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## Category: Uncategorized

### Irritating set of examples- I

I am trying to collect explicit examples for concepts and calculations. My hope is that this website becomes a useful repository of examples for anyone looking for them on the internet.

First some words of wisdom from the master himself, Professor Ravi Vakil: “Finally, if you attempt to read this without workign through a significant number of exercises, I will come to your house and pummel you with the EGA until you beg for mercy. As Mark Kisin has said, “You can wave your hands all you want, but it still won’t make you fly.”

1. Can we have two products of the same two objects, say $A$ and $B$, in the same category? This question is much more general than I am making it out to be. Can we have two distinct universal objects of the same kind in a category (although they may be isomorphic, and even through unique isomorphism)? The only example of the product of objects being isomorphic but not the same is the following: $A\times B$ and $B\times A$. These aren’t the same objects, but they’re isomorphic through unique isomorphism.

2. Groupoid- In the world of categories, a groupoid is a category in which all morphisms between objects are isomorphisms. An example of a groupoid, which is not a group, is the category $\mathfrak{Set}$ with the following restriction: $\text{Hom(A,B)}$ now only consists of isomorphisms, and not just any morphisms. This example, although true, is not very illustrative. This [link](http://mathoverflow.net/questions/1114/whats-a-groupoid-whats-a-good-example-of-a-groupoid) provides a much better demonstration of what is going on. Wikipedia says that the way in which a groupoid is different from a group is that the product of some pairs of elements may not be defined. The Overflow link suggests the same thing. You can’t take any any pair of moves that one may make on the current state of the jigsaw puzzle, and just compose them. The most important thing to note here is that the elements of the group do not correspond to objects of the categories. They correspond to morphisms between those objects. This is the most diabolical shift of perspective that one encounters while dealing with categories. Suddenly, morphisms encode much more information than you expect them to.

3. Algebraic Topology example: Consider a category in which points are objects of the category, and the paths between points, upto homotopy, are morphisms. This is a groupoid, as paths between points are invertible. The return path should not wrap around a wayward hole, obviously. One may consider the path as the same, just travelling in the opposite direction. The automorphism group of a point would be the fundamental group of paths centred at that point.

Another category that stems from Algebraic Topology is one in which all objects are topological spaces, and the morphisms between maps are the continuous maps between those spaces. Predictably, the isomorphisms are the homeomorphisms.

4. Subcategory: An example would be one in which objects are sets with cardinaly $1$, and morphisms would be the same as those defined in the parent category- $\mathfrak{Set}$.

5. Covariant functor: Consider the forgetful functor from $\mathfrak{Set}$ to $\mathfrak{Vec}_k$. The co-domain is bigger than the domain. One could think of this functor as an embedding.

A topological example is the following: one which sends the topological space $X$, with the choice of a point $x_0$, to the object $\pi(X,x_0)$. How does this functor map morphisms? It just maps paths in $X$ to their image under the same continous map. How do we know that the image is a path? This is easy to see. We can prove that we ultimately have just a continuous map from $[0,1]$ to that image, and we will be done. Do we have to choose a point in each topological space? Yes. What if we have the following two tuple $(X,x_0), (Y,y_o)$, such that $x_0$ is not mapped to $y_0$? Then there is no morhism between these two objects. In other words, the set of morphisms $\text{Hom}((X,x_0), (Y,y_o))$ consists of only those morphisms which map $x_0$ to $y_0$. An illustrative example is the following: $f_1: [0,1]\to (\cos (2\pi t),\sin (2\pi t))$ and $f_2: [0,1]\to (\cos (4\pi t),\sin (4\pi t))$. These are two different continuous maps between the same two topological spaces. They both map $0$ to the point $(1,0)$ in $S^1$, but they map a path starting and ending

Side note: Example of two homotopic paths being mapped to homotopic paths under a continuous map. Let $f: [0,1]\to S^1$ be the continuous map under consideration. Consider any path $p$ in $[0,1]$ which starts and ends at $0$. We know that this is homotopic to the constant path at $0$ (one may visualize the homotopy as shrinking this path successively toward $0$). Then the image of this homotopy is mapped to a path in $S^1$ that shrinks toward the constant map at $(1,0)$.

6. Contravariant functors: Mapping a vector space to its dual. This example is pretty self-explanatory.

7. Natural Transformation: A natural transformation is a morphism between functors. Abelianization is a common example of a natural transformation. The two functors, both of which are covariant, are $id$ and $id^{ab}$. The first one maps a group to itself, and the second one maps a group to its commutator. The resultant commutative diagram is easy to see too. The data of the natural transformation is just $m:G\to G^{ab}$ and $m:G'\to G'^{ab}$.

