\circ f$ is a regular map from $U$ to $\mathbb{A}^1$, and its zeros set is closed subset of $U$. And hence, $f$ is continuous.
(ii) $f$ induces the map $f^*$ from $\mathscr{O}_Y$ to $\mathscr{O}_X(U)$. It is quite obvious, because $\mathscr{O}_Y$ consists of polynomial maps from $Y$ to $\mathbb{A}^1$, and the composition $f\circ \varphi$ is in $\mathscr{O}_X(U)$, for all $\varphi\in\mathscr{O}_Y$.
(iii) For any open subset $V\subset Y$, and $\varphi\in\mathscr{O}_Y(V)$, $f^{-1}(V)$ is open in $U$, and by the definition of regular function, one can easily check that $\varphi\circ f\in \mathscr{O}_X(f^{-1}(V))$.
Conversely, we can see that the projection maps $p_i(i=1,...,n)$ from $Y$ to $k$ is polynomial maps. And hence, $p_i\in\mathscr{O}_Y$. By the previous proposition, one can see $p_i$ is the morphism from $Y$ to $\mathbb{A}^1$. Hence if $f$ is a morphism from $U$ to $Y$, by the composition property, $f\circ p_i$ are morphisms from $U$ to $\mathbb{A}^1$, and by Proposition 1.3, $\phi_i\in \mathscr{O}_X(U)$.
Hence, all morphisms from $U$ to $Y$ has the form $(f_1,...,f_n)$ where $f_1\in \mathscr{O}_X(U)$. In particular, the morphism from $X$ to $Y$ is just the polynomial mapping, because $\mathscr{O}_X= k[X]$ for all affine variety $X$. (Q.E.D)
Theorem 1.4. Let $X, Y$ be two affine varieties, then we have the bijective map between the two sets $\{\text{morphisms from } X \text{to } Y\}$ and $\{k\text{-algebra from } k[Y]\text{to } k[X]\}$, via the map $f \mapsto f^*$.
Proof. We already saw that how $f$ induces $f^*$. Let $f'$ be a ring homomorphism from $k[Y]$ to $k[X]$, and $p_i$'s are projections from $Y$ to $k$. Then $p_i\in k[Y]$ for all $i$. We next denote $f=(f'(p_1),...,f'(p_n))$ the map from $X$ to $Y$. Because each $f'(p_i)\in \mathscr{O}_X$, we have $f\in \mathscr{O}_Y$, as mentioned in Theorem 1.3.
We next prove that $f^*=f'$. It can be seen that $f^*(p_i)=f'(p_i)$, and all polynomial functions of $k[Y]$ are just multiplications and additions of projections. Both $f^*$ and $f'$ are ring homomorphism implies that they must be identical on all $k[Y]$.
From this, we can see the bijective map between the two sets. (Q.E.D)
Theorem 1.4 implies that studying polynomial mapping between affine varieties is equivalent to studying $k$-algebra homomorphism. It also has an important categorical meaning, when one considers the category of affine varieties, where objects are affine varieties, and morphisms are morphisms between their ringed spaces. The correspondence above induces the contravariant functor from the category of affine varieties to the category of $k$-algebra.
2. Product of affine varieties.
Let $X, Y$ be two affine varieties, as our old sense, where $X$ is defined by polynomials $f_1,...,f_p\in k[x_1,...,x_m]$, and $Y$ is defined by $g_1,...,g_q\in k[x_{m+1},...,x_n]$. Then the product of $X\times Y$ is defined by $f_1,...,f_p,g_1,...,g_q\in k[x_1,...,x_n]$. It can be seen that $X\times Y$ is a closed subset of $\mathbb{A}^n$, and its polynomial ring is $k[X\times Y]=k[x_1,...,x_n]/(f_1,...,f_p,g_1,...,g_q)$. Let us prove first
Theorem 2.1. $X\times Y$ is also an affine variety, which is the product of $X$ and $Y$ in the category of affine varieties. Besides, $k[X\times Y]=k[X]\otimes_k k[Y]$, and $\dim k[X\times Y]=\dim k[X]+\dim k[Y]$.
