Notes on divisors-2

Weil divisors v.s. Cartier divisors

Let’s continue the discussion of the last post. In this post, we shall establish a correspondence between Weil divisors and Cartier divisors for special schemes.

First let’s repeat what this correspondence is:

Suppose that $X$ is an integral, separated noetherian scheme such that all of its local rings are UFDs. Then the group $Div(X)$ of Weil divisors on $X$ is isomorphic to the group $\Gamma(X,\mathfrak{K}^*/\mathfrak{O}_X^*)$ of Cartier divisors. Besides, the principal Weil divisors correspond to the principal Cartier divisors.

This correspondence and the following proof are taken from Hartshorne’s book, ‘algebraic geometry’.

We shall not give a complete and rigorous proof to this statement. We only give a rough idea of what this correspondence is like. First note that $X$ is integral(reduced and irreducible), thus the sheaf $\mathfrak{K}$ is the constant sheaf of $K$ on $X$ where $K$ is the function field on $X$, which is also the local ring $\mathfrak{O}_{X,x_0}$ with $x_0$ the generic point of $X$. So, a Cartier divisor is given by a family $\{U_i,f_i\}$ where $U_i$ is an open cover of $X$ and $f_i\in \Gamma(U_i,\mathfrak{O}_X)=K^*$. We now construct a Weil divisor: for each prime divisor $Y$ of $X$, by definition, this means that $Y$ is an integral closed subscheme of $X$, now for each $f_i$, we set the coefficient of $Y$ to be $v_Y(f_i)$. Note that for any two $U_i,U_j$ such that $U_i\bigcap U_j$, then by definition, $f_i/f_j\in \Gamma(U_i\bigcap U_j,\mathfrak{O}_X)$, that is to say, $f_i/f_j$ is invertible on $U_i\bigcap U_j$, thus $v_Y(f_i)=v_Y(f_j)$, so the coefficient of $Y$ is well-defined. And we set the Weil divisor to be $D=\sum_Yv_Y(f_i)$. Note here we should guarantee that this is a finite sum. This can be deduced easily from the assumption that $X$ is noetherian(we will say about this later).

Now conversely, if $D$ is a Weil divisor on $X$, then for each point $x\in X$, $D$ gives a divisor on $Spec(\mathfrak{O}_{X,x})$. Now by assumption, $\mathfrak{O}_{X,x}$ is a UFD, we can show that $D_x$ is a principal divisor(we shall not prove this point for the present), that is, $D_x=(f_x)$ for some $f_x\in K^*$. This means that there is an open subset $x\in U$ of $X$ such that $f_x$ and $D$ agree on $U$. In this way, varying $x$, we get an open cover $U_i$ of $X$ with $f_i$ such that $f_i$ and $D$ agree on $U_i$. And we set $\{U_i,f_i\}$ to be the corresponding Cartier divisor. So, we finish the construction of the correspondence between Weil divisors and Cartier divisors.

It is not too hard to see that these two construction are inverse one to another.

Relation to invertible sheaves

Next we want to relate divisors to invertible sheaves.

Suppose that $X$ is a scheme and $\mathfrak{O}_X$ its structure sheaf. Then we define an invertible sheaf to be an $\mathfrak{O}_X$-module locally free of rank $1$.

Now let $D$ be a Cartier divisor on $X$ given by $\{U_i,f_i\}$, then we define the sheaf $\mathfrak{L}(D)$ to be the submodule of $\mathfrak{K}$ generated by $1/f_i$ on each $U_i$ for all $i$. We call $\mathfrak{L}(D)$ to be the sheaf associated to $D$.

The important result is that, this gives a one-to-one correspondence between the set of Cartier divisors and the set of invertible submodules of $\mathfrak{K}$. This can be seen like this: given a invertible submodule $\mathfrak{L}$ of $\mathfrak{K}$, for an open subset $U$ of $X$, $\mathfrak{L}(U)$ is a free module of $\mathfrak{O}_X(U)$ of rank $1$, which is also a submodule of $\mathfrak{K}(U)$. So, we take a generator $f_U$ of $\mathfrak{L}(U)$(it is not zero since $\mathfrak{L}(U)$ is of rank $1$, not $0$), then we get an open covering $U_i$ of $X$ such that on each $U_i$ there is an $f_i$. We can verify that these $f_i$ can glue together, and thus we can set $\{u_i,f_i\}$ to be the Cartier divisor. It can be verified that this process is converse to the above one, thus we showed that there is a correspondence between the set of Cartier divisors and the set of invertible submodules of $\mathfrak{K}$.

In fact, this is more than a bijection, this is a group homomorphism. Recall that we give a group structure to the set of invertible sheaves on $X$ by tensor product: suppose that $\mathfrak{L},\mathfrak{L}'$ are two invertible sheaves on $X$, then by definition, $\mathfrak{L}\bigotimes_{\mathfrak{O}_X}\mathfrak{L}'$ is again an invertible sheaf on $X$. Besides, $\mathfrak{L}^*=Hom_{\mathfrak{O}_X}(\mathfrak{L},\mathfrak{O}_X)$ is also an invertible sheaf on $X$, and $\mathfrak{L}\bigotimes \mathfrak{L}^*\simeq\mathfrak{O}_X$. We call this group the Picard group of $X$,

Charles Emile Picard(photo taken from here)

and denote it by $Pic(X)$. So the set of invertible submodules of $\mathfrak{K}$ is a subgroup of $Pic(X)$.

It is not hard to see that $\mathfrak{L}(D/D')=\mathfrak{L}(D)\bigotimes\mathfrak{L}(D')^*$, so this gives a group homomorphism between the group of Cartier divisors and the group of invertible submodules of $\mathfrak{K}$.

In general, the group of Cartier divisors is not equal to the Picard group of $X$. Yet, under some mild condition, we can show that, these two groups are isomorphic under the above morphism:

Suppose that $X$ is an integral scheme, then the group of Cartier divisors on $X$ is isomorphic to $Pic(X)$.

We will establish this isomorphism in the next post.