The double dual of a vector space is another example of a natural transformation. The dual would have worked too, except for the fact that the dual functor is contravariant. Note: one of the functors, in both these natural transformations, is the identity functor.

8. Equivalence of categories- This is exactly what you think it is. Two categories that are not equivalent are $\mathfrak{Grp}$ and $\mathfrak{Grp^{ab}}$. Too much information is lost while abelianizing the group, which cannot be regained easily.

9. Initial object- The empty set is the initial object in the category $\mathfrak{Set}$. Why not a singleton? Because the map from the initial object to any object also has to be unique. Moreover, a singleton will not map to an object- namely the empty set. And an initial object should map to all objects.

10. Final object- A singleton will be a good final object in the category $\mathfrak{Set}$.

11. Zero object- The identity element in the category $\mathfrak{Grp}$ would be such an object.

12. Localization through universal property: Consider $\Bbb{Z}$, with the multiplicative subset $\Bbb{Z}-\langle 7\rangle$. The embedding $\iota: \Bbb{Z}\to \Bbb{Q}$ ensures that every integer goes to an invertible element. Trivially, so does every element of $\Bbb{Z}-\langle 7\rangle$. Hence, there exists a unique map from $(\Bbb{Z}-\langle 7\rangle)^{-1}\Bbb{Z}$ to $\Bbb{Q}$. We can clearly see that this is overkill. Many more elements than just those of $\Bbb{Z}-\langle 7\rangle$ are mapped to invertible elements. The point is that there may be a ring $A$ such that only elements of $\Bbb{Z}-\langle 7\rangle$ are mapped to invertible elements in $A$. Hence, in that case too, there will exist a unique map from $(\Bbb{Z}-\langle 7\rangle)^{-1}\Bbb{Z}$ to $A$. Why do we care about there existing a map from some other object to rings which $\Bbb{Z}$ maps to at all? When we have a morphism $\phi:A\to B$, and we can say that there exists a map $A/S\to B$, where $S$ is a set of relations between elements of $A$, then we’re saying something special about the properties of elements in $B$ (at least the properties of elements mapped to by $S$).

### Prūfer Group

This is a short note on the Prūfer group.

Let $p$ be a prime integer. The Prūfer group, written as $\Bbb{Z}(p^\infty)$, is the unique $p$-group in which each element has $p$ different $p$th roots. What does this mean? Take $\Bbb{Z}/5\Bbb{Z}$ for example. Can we say that for any element $a$ in this group, there are $5$ mutually different elements which, when raised to the $5$th power, give $a$? No. Take $\overline{2}\in\Bbb{Z}/5\Bbb{Z}$ for instance. We know that only $2$, when raised to the $5$th power, would give $2$. What about $\Bbb{Z}/2^2\Bbb{Z}$? Here $p=2$. Does every element have two mutually different $2$th roots? No. For instance, $\overline{2}\in\Bbb{Z}/2^2\Bbb{Z}$ doesn’t. We start to get the feeling that this condition would only be satisfied in a very special kind of group.

The Prūfer $p$-group may be identified with the subgroup of the circle group $U(1)$, consisting of all the $p^n$-th roots of unity, as $n$ ranges over all non-negative integers. The circle group is the multiplicative group of all complex numbers with absolute value $1$. It is easy to see why this set would be a group. And using the imagery from the circle, it easy to see why each element would have $p$ different $p$th roots. Say we take an element $a$ of the Prūfer group. Assume that it is a $p^{n}$th root of $1$. Then its $p$ different $p$th roots are $p^{n+1}$th roots of $1$. It is nice to see a geometric realization of this rather strange group that seems to rise naturally from groups of the form $\Bbb{Z}/p^n\Bbb{Z}$.

### Sheafification

This is a blog post on sheafification. I am broadly going to be following Ravi Vakil’s notes on the topic.

Sheafification is the process of taking a presheaf and giving the sheaf that best approximates it, with an analogous universal property. In a previous blog post, we’ve discussed examples of pre-sheaves that are not sheaves. A classic example of sheafification is the sheafification of the presheaf of holomorphic functions admitting a square root on $\Bbb{C}$ with the classical topology.