Proof. We will prove first $X\times Y$ is irreducible closed subset of $\mathbb{A}^n$. Assume that $X\times Y=A_1\cup A_2$, where $A$ and $B$ are proper closed subset of $X\times Y$. For each $y\in Y$, we consider the map $\pi_y: X \rightarrow X\times Y$, that sends $x$ to $(x,y)$. It is a homeomorphism from $X$ to $X_y=X\times \{y\}$. Hence, $\pi_y^{-1}(A_i)$ is closed in $X$, and so is $\cap_{y\in Y}\pi_y^{-1}(A_i)$. This implies $Z_i=\{x\in X|{x}\times Y\in A_i\}$ is closed in $X$, and $X=Z_1\cup Z_2$. Because $X$ is irreducible, $Z_1=X$ or $Z_2=X$, this implies $X\times Y= A_1$ or $X\times Y= A_2$. Hence, $X\times Y$ is irreducible.
Let $\pi_X, \pi_Y$ is the projection map from $X\times Y$ to $X$ and $Y$, respectively, and $Z$ be an affine variety. Let $f_X:Z\rightarrow X$, and $f_Y: Z\rightarrow Y$ are polynomial mappings. Then we define $f: Z\rightarrow X\times Y$ that sends $z$ to $(f_X(z), f_Y(z))$. Then it is a polynomial mapping from $Z$ to $X\times Y$, and the uniqueness of $f$ now follows. Therefore, $X\times Y$ is the product of $X$ and $Y$ in the category of affine varieties.
Taking the contravariant functor as mentioned above, we obtain $k[X\times Y]$ is the coproduct of $k[X]$ and $k[Y]$ in the category of $k$-algebra. This implies $k[X\times Y]$ is the tensor product of $k[X]$ and $k[Y]$ over $k$.
The last part is left as an exercise (Hint: using the Noether's normalization theorem). (Q.E.D)
One of the usefulness of the concept of ringed space is that it allows us to extend the definition of affine variety.
Definition 2.2. Let $(X,\mathscr{O}_X)$ be a ringed space. Then $(X,\mathscr{O}_X)$ is called affine variety if it is isomorphic to a ringed space of an affine variety as the old sense.
Using Theorem 2.1 and our new definition above, we can consider the basis kind of open subsets is an affine variety.
Proposition 2.2. Let $X$ be an affine variety and $f\in k[X]$, then the open set $D(f)$ is an affine variety.
Proof. Consider the affine variety $Y=\{(x,t)\subset X\times \mathbb{A}^1|f(x)t = 1\}$. Assume $k[X] = k[x_1,...,x_n]/I$, then $k[Y]=k[x_1,...,x_n,t]/(I, tf-1)$. It can be seen that $k[Y]\cong k[X]_f$. Also, it is proved in our previous note that $\mathscr{O}_X(D(f))\cong k[X]_f$. Furthermore, it is obvious to see the map $g: D(f) \rightarrow Y$ that sends $y$ to $\bigg(y, \frac{1}{f(y)}\bigg)$ and its inverse $g^{-1}$ that sends $\bigg(y,\frac{1}{f(y)}\bigg)$ to $y$ are isomorphisms between the two ringed spaces. (Q.E.D)
From the proposition, from now on, we can consider $\mathbb{A}^1-\{0\}$ is an affine variety, because it is the open subset defined by $D(x=0)$. But it is false for $\mathbb{A}^2-\{0,0\}$.
Problem 2.3. Prove that $\mathbb{A}^2-\{0,0\}$ is not an affine variety.
But actually, we can glue two affine varieties to obtain $\mathbb{A}^2-\{0,0\}$, and it is tempting to study prevarieties, which are obtained by gluing finite affine varieties. This will be the content of our third part.
Hence, all morphisms from $U$ to $Y$ has the form $(f_1,...,f_n)$ where $f_1\in \mathscr{O}_X(U)$. In particular, the morphism from $X$ to $Y$ is just the polynomial mapping, because $\mathscr{O}_X= k[X]$ for all affine variety $X$. (Q.E.D)
Theorem 1.4. Let $X, Y$ be two affine varieties, then we have the bijective map between the two sets $\{\text{morphisms from } X \text{to } Y\}$ and $\{k\text{-algebra from } k[Y]\text{to } k[X]\}$, via the map $f \mapsto f^*$.