Notes on divisors-1

Introduction

Divisors are a very important tool in algebraic geometry. Let’s quote an example from Hartshorne’s ‘Algebraic Geometry’: let $C$ be a non-singular projective curve in the projective plane $X=\mathbb{P}^2_k$ over a field $k$. Now let $L$ be any line in $X$, and we write $L\bigcap C$ for their intersection, which consists of finitely many points counted with multiplicity. Now let’s vary the line $L$, then we obtain a family of finite sets $L\bigcap C$. It is not hard to see that we can construct the embedding of $C$ in $X$ using this family of sets $L\bigcap C$. This is a very powerful tool in studying this kind of embeddings.

Weil divisors

The simplest kind of divisors is perhaps the Weil divisors.

André Weil(photo from here)

Suppose that $X$ is a scheme, we say that $X$ is regular of co-dimensional one if all its local rings $\mathfrak{O}_{X,x}$ of dimensional $1$ is regular.

Let me explain these terms. Let $R$ be a ring, then for any prime ideal $\mathfrak{p}$ of $R$, its height $h(\mathfrak{p})$ is defined to be the largest integer $n$ such that there exists a strictly increasing sequence of prime ideals $\mathfrak{p}_0\subset \mathfrak{p}_1\subset...\subset\mathfrak{p}_n=\mathfrak{p}$. And the dimension $dim(R)$ of $R$ is defined to be the maximum of heights of its prime ideals. This is the so-called Krull dimension of $R$. For some rings $R$, this $dim(R)$ can be infinite, even if $R$ is a noetherian ring, $dim(R)$ can still be infinite, the classical examples are perhaps due to Nagata. Let $k$ be a field and consider $R=k[x_1,x_2,...,x_n,...]$ the polynomial ring on infinite variables. We set the prime ideals to be $\mathfrak{p}_n=R$ generated by the variables in the bracket. And we set $S=R-\bigcup_n \mathfrak{p}_n$, a multiplicative closed subset of $R$, then we take the localization $R'=S^{-1}R$. It can be shown that the maximal ideals of $R'$ are of the form $\mathfrak{p}_nR'$, so we have that $h(\mathfrak{p}_nR')\geq (n+1)^2-n^2$. Thus we have that $dim(R')=\infty$(cf. Eisenbud’s ‘Commutative algebra with a view towards algebraic geometry’, Exercise 9.6).

Yet the good news for algebraic geometers is that for noetherian local rings, their dimensions are always finite. We can find this result in Atiyah-Macdonald’s ‘commutative algebra’, the last chapter on dimension theory. The usefulness of the dimensions of a ring gives a good criterion for a variety to be non-singular at a point. For that, we need the concept of regular rings. Let $R$ be a noetherian local ring with maximal ideal $\mathfrak{m}$ and residue field $k=R/\mathfrak{m}$, then we say that $R$ is regular if $dim_k(\mathfrak{m}/\mathfrak{m}^2)=dim(R)$.

Let $Y\subset\mathbb{A}^n_k$ be an affine variety and $P$ is a point in $Y$, then $Y$ is non-singular at $P$ if and only if $\mathfrak{O}_{Y,P}$ is regular.

The importance of this result is that it gives an intrisinc description of singular points of a variety. Of course, this criterion can be generalized to other situations since it is a local criterion.

Now let’s return to our origin: we say that a scheme $X$ is regular of co-dimension one if all its local rings $\mathfrak{O}_{X,x}$ of dimension $1$ are regular rings. So it is clear that nonsingular varieties are regular of co-dimension $1$(considering its associated scheme). Another important class of examples is noetherian normal schemes. Recall that a normal scheme $X$ is a scheme such that all of its local rings $\mathfrak{O}_{X,x}$ are integrally closed. We can use Proposition 9.2 of Atiyah-Macdonald’s ‘commutative algebra’ to show that if a noetherian local ring of dimension one is integrally closed, then it is regular. Note that the word ‘co-dimension’ refers to the dimension of the local rings.

In this post we shall assume that a scheme $X$ is noetherian integral separated, regular of co-dimension one. We will see why we make these assumptions on $X$.

A prime divisor of $X$ is a closed integral subscheme $Y$ of $X$ of co-dimension $1$(here integral scheme correspond to algebraic varieties, ‘integral’ means reduced and irreducible). Then a Weil divisor of $X$ is define to be an element of the free abelian group on the basis of all prime divisors of $X$. If $D=\sum_in_iY_i$ is a Weil divisor on $X$, we say that $D$ is effective if $n_i\geq 0$ for all $i$.

Note that if $Y$ is a prime divisor of $X$, then by definition, $Y$ is irreducible, thus it has a (unique) generic point, say, $y_0\in Y$. We have that $\mathfrak{O}_{Y,y_0}$ is a discrete valuation ring(this also comes from Atiyah-Macdonald’s book, prop 9.2) with quotient field $K$, and we write $v_Y$ for this valuation.

We can show that under the assumptions, $K$ is equal to the function field on $X$. Now for any $f\in K^*$, which is also an element in the quotient field of $\mathfrak{O}_{X,y_0}$ where $x$ is the generic point of a prime divisor $Y$, we set $(f)=\sum_Yv_Y(f)Y$, which is thus a Weil divisor on $X$(if we admit the result that $v_Y(f)=0$ for almost all such $Y$). We call $(f)$ a principal Weil divisor on $X$, and we say that two Weil divisors are linearly equivalent if their difference is a principal Weil divisor.

We will say something more on the properties of Weil divisors later on. For the present, we will turn to the definition of Cartier divisors.

Pierre Emil Jean Cartier(photo from here)

Cartier divisors

Suppose again that $X$ is a scheme and $U=Spec(A)\subset X$ is an affine open subset of $X$. We write $S$ to be the subset of $A$ consisting of non-zero divisors and $K(U)=S^{-1}A$, which we call the total quotient ring of $A$(or the total fraction ring of $A$). For general open subset $U\subset X$, we set $S$ to be the subset of $\Gamma(U,\mathfrak{O}_X)$ consisting of elements $f\in \Gamma(U,\mathfrak{O}_X)$ such that $f_x$ is not a zero-divisor for each $x\in U$. Then we define a presheaf of rings on $X$ to be: for each $U$, the ring is $S^{-1}\Gamma(U,\mathfrak{O}_X)$. Then we call the associated sheaf $\mathfrak{K}$ to be the sheaf of total quotient rings of $\mathfrak{O}_X$. And $\mathfrak{K}^*$ is the sheaf of multiplicative groups, which consists of invertible elements in the sheaf $\mathfrak{K}$. Similarly, we have $\mathfrak{O}_X^*$.