Let $\mathcal{F}$ be a presheaf. Then the morphism of presheafs $\mathcal{F}\to\mathcal{F}^{sh}$ is a sheafification of $\mathcal{F}$ if $\mathcal{F}^{sh}$ is a sheaf, and for any presheaf morphism $\mathcal{F}\to \mathcal{G}$, where $\mathcal{G}$ is a sheaf, there exists a unique morphism $\mathcal{F}^{sh}\to \mathcal{G}$ such that the required diagram commutes. What this means is that $\mathcal{F}^{sh}$ is the “smallest” or “simplest” sheaf containing the presheaf $\mathcal{F}$.

Because of the uniqueness of the maps, it is easy to see that the sheafification is unique upto unique isomorphism. This is just another way of saying that all sheafifications are isomorphic, and that there is only one (one each side) isomorphism between each pair of sheafifications. Also, sheafification is a functor. This is because if we have a map of presheaves $\phi:\mathcal{F}\to \mathcal{G}$, then this extends to a unique map $\phi':\mathcal{F}^{sh}\to\mathcal{G}^{sh}$. How does this happen? Let $g:\mathcal{G}\to\mathcal{G}^{sh}$. Then $g\circ\phi:\mathcal{F}\to\mathcal{G}^{sh}$ is a map from $\mathcal{F}$ to a sheaf. Hence, there exists a unique map from $\mathcal{F}^{sh}\to\mathcal{G}^{sh}$, as per the definition of sheafification. Hence, sheafification is a covariant functor from the category of presheaves to the category of sheaves.

We now show that any presheaf of sets (groups, rings, etc) has a sheafification. If the presheaf under consideration is $\mathcal{F}$, then define for any open set $U$, define $\mathcal{F}^{sh}$ to be the set of all compatible germs of $\mathcal{F}$ over $U$. What exactly are we doing? Are we just taking the union of all possible germs of that presheaf? How does that make it a sheaf? This is because to each open set, we have now assigned a unique open set. These open sets can easily be glued, and uniquely too, to form the union of all germs at each point of $\mathcal{F}$. Moreover, the law of the composition of restrictions holds too. But why is this not true for every presheaf, and just the presheaves of sets? Are germs not defined for a presheaf in general?

A natural map of presheaves $sh: \mathcal{F}\to\mathcal{F}^{sh}$ can be defined in the following way: for any open set $U$, map a section $s\in \mathcal{F}(U)$ to the set of all germs at all points of $U$, which in other words is just $\mathcal{F}^{sh}(U)$. We can see that all the restriction maps to smaller sets hold. Moreover, $sh$ satisfies the universal property of sheafification. This is because $sh$ can be extended to a unique map between $\mathcal{F}^{sh}$ and $\mathcal{F}^{sh}$: the unique map is namely the identity map.

We now check that the sheafification of a constant presheaf is the corresponding constant sheaf. We recall that the constant sheaf assigns a set $S$ to each open set. Hence, each germ at each point is also precisely an element of $S$, which implies that the sheaf too is just the set of all elements of $S$: in others words, just $S$. The stalk at each point is also just $S$, which implies that this is a constant sheaf.

What is the overall picture that we get here? Why is considering the set of all germs the “best” way of making a sheaf out of a pre-sheaf? I don’t know the exact answer to this question. However, it seems that through the process of sheafification, to each open set, we’re assigning a set that can be easily and uniquely glued. It is possible that algebraic geometers were looking for a way to glue the information encoded in a presheaf easily, and it is that pursuit which led to this seemingly arbitrary method.

### Toric Varieties: An Introduction

This is a blog post on toric varieties. We will be broadly following Christopher Eur’s Senior Thesis for the exposition.

A toric variety is an irreducible variety with a torus as an open dense subset. What does a dense subset of a variety look like? For instance, in $\Bbb{R}$ consider the set of integers. Or any infinite set of points for that matter. The closure of that set, under the Zariski Topology, is clearly the whole real line. Hence, a dense set under the Zariski topology looks nothing like a dense set under the standard topology.

An affine algebraic group $V$ is a variety with a group structure. The group operation is given by $\phi:V\times V\to V$, which is interpreted as a morphism of varieties (remember that the cartesian product of two varieties is a variety). The set of algebraic maps of two algebraic groups $V,W$, denoted as $\text{Hom}(V,W)$ is the set of group homomorphisms between $V$ and $W$ which are also morphisms between varieties. Are there variety morphisms which are not group homomorphisms? Yes. Consider the morphism $f:\Bbb{R}\to \Bbb{R}$ defined as $x\to x+1$.