Proof. We already saw that how $f$ induces $f^*$. Let $f'$ be a ring homomorphism from $k[Y]$ to $k[X]$, and $p_i$'s are projections from $Y$ to $k$. Then $p_i\in k[Y]$ for all $i$. We next denote $f=(f'(p_1),...,f'(p_n))$ the map from $X$ to $Y$. Because each $f'(p_i)\in \mathscr{O}_X$, we have $f\in \mathscr{O}_Y$, as mentioned in Theorem 1.3.
We next prove that $f^*=f'$. It can be seen that $f^*(p_i)=f'(p_i)$, and all polynomial functions of $k[Y]$ are just multiplications and additions of projections. Both $f^*$ and $f'$ are ring homomorphism implies that they must be identical on all $k[Y]$.
From this, we can see the bijective map between the two sets. (Q.E.D)
Theorem 1.4 implies that studying polynomial mapping between affine varieties is equivalent to studying $k$-algebra homomorphism. It also has an important categorical meaning, when one considers the category of affine varieties, where objects are affine varieties, and morphisms are morphisms between their ringed spaces. The correspondence above induces the contravariant functor from the category of affine varieties to the category of $k$-algebra.
2. Product of affine varieties.
Let $X, Y$ be two affine varieties, as our old sense, where $X$ is defined by polynomials $f_1,...,f_p\in k[x_1,...,x_m]$, and $Y$ is defined by $g_1,...,g_q\in k[x_{m+1},...,x_n]$. Then the product of $X\times Y$ is defined by $f_1,...,f_p,g_1,...,g_q\in k[x_1,...,x_n]$. It can be seen that $X\times Y$ is a closed subset of $\mathbb{A}^n$, and its polynomial ring is $k[X\times Y]=k[x_1,...,x_n]/(f_1,...,f_p,g_1,...,g_q)$. Let us prove first
Theorem 2.1. $X\times Y$ is also an affine variety, which is the product of $X$ and $Y$ in the category of affine varieties. Besides, $k[X\times Y]=k[X]\otimes_k k[Y]$, and $\dim k[X\times Y]=\dim k[X]+\dim k[Y]$.
Proof. We will prove first $X\times Y$ is irreducible closed subset of $\mathbb{A}^n$. Assume that $X\times Y=A_1\cup A_2$, where $A$ and $B$ are proper closed subset of $X\times Y$. For each $y\in Y$, we consider the map $\pi_y: X \rightarrow X\times Y$, that sends $x$ to $(x,y)$. It is a homeomorphism from $X$ to $X_y=X\times \{y\}$. Hence, $\pi_y^{-1}(A_i)$ is closed in $X$, and so is $\cap_{y\in Y}\pi_y^{-1}(A_i)$. This implies $Z_i=\{x\in X|{x}\times Y\in A_i\}$ is closed in $X$, and $X=Z_1\cup Z_2$. Because $X$ is irreducible, $Z_1=X$ or $Z_2=X$, this implies $X\times Y= A_1$ or $X\times Y= A_2$. Hence, $X\times Y$ is irreducible.
Let $\pi_X, \pi_Y$ is the projection map from $X\times Y$ to $X$ and $Y$, respectively, and $Z$ be an affine variety. Let $f_X:Z\rightarrow X$, and $f_Y: Z\rightarrow Y$ are polynomial mappings. Then we define $f: Z\rightarrow X\times Y$ that sends $z$ to $(f_X(z), f_Y(z))$. Then it is a polynomial mapping from $Z$ to $X\times Y$, and the uniqueness of $f$ now follows. Therefore, $X\times Y$ is the product of $X$ and $Y$ in the category of affine varieties.
Taking the contravariant functor as mentioned above, we obtain $k[X\times Y]$ is the coproduct of $k[X]$ and $k[Y]$ in the category of $k$-algebra. This implies $k[X\times Y]$ is the tensor product of $k[X]$ and $k[Y]$ over $k$.
The last part is left as an exercise (Hint: using the Noether's normalization theorem). (Q.E.D)
One of the usefulness of the concept of ringed space is that it allows us to extend the definition of affine variety.