Now we define a Cartier divisor to be a global section of the sheaf $\mathfrak{K}^*/\mathfrak{O}_X^*$ the sheaf of multiplicaitve groups(since these groups are abelian, we shall write them additively if there is no ambiguity). The set of Cartier divisors on $X$ clearly forms an abelian group. A Cartier divisor is principal if it is in the image of the map $\Gamma(X,\mathfrak{K}^*)\rightarrow\Gamma(X,\mathfrak{K}^*/\mathfrak{O}_X^*)$. Two Cartier divisors are called linearly equivalent if they differ by a principal Cartier divisor.

Weil divisors v.s. Cartier divisors

At first sight, there is no relation between Weil divisors and Cartier divisors. Yet the following proposition tells us that in special cases these two are in fact the same thing.

Suppose $X$ is an integral, separated noetherian scheme. Suppose also that all its local rings are UFDs, then the abelian group $Div(X)$ of Weil divisors on $X$ and the group of Cartier divisors $\Gamma(X,\mathfrak{K}^*/\mathfrak{O}_X^*)$ are isomorphic. Besides, the principal Weil divisors corresponds to the principal Cartier divisors.

We shall prove this proposition in the next post.

notes on topological K-theory

This post is about topological K-theory. As for general information on this branch of mathematics, let’s quote some phrases from wikipedia: K-theory, roughly speaking, is the study of certain invariants of large matrices. In this post we will touch a very little bit of this K-theory. The materials of this post are taken from Husemoller’s book, ‘Fibre Bundles'(GTM82) and Hatcher’s bool, ‘algebraic topology’.

Preliminaries

Here we list some foundational concepts for understanding K-theory.

Homotopy is the most basic one among these concepts. For any two topological spaces $X$ and $Y$ and any two continuous maps $f,g:X\rightarrow Y$, we say that $f$ is homotopic to $g$ is there is a continuous map $F:X\times[0,1]\rightarrow Y$ such that $F(x, 0)=f(x),F(x,1)=g(x),\forall x\in X$, here the unit interval $[0,1]$ is given the Euclidean topology. In other words, $f$ is homotopic to $g$ if $f$ can vary continuously to $g$. It is not difficult to verify that homotopy is an equivalence relation. One major concern of algebraic topology is to classify topological spaces up to homotopy. The category of topological spaces with maps up to homotopy equivalence is a quotient category of the category of topological spaces with continuous maps.

Sometimes we study also based spaces and maps. A based space is a topological space with a point in it, $(X,x_0)$. A map between two based spaces $(X,x_0),(Y,y_0)$ is a continuous map $f,X\rightarrow Y$ such that $f(x_0)=y_0$. The category of based topological spaces with maps up to homotopy equivalence is a quotient category of the topological spaces with continuous maps.

Next let’s define homotopy groups. A loop in a based topological space $(X,x_0)$ is a continuous map $f:[0,1]\rightarrow (X,x_0)$ such that $f(0)=f(1)=x_0$. Up to homotopy, a loop has an inverse, two loops can compose, and the constant map makes the set of loops in $(X,x_0)$ into a group, which we write as $\pi_1(X,x_0)$. We write $\Omega(X)$ for the space of loop spaces. This is a subspace of the map space $Map(I,X)$, to which we give the compact open topology. This is again a based space with base point the constant map from $I$ to $(X,x_0)$.

In order to introduce higher order homotopy groups, we need some operations in the category of (based) spaces. For two based spaces $(X,x_0),(Y,y_0)$, we define their based union to be $X\vee Y$ to be $X\sqcup Y/(x_0\sim y_0)$(with base point $x_0=y_0$), the is wedge product to be $X\wedge Y=X\times Y/((x,y_0)\sim(x_0,y_0)\sim(x_0,y))$ with base point $(x_0,y_0)$. These two operations are really the analog of the disjoint union and Cartesian product in the category of topological spaces. We write $C(X)$ for the wedge product of $(X,x_0)$ and $(I,0)$ where $I=[0,1]$ and $S(X)$ the wedge product of $(X,x_0)$ and $(S^1,0)$ where $S^1=I/(0\sim1)$. $C(X)$ is called the cone of $X$, and $S(X)$ is called the suspension of $X$. It is remarkable to see that $S^{n+1}=S(S^n)$, that is, the $n+1$-sphere is the wedge product of an $n$-sphere. It is not hard to imagine this identification, yet it takes some effort to write it down.

Next we say that a multiplication on a based space $(Y,y_0)$ is a map of based spaces $f:Y\times Y\rightarrow Y$. Note that this $f$ induces in an obvious way a map $f_X: Map_0(X,Y)\times Map_0(X,Y)\rightarrow Map_0(X,Y)$ where $Map_0(X,Y)$ is the spaces of maps of based spaces $(X,x_0),(Y,y_0)$ given the subspace topology induced from $Map(X,Y)$, the space of continuous maps from $X$ to $Y$, which is given the compact open topology. A co-multiplication on $(X,x_0)$ is a map $g:X\rightarrow X\vee X$. Again this $g$ induces a map $g^X:Map_0(X,Y)\times Map_0(X,Y)\rightarrow Map_0(X,Y)$(indeed, for any two $i,j:(X,x_0)\rightarrow(Y,y_0)$, we can define $g^Y(i,j)\in Map_0(X,Y)$ as follows: for any $x\in X$, if $g$ takes $x$ to the first copy of $X$ in $X\vee X$, then we set $g^Y(i,j)(x)=i(x)$, otherwise $g^Y(i,j)(x)=j(x)$. The importance of based spaces’ assumption is that, at the point $x_0$, $g^Y(i,j)$ is well defined).