The most important example for us is $(\Bbb{C}^*)^n\simeq \Bbb{C}^n-V(x_1x_2\dots x_n)$. This is the same as removing all the hyperplanes $x_i=0$ from $\Bbb{C}^n$. Again, this is the same as $V(1-x_1x_2\dots x_n y)\subset \Bbb{C}^{n+1}$, which is the same as embedding a variety in a higher dimensional space. The coordinate ring of $(\Bbb{C}^*)^n$ looks like $\Bbb{C}[x_1^{\pm},x_2^{\pm},\dots,x_n^{\pm}]\simeq \Bbb{C}[\Bbb{Z}^n]$. Why does the coordinate ring look like this? This is because in the ring $\Bbb{C}[x_1,x_2,\dots,x_n,y]/(1-x_1x_2\dots x_ny)$, all the $x_i's$ become invertible (in general, all of the $n+1$ variables become invertible. However, $y$ can be expressed in terms of the $x_i$‘s).

A torus is an affine variety isomorphic to $(\Bbb{C^*})^n$ for some $n$, whose group structure is inherited from that of $(\Bbb{C^*})^n$ through a group isomorphism.

Example: Let $V(x^2-y)\subset \Bbb{C}^2$, and consider $V_{xy}=V\cap (\Bbb{C}^*)^n$. We will now establish an isomorphism between $\Bbb{C}^*$ and $V_{xy}$. Consider the map $t\to (t,t^2)$ from $\Bbb{C}^*$ to $V_{xy}$. This map is bijective. How? If $t$ is non-zero, then so is each coordinate of $(t,t^2)$. Also, each point in $X_{xy}$ looks like $(t,t^2)$, where $t$ is a non-zero number, and each such point has been mapped to by $t\in\Bbb{C}^*$. Hence, we have a bijection. How does $V_{xy}$ inherit the group structure of $\Bbb{C}^*$? By the following relation: $(a,a^2).(b,b^2)=(ab,(ab)^2)$. Remember that $V_{xy}$ had no natural group structure before. Now it has one.

A map $\phi:(\Bbb{C}^*)^n\to (\Bbb{C}^*)^m$ is algebraic if and only if the map $\phi^*: \Bbb{C}[y_1^{\pm},y_2^{\pm},\dots,y_m^{\pm}]\to \Bbb{C}[x_1^{\pm},x_2^{\pm},\dots,x_n^{\pm}]$ is given by $y_i\to x^{\alpha_i}$ for $\alpha_i\in \Bbb{Z}^n$. In other words, the maps correspond bijectively to lattice maps $\Bbb{Z}^m\to \Bbb{Z}^n$. What does this mean? The condition that the variety morphism also be a group homomorphism was surely expected to place certain restrictions on the the nature of the nature of the morphism. The way that this condition places restrictions is that a unit can only map to a unit. And the only units in $\Bbb{C}[x_1^{\pm},x_2^{\pm},\dots,x_n^{\pm}]$ are monomials times a constant. Why’s that? Why isn’t an expression of the form $x_1+x_2$, for instance, a unit? Because $\Bbb{C}[x_1^{\pm},x_2^{\pm},\dots,x_n^{\pm}]$ is not a field! It is just a polynomial ring in which the variables happen to be invertible. Polynomials in those variables need not be! This is not the same as the rational field corresponding to the polynomial ring $\Bbb{C}[x_1,x_2,\dots,x_n]$. Returning to the proof, the constant is found to be $1$, and one side of the theorem is proved. The converse is trivial.

A character of a Torus $T$ is an element $\chi\in\text{Hom}(T,\Bbb{C}^*)$. An analogy that immediately comes to mind is that of a functional on an $n$-dimensional vector space. Characters are important in studying toric varieties.

### Birational Geometry

This is a blog post on birational geometry. I will broadly be following this article for the exposition.

A birational map $f:X\to Y$ is a rational map such that its inverse map $g:Y\to X$ is also a rational map. The two (quasiprojective) varieties $X$ and $Y$ are known as birational varieties. An example is $X=Y=\Bbb{R}\setminus \{0\}$, and $f=g: x\to \frac{1}{x}$.

Varieties are birational if and only if their function fields are isomorphic as extension fields of $k$. What are function fields? A function field of a variety $X$ is the field of rational functions defined on $X$. In a way, it is the rational field of the coordinate ring on $X$. But what about the functions which are $0$ on some part of $X$, although not all of it? They can still be inverted. In the complex domain, such functions are called meromorphic functions (isolated poles are allowed).