Definition 2.2. Let $(X,\mathscr{O}_X)$ be a ringed space. Then $(X,\mathscr{O}_X)$ is called affine variety if it is isomorphic to a ringed space of an affine variety as the old sense.
Using Theorem 2.1 and our new definition above, we can consider the basis kind of open subsets is an affine variety.
Proposition 2.2. Let $X$ be an affine variety and $f\in k[X]$, then the open set $D(f)$ is an affine variety.
Proof. Consider the affine variety $Y=\{(x,t)\subset X\times \mathbb{A}^1|f(x)t = 1\}$. Assume $k[X] = k[x_1,...,x_n]/I$, then $k[Y]=k[x_1,...,x_n,t]/(I, tf-1)$. It can be seen that $k[Y]\cong k[X]_f$. Also, it is proved in our previous note that $\mathscr{O}_X(D(f))\cong k[X]_f$. Furthermore, it is obvious to see the map $g: D(f) \rightarrow Y$ that sends $y$ to $\bigg(y, \frac{1}{f(y)}\bigg)$ and its inverse $g^{-1}$ that sends $\bigg(y,\frac{1}{f(y)}\bigg)$ to $y$ are isomorphisms between the two ringed spaces. (Q.E.D)
From the proposition, from now on, we can consider $\mathbb{A}^1-\{0\}$ is an affine variety, because it is the open subset defined by $D(x=0)$. But it is false for $\mathbb{A}^2-\{0,0\}$.
Problem 2.3. Prove that $\mathbb{A}^2-\{0,0\}$ is not an affine variety.
But actually, we can glue two affine varieties to obtain $\mathbb{A}^2-\{0,0\}$, and it is tempting to study prevarieties, which are obtained by gluing finite affine varieties. This will be the content of our third part.
Yesterday, I presented a wrong proof for the last part of Theorem 2.1. The problem now reduce to check $\dim k[X]\otimes_k k[Y]=\dim k[X] + \dim k[Y]$. I now come up with another proof. Please help me check it.
ReplyDeleteFirst, both $k[X], k[Y]$ are finitely generated $k$-algebra. Using the Nother's normalization theorem, let $A,B$ be normalizations of $k[X],k[Y]$, respectively. Then $A, B$ is isomorphic to polynomial rings $k[x_1,...,x_n],k[y_1,...,y_n]$ over $k$, respectively. It is well-known that $A\otimes_k B\cong k[z_1,...,z_{m+n}]$. And $\dim A\otimes B=\dim A+\dim B$.
The Noether's normalization theorem also implies that $k[X]$ (similar for $k[Y]$) is integral over $A$, and hence, $\dim A=\dim k[X]$. We now prove that $A\otimes_k B$ is Noether's normalization of $k[X]\otimes_K k[Y]$. It is suffice to check $k[X]\otimes_K k[Y]$ is integral over $A\otimes_k B$.
Let $a\otimes y$ is an element in $k[X]\otimes_k k[Y]$, where $a\in A, y\in k[Y]$, then there exists $b_i\in B$, such that $y^n+b_{n-1}y^{n-1}+...+b_0=0$. And hence, $a^n\otimes y^n+a^n\otimes b_{n-1}y^{n-1}+...+a^n\otimes b_0= (a\otimes y)^n+(a\otimes b)(a \otimes y)^{n-1}+...+a^n\otimes b_0=0$. Hence, $a\otimes y$ is integral over $A\otimes B$. Similarly, $x\otimes b$ is integral over $A\otimes B$, for all $x\in k[X], b\in B$.
Because $k[X]$ is finitely generated $k$-algebra, it is also finitely generated $A$-module. Hence, there exists $x_i\in k[X]$, such that for all $x\in k[X]$, there exists $a_i\in A$, we have $x=\sum_i a_ix_i$. Similarly, there exists $y_j\in k[Y]$, such that, for all $y\in k[Y]$, there exists $b_j\in B$, we have $y=\sum_j b_jy_j$. From this, $x\otimes y=(\sum_i a_ix_i)\otimes (\sum_j b_jy_j)$, which is the combination of the form $\sum (a\otimes y)(x \otimes b)$, which is clear integral in $A\otimes_k B$.
Hence, the Noether's normalization of $k[X]\otimes k[Y]$ is $A\otimes B$. And our result now follows.