Are there some examples of multiplications and co-multiplications? In fact the loop space $\Omega(Y)$ is a multiplication space, note that for any $\alpha,\beta\in \Omega(Y)$, the natural composition of $\alpha,\beta$ gives the multiplication $f: \Omega(Y)\times \Omega(Y)\rightarrow \Omega(Y), (\alpha,\beta)\mapsto \alpha\circ\beta$. The suspension $S(X)$ is a co-multiplicaiton space, $g:S(X)\rightarrow S(X)\vee S(X)$ takes $(x,t)\in S(X)=X\wedge S^1$ to $((x,2t),*)$ if $0\leq t\leq 1/2$ and to $(*,(x,2t-1))$ otherwise, where $*$ is the base point of $S(X)$.

For any $(X,x_0)$ with a co-multiplication or $(Y,y_0)$ with multiplication, we see easily that $Map_0(X,Y)$ has a monoid structure, yet in general this is not a group. We call $(X,x_0)$ a co-$H$-space if $Map_0(X,Y)$ is a group for any based space $(Y,y_0)$, and we call $(Y,y_0)$ an $H$-space if $Map_0(X,Y)$ is a group for any $(X,x_0)$. We can verify that $S(X)$ is an $H$-space and $\Omega(Y)$ is a co-$H$-space. At first sight, this is quite magical, at least this is so for me. Yet let’s recall how to show that $S^1$ is an $H$-space. To construct the inverse of a loop in $(X,x_0)$, we use essential the identification $S^1=I/\partial I$. In general, this is also the case. Recall that $S(X)=X\wedge S^1$, a quotient space of $X\times I$, for any $\alpha\in Map_0(S(X),Y)$, we define $\alpha'(x,t)=\alpha(x,1-t)\in Y$. Now use the co-multiplication structure $g:S(X)\rightarrow S(X)\vee S(X)$, and we write $g^Y(\alpha,\alpha')=\beta$, then $\beta(x,t)=\alpha(x,2t)$ if $0\leq t\leq 1/2$, and $\beta(x,t)=\alpha'(x,2t-1)=\alpha(x,2-2t)$ otherwise. We need to show that $\beta$ is homotopic to the constant map $const: S(X)\rightarrow Y,(x,t)\mapsto y_0$. This is not hard to verify, just as in the case of $S^1$. We omit the case that $\Omega(Y)$ is an $H$-space.

Now finally we can define the homotopy group. The $n$-th homotopy group($n>0$) of a based space $(X,x_0)$ is $\pi_n(X,x_0)=Map_0(S^n,X)$.

One surprising result is that $\pi_n(X,x_0)$ is abelian for all $n>1$. The usefulness of homotopy groups comes from Whitehead’s theorem:

For two CW-complexes $X, Y$, if the map $f:X\rightarrow Y$ induces isomorphisms $f_*: \pi_n(X)\rightarrow \pi_n(Y)$ for all $n$, then $f$ is a homotopy equivalence.

So this theorem justifies the importance of the homotopy groups, yet they are very hard to compute.

In the next post we shall say something about fibre bundles, which turns out to be a fundamental tool in the development of $K$-theory.

Notes on quasi-coherent sheaves and coherent sheaves-1

Sheaves are great generalizations of function spaces on a topological space. Yet not all sheaves have nice properties, especially when it comes to computing their topological invariants, the homological groups and cohomological groups. So we should try to find some nicely-behaved sheaves. It was perhaps due to J-P.Serre in his article ‘Faiseaux Algèbriques Cohérents’ that we found at last some class of sheaves which indeed have nice properties, it is the quasi-coherent sheaves and coherent sheaves that we are going to talk about in this post. In this post, we will give the definitions and basic properties of these sheaves. The references for this post are Hartshorne’s Algebraic Geometry and Qing Liu’s Algebraic Geometry and Arithmetic Curves as well as some posts from the Internet.

Sheaf Modules

Given a topological space $X$ and a sheaf $\mathfrak{O}_X$ of (locally) ringed space, just as in the case of given a ring, where we would like to consider all the modules over this ring, here we also would like to consider the category of $\mathfrak{O}_X$-modules. In other words, an $\mathfrak{O}_X$-module is a sheaf $\mathfrak{F}$ over $X$ such that for any open subset $U$ of $X$, $\mathfrak{F}(U)$ is a module over $\mathfrak{O}_X(U)$. We define morphisms between $\mathfrak{O}_X$-modules in an obvious way, thus we get a category $Mod(\mathfrak{O}_X)$, and when there is no ambiguity, we write simply $Mod(X)$.

It is not difficult to verify that $Mod(X)$ is an abelian category. So there is a natural question arising: does this category admit enough projective or injective objects? Very interestingly, Grothendieck showed in his famous Tohoku paper ‘Sur quelques points d’algèbre homologique’ that $Mod(X)$ indeed has enough injectives. Nowadays the proof of this result becomes fair standard, and let’s copy such one from Hartshorne’s book:

For each $x\in X$, we consider the stalk $\mathfrak{F}_x$, which has a morphism $\mathfrak{F}_x\rightarrow I_x$ where $I_x$ is an injective module over $\mathfrak{O}_{X,x}$(interestingly, the existence of enough injectives in a category of modules over a ring is more difficult to prove than that of enough projectives). Now that there is a canonical injection $j_x:\{x\}\rightarrow X$ for each point $x\in X$, thus if we view $I_x$ also as the sheaf on the single point space $\{x\}$, then the push-forward map gives that $j_{x,*}(I_x)$ is a sheaf on $X$. Now we take the products $\mathfrak{I}=\prod_{x\in X}j_{x,*}(I_x)$, which is thus a sheaf over $X$. Next we show that $\mathfrak{F}$ maps injectively into $\mathfrak{I}$ and $\mathfrak{I}$ is injective. Let’s first show that second point. Now that $\mathfrak{I}$ is a product of these sheaves $I_x$, thus for any sheaf $\mathfrak{G}$ over $X$, we have that $Hom_{\mathfrak{O_X}}(\mathfrak{G},\mathfrak{I})=\prod_{x\in X}Hom_{\mathfrak{O}_X}(\mathfrak{G},j_{x,*}(I_x))$. Yet one has that $Hom_{\mathfrak{O}_X}(\mathfrak{G},j_{x,*}(I_x))=Hom_{\mathfrak{O}_{X,x}}(\mathfrak{G}_x,I_x)$ since $I_x=\mathfrak{I}_x$. Now we can in passing show the first point: take $\mathfrak{G}=\mathfrak{F}$, since for each $x\in X$, there is an injective map $\mathfrak{F}_x\rightarrow\mathfrak{I}_x$ by construction, thus there is a morphism $\mathfrak{F}\rightarrow\mathfrak{I}$, which is thus also injective. Note that $Hom_{\mathfrak{O_X}}(\mathfrak{G},\mathfrak{I})=\prod_{x\in X}Hom_{\mathfrak{O}_X}(\mathfrak{G},j_{x,*}(I_x))$, the functor of taking stalks for all $x\in X$, $\mathfrak{F}\mapsto\mathfrak{F}_x$ is exact(we emphasize on the word ‘all’), and at each stalk, the functor $\mathfrak{G}_x\mapsto Hom_{\mathfrak{O}_{X,x}}(\mathfrak{G}_x,I_x)$ is also exact since $I_x$ is injective, thus the functor $Hom_{\mathfrak{O}_X}(\cdot, \mathfrak{I})$ is exact, thus $\mathfrak{I}$ is injective.

So we have that any $Mod(X)$ has enough injectives, yet the problem of enough projectives is more subtle: not every $Mod(X)$ has enough projectives. Perhaps it is better to give some counterexamples(these are taken from the answers in this post). Moreover, it is easy to see that above argument does not apply to the projective case, the essential obstacle is that, $Hom_{\mathfrak{O}_X}(\mathfrak{G},j_{x,*}(I_x))=Hom_{\mathfrak{O}_{X,x}}(\mathfrak{G}_x,I_x)$. Note that $j_x^{-1}$ is left adjoint to the functor $j_{x,*}$, which gives the above identity. So the existence of enough injectives comes essentially from the point that the stalk functor at one point is left adjoint to the extension by zero functor. And that is why we can use this argument to show the existence of enough projectives.

Now there is a natural functor, the global section functor, $\Gamma(X,\cdot):Mod(X)\rightarrow Mod(\mathfrak{O}_X(X))$, from $Mod(X)$ to the category of $\mathfrak{O}_X(X)$-modules.

Unfortunately, this functor is not an exact functor(in some sense, this is also formate for us, since otherwise there would not be such a rich theory of sheaf cohomology). Suppose that $X=\mathbb{A}^1_k$ the affine line over a field $k$, and $P,Q\in X$ two distinct closed points and $U=X-\{P,Q\}$, the open complement of them in $X$. Now we take the constant sheaf $\mathfrak{O}_X=\mathbb{Z}_X$ over $X$. Now we set $\mathbb{Z}_U=i_!(\mathbb{Z}_X|_U)$ where $i:U\rightarrow X$ is the inclusion map and $\mathbb{Z}_Y=j_*(\mathbb{Z}_X|_Y)$ where $j:Y=\{P,Q\}\rightarrow X$ is also the inclusion. We have an exact sequence $0\rightarrow \mathbb{Z}_U\rightarrow\mathbb{Z}_X\rightarrow\mathbb{Z}_Y\rightarrow0$. Yet consider what is $\mathbb{Z}_Y$ now: by definition, $\mathbb{Z}_Y$ is a sheaf on $X$ such that for each open subset $V\subset X$, there is $\mathbb{Z}_Y(V)=\mathbb{Z}_X|_Y(j^{-1}(V))$. If $P\in V,Q\not\in V$, we have that $\mathbb{Z}_Y(V)=\mathbb{Z}$, similar for the case $P\not\in V,Q\in V$. If $P,Q\in V$, then $\mathbb{Z}_Y(V)=\mathbb{Z}^2$. Taking global section, we see that $\mathbb{Z}_Y(X)=\mathbb{Z}^2$, $\mathbb{Z}_X(X)=\mathbb{Z}$, $\mathbb{Z}_U(X)=0$, thus the global section sequence $0\rightarrow 0\rightarrow \mathbb{Z}\rightarrow\mathbb{Z}^2\rightarrow0$ is no longer exact.

So now we may wonder what conditions we should pose on the underlying topological space or on the sheaves to ensure that $H^i(X,\mathfrak{F})=0$ for $i>0$, where we write $H^i(X,\mathfrak{F})$ for the $i$-th right derived functor of the global section functor.

Quasi-coherent sheaves

Quasi-coherent sheaves partially answers this question. Now let’s give the definition:

Suppose $\mathfrak{F}$ is an $\mathfrak{O}_X$-module, we say that $\mathfrak{F}$ is generated by its global sections at $x\in X$ if the canonical morphism $\mathfrak{F}(X)\bigotimes_{\mathfrak{O}_X(X)}\mathfrak{O}_{X,x}\rightarrow \mathfrak{F}_x$ is surjective. We say that $\mathfrak{F}$ is generated by its global sections if it is so at each point $x\in X$.

It is not hard to see that $\mathfrak{F}$ is generated by its global sections if and only if there is an index set $I$ and a surjective morphism $\mathfrak{O}_X^{\bigoplus |I|}=\mathfrak{O}_X^{(I)}\rightarrow\mathfrak{F}\rightarrow0$.

We say that $\mathfrak{F}$ is quasi-coherent if for each $x\in X$, there is an open neighborhood $x\in U\subset X$ such that there is an exact sequence $\mathfrak{O}_X^{(J)}|_U\rightarrow\mathfrak{O}_X^{(I)}|_U\rightarrow \mathfrak{F}|_U\rightarrow0$.

So this means that a quasi-coherent sheaf $\mathfrak{F}$ locally admits free resolutions(of exact sequence of length $2$).

Quasi-coherent sheaves behave quite nicely over an affine scheme. This can be seen from the result below.