A variety $X$ is called rational if it is birational to affine space of some dimension. For instance, take the circle $x^2+y^2=1$. This is birational to the affine space $\Bbb{R}$. Consider the map $\Bbb{R}\to \Bbb{R}^2: t\to (\frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2})$. This is a rational map, for which the inverse is $(x,y)\to (1-y)/x$.

In general, a smooth quadric hypersurface (degree 2) is rational by stereographic projection. How? Choose a point on the hypersurface, say $p$, and consider all lines through $p$ to the various other points on the hypersurface. Each such line goes to a point in $\Bbb{P}^n$. Note that this map is not defined on the whole of the hypersurface. How do we know that the line joining $p$ and point does not pass through another point on the hypersurface? This is precisely because this is a quadric surface. A quadratic equation can only have a maximum of two distinct solutions, and one of them is already $p$.

Now we state some well-known theorems. Chow’s Theorem states that every algebraic variety is birational to a projective variety. Hence, if one is to classify varieties up to birational isomorphism, then considering only the projective varieties is sufficient. Then Hironaka further went on to prove that every variety is birational to a smooth projective variety. Hence, we now have to classify a much smaller set of varieties. In dimension, $1$, if two smooth projective curves are birational, then they’re isomorphic. However, this breaks down in higher dimensions due to blowing up. Due to the blowing up construction, every smooth projective variety of at least degree $2$ is birational to infinitely many “bigger” varieties with higher Betti numbers. This leads to the idea of minimal models: is there a unique “simplest variety” in each birational equivalence class? The modern definition states that a projective variety is minimal if the canonical bundle on each curve has non-negative degree. It turns out that blown up varieties are never minimal.

This is going to be a blog post on Filtrations and Gradings. We’re going to closely follow the development in Local Algebra by Serre.

A filtered ring is a ring with the set of ideals $\{A_n\}_{n\in\Bbb{Z}}$ such that $A_0=A$, $A_{n+1}\subset A_n$, and $A_pA_q=A_{p+q}$. An example would be $A_n=(2^n)$, where $(2^n)$ is the ideal generated by $2^n$ in $\Bbb{Z}$.

Similarly, a filtered module $M$ over a filtered ring $A$ is defined as a module with a set of submodules $\{M_n\}_{n\in\Bbb{N}}$ such that $M_0=M$, $M_{n+1}\subset M_n$, and $A_pM_q\subset M_{p+q}$. Why not just have $M_pM_q\subset M_{p+q}$? This is because multiplication between elements of a module may not be defined. An example would be the module generated by by the element $v$ over $\Bbb{Z}$, where $M_n=2^n M$.

Filtered modules form an additive category $F_A$ with morphisms $u:M\to N$ such that $u(M_n)\subset N_n$. A trivial example is $\Bbb{Z}\to\Bbb{Z}$, defined using the grading above, and the map being defined as $x\to -x$.

If $P\subset M$ is a submodule, then the induced filtration is defined as $P_n=P\cap M_n$. Is every $P_n$ a submodule of $P$? Yes, because every $M_n$ is by definition a submodule of $M$, and the intersection of two submodules ($M_n$ and $P$ in particular) is always a submodule. Simialrly, the quotient filtration $N=M/P$ is also defined. As the quotient of two modules, the meaning of $M/P$ is clear. However, what about the filtration of $M/P$? Turns out the filtration of $N=M/P$ is defined the following way: $N_n=(M_n+P)/P$. We need to have $M_n+P$ as the object under consideration because it is not necessary that $M_n\in P$.

An important example of filtration is the $m$-adic filtration. Let $m$ be an ideal of $A$, and let the filtration of $A$ be defined as $A_n=m^n$. Similarly, for a module $M$ over $A$, the $m$-adic filtration of $M$ is defined by $M_n=m^nM$.

Now we shall discuss the topology defined by filtration. If $M$ is a filtered module over the filtered ring, then $M_n$ form a basis for neighbourhoods around $0$. This obviously is a nested set of neighbourhoods, and surely enough the intersection of a finite number of neighbourhoods is also a neighbourhood, and so is the union of any set of neighbourhoods. Hence, the usual topological requirements for a basis is satisfied. But why $0$?

Proposition: Let $N$ be a submodule of a filtered module $M$. Then the closure of $\overline{N}$ of $N$ is defined as $\bigcap(N+M_n)$. How does this work? If one were to hand wave a bit, we are essentially finding the intersection of all neighbourhoods of $N$. Remember that each $M_n$ is a neighbourhood of $0$. We’re translating each such neighbourhood by $N$, which is another way of saying we’re now considering all neighbourhoods of $N$. And then we find the intersection of all such neighbourhoods to find the smallest closed set containing $N$. There is an analogous concept in metric spaces- the intersection of all open sets containing $[0,1]$, for instance, is the closed set $[0,1]$. The analogy is not perfect, as the intersection of all neighbourhoods of $(0,1)$ is $(0,1)$ itself, which is not a closed set. But hey. We at least have something to go by.

Corollary: $M$ is Hausdorff if and only if $\cap M_n=0$.

### Invertible Sheaves and Picard Groups

This is a blog post on invertible sheaves, which form elements (over a fixed algebraic variety) of the Picard Group. The group operation here is the tensor product. We will closely follow the developments in Victor I. Piercey’s paper.

We will develop invertible sheaves on algebraic varieties. However, instead of studying sheaves over varieties, we will be studying the algebraic analogues of these geometric entities- we’ll be studying modules over coordinate rings.

First we discuss what it means for a module to be invertible over a ring. Over a ring $A$, a module $I$ is invertible if it is finitely generated and if for any prime ideal $p\subset A$, we have $I_p\simeq A_p$ as $A_p$-modules. Here $A_p$ is the localization of the ring $A$ with respect to the prime ideal $p$, and $I_p$ is just the ideal over the localized ring $A_p$. What does the expression $I_p\simeq A_p$ mean? One way that this condition is easily seen to be satisfied is that $I$ is generated by a single element over $A$. I can’t think of any other ways right now. It is perhaps fitting that the article says next that this condition implies that $I_p$ is locally free of rank $1$.

The reason that the notation $I$ is chosen for an invertible module is that we shall soon see that every invertible module is isomorphic to an invertible ideal. How does one see that? An ideal of a ring is definitely a module over that ring. Assuming that the ideal is a principal ideal and the module under consideration is also generated by a single element, all we need to do is to map the generator of the module to the generator of the ideal. The reason we can assume that the ideal is principal and that the module is generated by a single element is that we want both the modules to be locally of rank $1$, and this is the easiest way of doing so.

We will now discuss an ideal of a module that is locally free, but not principal. Let $A=\Bbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$. It is easy to see that this ideal is not principal. Also, $A/I=F_2$. Hence, $I$ is maximal in $A$. Now if $I\not\subset p$, where $p$ is the prime ideal under consideration, then $I\cap A\setminus p\neq\emptyset$. Hence, $I_p=A_p$. This is because there is an element of $I$ which has been inverted, which causes the ideal to be equal to the ring. We therefore assume that $I\subset p$. As $I$ is maximal, we conclude that $I=p$. We observe that $3$ is not in $I$, and hence invertible in $A_p$ (which can now be written as $A_I$). Now $2$, which is one of the generators of $I$, is written as an element of $I_p$ (it is written as $\{(1+\sqrt{-5})(1-\sqrt{-5})\}/3$). This shows that the ideal can be generated by a single element in $A_p$, which makes it isomorphic to $A_p$.

The isomorphism classes of invertible modules over the ring $A$ form the Picard group. The identity element is the isomorphism class of $A$ over itself. Given an invertible module $I$, its inverse is the module $I^*=\text{Hom}(I,A)$. Why is this the inverse element? This is because there is a natural map $I^*\otimes I\to A$, which is defined as $\psi\otimes a=\psi(a)$. As the isomorphism class of $A$ is the identity element, this is a map of the product of two elements to the identity, which makes one the inverse of the other. What about $I\otimes I^*$? Shouldn’t we have a two-sided inverse? Remember that in general, for any two modules $M$ and $N, M\otimes N\simeq N\otimes M$. Hence, we can define $a\otimes \psi$ to be the same as $\psi\otimes a$, and get away with it.

Theorem 1: If $I$ is an $A$-module, then $I$ is invertible if and only if the natural map $\mu:I^*\otimes I\to A$ is an isomorphism.

The proof and subsequent theorems in the paper will be discussed in a later blog post.

### Tight Closure

This is a small introduction on tight closure. This is an active field of research in commutative algebra, and this is essentially a survey article. This article will closely follow the paper “An introduction to tight closure” by Karen Smith.