Let’s fix a commutative ring $R$ and set $X=Spec(R)$. Now if $M$ is a module over $R$, then we can give a natural $\mathfrak{O}_X$-module $M^{\sim}$ constructed as follows: for each affine open subset $D(r)$ of $X$ with $r\in R$, we set $M^{\sim}(D(r))=M_r=M\bigotimes_RR_r$. For general open subsets, we take the direct limit. It is easy to see that $M^{\sim}(X)=M$, and $(M^{\sim})_{\mathfrak{p}}=M_{\mathfrak{p}}$ for any $\mathfrak{p}\in X$. Clearly, we have that $(\bigoplus_iM_i)^{\sim}\simeq \bigoplus_iM^{\sim}_i$ where each $M_i$ is an $R$-module. One surprising result is that

A sequence of $R$-modules $L\rightarrow M\rightarrow N$ is exact if and only if $L^{\sim}\rightarrow M^{\sim}\rightarrow N^{\sim}$ is so. This is really an important proposition, and let’s give a proof of it.

Suppose that $L\rightarrow M\rightarrow N$ is exact, then we know that $R_{\mathfrak{p}}$ is flat, thus the sequence $L\bigotimes_RR_{\mathfrak{p}}\rightarrow M\bigotimes_RR_{\mathfrak{p}}\rightarrow N\bigotimes_RR_{\mathfrak{p}}$ is again exact, yet the latter is just $(L^{\sim})_{\mathfrak{p}}\rightarrow(M^{\sim})_{\mathfrak{p}}\rightarrow(N^{\sim})_{\mathfrak{p}}$, which shows that $L^{\sim}\rightarrow M^{\sim}\rightarrow N^{\sim}$ is indeed exact. Conversely, suppose that $L^{\sim}\rightarrow M^{\sim}\rightarrow N^{\sim}$ is exact, then at stalk level, we have that $(L^{\sim})_{\mathfrak{p}}\rightarrow(M^{\sim})_{\mathfrak{p}}\rightarrow(N^{\sim})_{\mathfrak{p}}$ is also exact, so is $L\bigotimes_RR_{\mathfrak{p}}\rightarrow M\bigotimes_RR_{\mathfrak{p}}\rightarrow N\bigotimes_RR_{\mathfrak{p}}$. Now consider the original sequence $L\xrightarrow{f} M\xrightarrow{g} N$ where $g\circ f=0$ since the sheaf sequence is exact. We have that $(Ker(g)/Im(f))_{\mathfrak{p}}=0$ for all $\mathfrak{p}\in Spec(R)$. This shows that $Ker(g)/Im(f)=0$, thus $L\rightarrow M\rightarrow N$ is indeed exact.

So in the case of affine schemes, for this type of sheaves, we see that the global section functor is exact.

Then what is the relation of this kind of sheaves with quasi-coherent shaves that we have just introduced? Well, note that if $M$ is an $R$-module, then it admits a free presentation, $R^{(J)}\rightarrow R^{(I)}\rightarrow M\rightarrow 0$ is an exact sequence. And therefore $(R^{(J)})_r\rightarrow (R^{(I)})_r\rightarrow M_r\rightarrow 0$ is again exact since $R_r$ is flat module over $R$ for each $r\in R$. So, this means that for each $\mathfrak{p}\in X$, there is an open neighborhood $\mathfrak{p}\in D(r)\subset X$ such that $(R^{(J)})^{\sim}|_{D(r)}\rightarrow (R^{(I)})^{\sim}|_{D(r)}\rightarrow M|_{D(r)}\rightarrow 0$ is exact since $M|_{D(r)}=M_r$, similar for the other two.

In fact, we can show that quasi-coherent sheaves locally look like this, that is:

Proposition: let $X$ be a scheme(not necessarily affine), and $\mathfrak{F}$ be an $\mathfrak{O}_X$-module. Then $\mathfrak{F}$ is quasi-coherent if and only for each affine open subset $U\subset X$, there is $\mathfrak{F}(U)^{\sim}\simeq \mathfrak{F}|_U$.

This is really exciting news: quasi-coherent sheaves locally look all like sheaf modules constructed as above, which, as we have seen, have very nice properties.

Let’s also try to give a proof of this result. For this, we need a lemma:

Lemma: let $X$ be a scheme which is Noetherian or separated and quasi-compact, $\mathfrak{F}$ be a quasi-coherent sheaf over $X$, then for each $f\in \mathfrak{O}_X(X)$, the canonical morphism

$\mathfrak{F}(X)_f=\mathfrak{F}(X)\bigotimes_{\mathfrak{O}_X(X)}\mathfrak{O}_X(X)_f\rightarrow \mathfrak{F}(X_f)$

is an isomorphism, where $X_f=\{x\in X|f_x\in \mathfrak{O}_{X,x}^*\}$.

Note that since $\mathfrak{F}$ is quasi-coherent, for each $x\in X$, there is an affine open subset $x\in U\subset X$ such that there is a free representation $\mathfrak{O}_X^{(J)}|_U\rightarrow \mathfrak{O}_X^{(I)}|_U\xrightarrow{f} \mathfrak{F}|_U\rightarrow 0$. Taking global sections, we set $M=f(U)$, then we have that $\mathfrak{O}_X^{(J)}|_U\rightarrow \mathfrak{O}_X^{(I)}|_U\xrightarrow{f} M^{\sim}\rightarrow 0$ is exact, which implies that $M^{\sim}\simeq \mathfrak{F}|_U$, thus we get that $M\simeq \mathfrak{F}(U)$. In other words, for each $x\in X$, there is an affine open subset $x\in U\subset X$ such that $\mathfrak{F}|_U\simeq \mathfrak{F}(U)^{\sim}$. By assumption, $X$ is quasi-compact, thus we can find a finitely many such $U_i$ covering $X$. Moreover, we set $V_i=U_i\bigcap X_f$. Let’s spell out what $V_i$ is. Recall that $X_f$ consists of points $x\in X$ such that $f_x$ is invertible in $\mathfrak{O}_{X,x}$, in other words, if $x$ is viewed as a prime ideal in $R_i$ where $U_i=Spec(R_i)$, then $f|_{U_i}\not\in x$. So we see that $V_i=D(f|_{U_i})$. Now for each $i$, we have maps $\mathfrak{F}(U_i)_f=\mathfrak{F}(U)\bigotimes_{\mathfrak{O}_X(U_i)}\mathfrak{O}_X(U_i)_f\rightarrow \mathfrak{F}(V_i)=\mathfrak{F}(D(f|_{U_i}))$, this is an isomorphism since $\mathfrak{F}|_{U_i}\simeq \mathfrak{F}(U_i)^{\sim}$. Now using the routine commutative diagram,

The horizontal sequences are exact according to the definition of scheaves, and the first, the third and the fourth vertical arrows are isomorphisms(we should ensure that $U_i\bigcap U_j$ is again affine, which indeed is the case since $X$ is Noetherian or separated, so we used the hypothesis in an essential way), and thus we get that the second vertical arrow is also an isomorphism, which finishes the proof of this lemma.