Definition: Let $R$ be a Noetherian domain of prime characteristic $p$ (not that in general, $p$ need not be prime). Let $I\subset R$ be an ideal with generators $(y_1,y_2,\dots,y_r)$ Then an element $z$ is defined to be in the tight closure $I^*$ if $\exists c\in R$ such that $cz^{p^e}\in (y_1^{p^e},y_2^{p^e},\dots,y_r^{p^e})$.

What does this condition even mean? Let the ring under consideration be $\Bbb{Z_3}[x,y,z]$, and let the ideal $I$ be $(x,y)$. Does the tight closure $I^*$ contain $I$? For example, $x+y\in (x,y)$. Then is it true that $(x+y)^9\in (x^9, y^9)$? Yes! Because remember that the ring has characteristic $3$. Hence all the other terms in the binomial expansion are $0$. In general, $I\subset I^*$. It is easy to see why. What is an example of an element outside of $I$ that belongs to $I^*$? Clearly, $z\notin I^*$? Why? Why can we not have a value for $c$ such that $cz^{p^e}\in (x^{p^e}. y^{p^e})$? For example, $c=x$. However, the value for $c$ should remain the same for all prime powers $p^e$. Clearly, there is no such $c$.

Is $I^*$ an ideal? Yes. This is part is quite clear.

Properties of tight closure:

1. If $R$ is regular, then all ideals of $R$ are tightly closed. In fact, one of the most important uses of tight closure is to compensate for the fact that the ring under consideration may not be regular.

2. If $R\hookrightarrow S$ is an integral extension, then $IS\cap R\subset I^*$ for all ideals $I\subset R$. What does this condition mean? You’re multiplying $I$ with an ideal outside of $R$. It might create elements in $R$ that are outside of $I$, and even outside of $I^*$. The former is possible, but the latter is not.

3. If $R$ is local, with system of parameters $x_1,x_2,\dots,x_d$, then $(x_1,x_2,\dots,x_i):x_{i+1}\subset (x_1,x_2,\dots,x_i)^*$. This means that we start building an ideal with the element $x_1$, and then every subsequent element that we add is present in the closure of the pre-existing ideal. Hence, it is like we’re building an ideal up from $(x_1)$ to $(x_1)^*$.

4. If $\mu$ denotes the minimal number of generators of $I$, then $\overline{I^\mu}\subset I^*\subset\overline{I}$. Here $\overline{I}$ denotes the integral closure of $I$. Note that the number of generators of an ideal is generally not well defined. For instance the ideal $(x)\subset \Bbb{Q}[x]$ can also be written as $(x^2+2x,x^2)$. However, the minimal number of generators is well-defined, as we’re talking about a Noetherian ring. Hence, every ideal has a finite number of generators. Note that $I^*\subset \overline{I}$ is easy to see. For instance, let $I=(x,y)$ in $\Bbb{R}[x,y]$. Then $a\in I^*$ implies that $ca^{p^e}\in (x^{p^e},y^{p^e})$. Hence, $ca^{p^e}-(\text{polynomial in }x^{p^e}\text{ and }y^{p^e})=0$. This implies that $a$ is integral over $I$, and hence $I^*\subset \overline{I}$. What about $\overline{I^\mu}\subset I^*$?

5. If $\phi:R\to S$ is any ring map, $I^*S\subset (IS)^*$. Here $IS$ is actually $\phi(I)S$. This property is labelled as “persistence” in the paper. I suppose what this means is that it is good to persist (find the closure *after* you find the image) rather than throw up your hands at the beginning (finding the closure right at the beginning).

But I’m probably just putting words into Karen’s mouth. What do I know.

It seems to me that a tight closure is a “tighter” form of closure; tighter than integral closure for instance. And for a lot of analytic requirements, it is just the right size; integral closure would be too big.