Return to our original proposition. Suppose that $U$ is an affine open subset of $X$, then for each $f\in \mathfrak{O}_X(U)$, we have that $\mathfrak{F}(U)_f\simeq \mathfrak{F}(D(f))$ by the above lemma, thus $\mathfrak{F}(U)^{\sim}\simeq \mathfrak{F}|_U$. The converse is clear.

At last, let’s give a proposition which reveals the exactness of the global section functor in a larger extent:

Proposition: let $X$ be an affine scheme, and $0\rightarrow \mathfrak{F}\rightarrow\mathfrak{G}\rightarrow\mathfrak{H}\rightarrow 0$ be an exact sequence of sheaves over $X$. Suppose that $\mathfrak{F}$ is quasi-coherent, then the sequence

$0\rightarrow \mathfrak{F}(X)\rightarrow\mathfrak{G}(X)\rightarrow\mathfrak{H}(X)\rightarrow 0$

is exact.

We will show this result in the next post, which will use Cech cohomology.

Notes on semisimple rings and modules

In this post I want to say something on semisimple rings and semisimple modules.

Basic settings

Suppose that $R$ is a ring with unit and $M$ is a module over $R$. We say that $M$ is a simple module if it does not contain other submodule other than $0$ and $M$. Then what is a semisimple module over $R$? We say that $M$ is semisimple if any submodule $M'$ of $M$ is a direct summand of $M$. Note that this is a very strong condition on the module. Recall that this kind of phenomena occurs most naturally to vector spaces over a field. So, in this way, we can say that semisimple modules are the modules most similar to vector spaces. Besides, it is easy to see that a simple module is automatically a semisimple module.

Let’s then define what a semisimple ring is. We say that $R$ is a semisimple ring if the $R$-module $R$(the ring $R$ acts by left multiplication on the abelian group $R$) is a semisimple module over $R$. So here we should really be careful about semisimple rings because we are apt to thinking that $R$ is a semisimple ring if any subring of $R$ is a direct summand of $R$, which is completely false according to the definition. In fact, $R$ is a semisimple ring if any ideal $R$ is a direct summand of the module $R$.

Now it is not difficult to see that: $R$ is a semisimple ring if and only if any module $M$ over $R$ is semisimple. The ‘if’ part is clear. For the ‘only if’ part, note that $M$ is a quotient of some $R^n$, which is again semisimple. Now consider the exact sequence

$0\rightarrow Ker(p)\rightarrow R^n\xrightarrow{p}M\rightarrow 0$

Note that $Ker(p)$ is a submodule of $R^n$, thus is a direct summand of $R^n$, so $M$ is also a direct summand of $R^n$, thus is also semisimple.

Here we have used a crucial property of semisimple modules over $R$: if $M$ is semisimple, then its submodules and quotient modules are again semisimple.

So in this way, a semisimple ring $R$ looks very like a field in that all the modules over it are semisimple.

Structure theory of semisimple rings

What is fascinating about the theory of semisimple rings is that there is a nice decomposition of a semisimple ring into some well known algebraic objects. Let’s briefly have a look at this part of the theory.

Suppose that $R$ is a semisimple ring and $M$ is a simple $R$-module. Now choose any element $m\in M$, we have a morphism of $R$-modules

$f_m:R\rightarrow M,r\mapsto rm$

It is easy to see that $Im(f_m)$ is a submodule of $M$, and since $M$ is simple, thus either $Im(f_m)=M$ or $Im(f_m)=0$. If $m\neq 0$, then $1_Rm=m\in Im(f_m)$, this shows that $Im(f_m)=M$ for $m\neq 0$. Thus $M$ is a quotient of $R$, and since the latter is a semisimple $R$-module, thus $R\simeq M\bigoplus Ker(f_m)$. Note that this is an isomorphism of $R$-modules, thus we can view $M$ as an ideal of $R$. In this way, we find that all simple modules come from ideals of $R$, so now we shall restrict our attention to simple ideals of $R$(which is defined in an obvious way to be an ideal of $R$, which, viewed as an $R$-module, is a simple module).

Note that if we do not assume that $R$ is commutative, and we consider for the present the left $R$-modules, then the resulting fact is to consider all the left simple ideals of $R$.

We write $\mathfrak{I}$ for the set of equivalence classes of left simple ideals of $R$. For any $I\in \mathfrak{I}$, we set $M_I$ to be the (direct) sum of the left ideals of $R$ isomorphic to $I$ as $R$-modules. We next show that $M_I$ is a two-sided ideal of $R$. For any two representatives $I,J\in\mathfrak{I}$, using Schur’s lemma, we see that $IJ=0$ if $I\not\simeq J$. Thus we have that $M_IM_J=0$ in this case. So, we see that these $M_I$ are two-sided ideals. Now that $R$ is semisimple, we can write $R=\sum_{I\in\mathfrak{I}}M_I$. Using the (finite) decomposition of $1_R=\sum_I e_I$, we see that $R=\sum_IM_I$ is in fact a finite sum and $R=\bigoplus_I M_I$. The by-product of this process is the elements $e_I$ for each $I\in\mathfrak{I}$(which is now a finite set). These elements have very nice properties(suppose now $I,J\in\mathfrak{I}$ are isomorphism classes, not representatives of these classes):

$e_Ie_J=0,I\neq J; e_Ie_I=e_I;\sum_I e_I=1_R$

Note that these properties give us another description of the above decomposition $R=\bigoplus_I Re_I$. This is an isomorphism of $R$-modules. Yet we have something more: each $Re_I$ is itself a ring. Let’s examine this nice result. Note that we have $1_R=\sum_I e_I$, thus for any $r\in R$, $r=1_Rr=\sum_Ie_Ir=e1_R=\sum_Ire_I$. Since this is a direct sum, thus according to the uniqueness of the decomposition, we get that $e_Ir=re_I$ for all $I\in\mathfrak{I}$. Now for any $re_I,r'e_I\in Re_I$, we have that $(re_I)(r'e_I)=r(e_Ir')e_I=r(r'e_I)e_I=rr'e_I$, which, combined with some other simple verifications, shows that $Re_I$ is indeed a ring. So, in this sense,

$R\simeq \bigoplus_I Re_I$

is also an isomorphism of rings(all of these use essentially the fact that $R$ is semisimple).

Simple rings and their structures

So to study the properties of $R$, it suffices to study these $M_I$. These $M_I$ are in fact simple rings in the following sense: a ring $A$ is said to be simple if $A$ is semisimple and it contains exactly one isomorphism class of left simple ideals. We will write a simple ring as $M_I$, just to emphasize that fact that the isomorphism class of left simple ideals is $I$.

Next we will start from $I$ to construct a division ring. This is one of the most interesting part of the theory. We fix a left simple ideal $I\neq0$ and we set $E=End_{M_I}(I)$. We next show that $E$ is this division ring. Indeed, for any $0\neq f\in E$, since $I$ is simple, thus according to Schur’s lemma, $f$ is an isomorphism. Clearly $E$ has a ring structure, so we are done: $E$ is indeed a division ring.

Now let’s give $I$ an $E$-module structure: for any $m\in I,f\in E=End_{M_I}(I)$, we define $f\cdot m=f(m)$. It is straightforward that this defines an $E$-module structure on $I$. Note that $E$ is already a division ring, so as an $E$-module, $I$ behave very much like a vector space over a field: we have the concept of dimensions. We will come to this point later.

Note until present we only consider one single left simple ideal of $M_I$. Next we consider the whole ring $M_I$, which is a direct sum of the left simple ideals, thus is also an $E$-module. Surprisingly, it has a much nicer description: $M_I$ is isomorphic to the endomorphism ring $E'=End_E(I)$.

Let’s first give a map from one to the other. For any $m\in M_I$, we set $f_m:I\rightarrow I, r\mapsto mr$ simply the left multiplication, which indeed maps $I$ to $I$ since $I$ is a left ideal. We should verify that $f_m\in End_E(I)$. For any $h\in E=End_{M_I}(I)$, we should show that $h\cdot(f_m(r))=f_m(h\cdot r)$. Indeed, $h\cdot(f_m(r))=h(mr)$, now $h$ is a $M_I$-linear map, thus $h(mr)=m(h(r))=f_m(h(r))=f_m(h\cdot r)$, so we are done: $f_m$ is indeed an element in $End_E(I)$. Next we have to show that this map

$\phi: M_I\rightarrow End_E(I),m\mapsto f_m$

is an isomorphism. Note that $Ker(\phi)$ is a two-sided ideal of $M_I$, thus as the above argument, $M_I=Ker(\phi)\bigoplus M_I/Ker(\phi)$.where both summands have ring structures and contain different isomorphism classes of left simple ideals, thus we must have $Ker(\phi)=0$ or $Ker(\phi)=M_I$. If it is the second case, $I=1\cdot I=0$, which is impossible since $I$ is non-zero, thus $Ker(\phi)=0$, showing that $\phi$ is injective. It is also easy to show that $\phi$ is surjective, thus we get an isomorphism

$M_I\simeq End_E(I)$

Note that if $E$ is actually a field(not simply a division ring), then $I$ is a vector space over $E$. Let’s suppose that $dim_E(I)<\infty$, then this result shows that the simple ring $M_I$ is the matrix ring $End_E(I)$. This is really a nice description of simple rings.

So, let’s return to the semisimple case: suppose again $R$ is a semisimple ring, then $R\simeq \bigoplus_I M_I$, each summand is a matrix algebra over some division ring $E_I=End_{M_I}(I)$.

Now we want to concentrate ourselves to some special (semi)simple rings: rings that are algebras over some field.

One special case

Let’s fix a field $k$. Suppose $R$ is a finite dimensional $k$-algebra and a semisimple ring, then each $M_I$ is a finite dimensional $k$-algebra and a simple ring. Now each $E_I=End_{M_I}(I)$ is a division ring and at the same time a $k$-algebra, which is finite dimensional since $End_{M_I}(I)\subset End_k(I)$.

Now let’s make a simple yet illuminating calculation. Suppose that $M_I\simeq I^{\bigoplus d_I}, n_I=dim_{E_I}(I)$. Then we have that

$dim_k(M_I)=d_Idim_k(I)=d_In_Idim_k(E_I)$

Yet $M_I\simeq End_{E_I}(I)$, thus

$dim_k(M_I)=n_I^2dim_k(E_I)$

Combining these two results, we get that

$n_Id_I=n_I^2\implies n_I=d_I$

So we have that $M_I\simeq M_{n_I}(E_I)$, the matrix ring with coefficients in $E_I$. And in general, $R\simeq\bigoplus_I M_{n_I}(E_I)$.

If we suppose that $k$ is algebraically closed, we should have that $k=E_I$ for all $I$.

One simple application

As an application of this result, we consider the representation theory of a finite group $G$ with $char(k)\not| |G|$. We set $R=k[G]$, the group algebra of dimension $dim_k(R)=|G|$. We can show that $R$ is semisimple(we will talk about this proof in another post). We supple also that $k$ is algebraically closed, then according to the above result, we have

$k[G]\simeq\bigoplus M_{n_I}(k)$

In other words, we have $|G|=\sum_In_I^2$. Besides, each $I$ is an irreducible representation of $G$, so in the end we get a fundamental result in the representation theory of finite groups

$|G|=\sum_Vdim_k(V)^2$

where $V$ runs through all the irreducible representations of $G$.