### Notes on Speyer’s paper titled “Some Sums over Irreducible Polynomials”

Let $\mathcal{P}$ be the set of irreducible polynomials over $F_2[T]$. Then $\sum\limits_{P\in \mathcal{P}}\frac{1}{1-P}=0$. The paper lists certain examples of $\frac{1}{1-P}$ below. These are all expanded as geometric series. As one can see only $P=T, T+1$ contribute to the coefficient of $T^{-1}$ in the sum $\sum\limits_{P\in \mathcal{P}}\frac{1}{1-P}=0$. Why don’t the other irreducible polynomials do the same? This is because these are the only two linear polynomials in $F_2[T]$. All other polynomials are of higher degree. Moreover, all other irreducible polynomials have the constant term $1$; otherwise they would be reducible, as $T$ would be a common factor. Hence $\frac{1}{P-1}$ would be of the form $\frac{1}{T^{a_1}+T^{a_2}+\dots+T^{a_n}}$, where $a_1>1$. Now divide both the numerator and denominator by $T^{a_1}$. So we get an expression of the form $\frac{1}{T^{a_1}}(\frac{1}{1+T^{a_2-a_1}+T^{a_3-a_1}+\dots+T^{a_n-a_1}})$. As $a_i-a_1<0$ for all $i\neq 1$, this is a power series expansion in negative powers of $T$. Also, as $a_1\geq 2$, all such negative exponents will be less than $-1$. This proves that only the polynomials $T$ and $T+1$ contribute to the coefficient of $T^{-1}$ in $\sum\limits_{P\in \mathcal{P}}\frac{1}{1-P}=0$.

We now try and understand Theorem 1.1 in this paper. Let $\mathcal{P_1}$ be the set of monic irreducible polynomials in $F_{2^n}[T]$. Then $\sum\limits_{P\in \mathcal{P_1}}\frac{1}{P^k-1}\in F_{2^n}(T)$ for any $k\equiv 0(\mod 2^n-1)$.

A corollary of this is that $\sum\limits_{P\in \mathcal{P}}\frac{1}{P^k-1}$ is in $F_{2^n}(T)$

Proof of corollary: We have rewritten $\sum\limits_{P\in \mathcal{P}}\frac{1}{P^k-1}$ as $\sum\limits_{P\in \mathcal{P}_1}\sum\limits_{a\in \Bbb{F}_q^\times}\frac{1}{(aP)^k-1}$, where $q=2^n$. Why can we do that? This is because for any $a\in\Bbb{F}_q^\times, a^{q-1}=1$. Hence, we’re essentially counting the same thing as before. Aren’t we counting each term $|\Bbb{F}_q^\times|$ times? Also, every irreducible polynomial is of the form $aP$ for some $P\in\mathcal{P_1}$. Now consider the identity $\sum\limits_{a\in \Bbb{F}_q^\times}\frac{1}{(aX)^k-1}=\frac{1}{(X)^{lcm(k,q-1)}-1}$ in $\Bbb{F}_q(U)$. Why is this true? This is because $\frac{1}{(aX)^k-1}$ can be written as $\sum\limits_{j=1}^\infty\frac{1}{(ax)^{kj}}$ (just multiply and divide $\frac{1}{(aX)^k-1}$ by $\frac{1}{({aX})^k}$).

Now, as $\sum\limits_{a\in\Bbb{F}_q}a^m=1$ if $m\equiv 0 \mod q-1$, and $\sum\limits_{a\in\Bbb{F}_q}a^m=0$ otherwise. This is because if $m\equiv 0 \mod q-1$, then $\sum a^m$ is essentially adding $1$ to itself $q-1$ times. As the characteristic of the field is $2$, and as $q-1$ is essentially $2^m-1$, this sum is equal to the inverse of $1$, which is exactly $1$. When $q\not\equiv 0\mod q-1$, then $\sum a^m=0$. This can be verified independently.

### Introduction to Schemes

This is a short introduction to Scheme Theory, as modeled on the article by Brian Lawrence.

A variety here is a zero set that can be covered by a finite number of affine varieties. Hence, a morphism between varieties can be considered to be a bunch of affine morphisms, as long as they agree on the intersections.

We need a shift in perspective. What this means is that we need to start thinking about the coordinate ring rather than the points themselves.

Now let us think about the following example: the coordinate ring of $y=0$ in $K^2$ is $K[x,y]/(y)$. However, the coordinate ring of $y^2=0$ is also $K[x,y]/(y)$; it is not $K[x,y]/(y^2)$. The reasons for this can be worked out easily. Hence, the variety in this case is not accurately recovered from the coordinate ring. We started off with the variety $y^2=0$, and got back $y=0$. We need a new concept, which would allow us to accurately get back the variety from the coordinate ring- something that would allow nilpotents.

An affine scheme, written as $\text{Spec }A$, is the data of a ring $A$. A morphism of affine schemes $\text{Spec } A\to \text{Spec }B$, is a morphism of rings $B\to A$. An affine scheme over a field $k$ is a scheme $\text{Spec }A$ where $A$ is equipped with a $k$-Algebra structure.

Why are morphisms defined backwards here? In other words, why is $\text{spec }A\to \text{spec }B$ defined as $B\to A$? This is because $A,B$ are the coordinate rings. Let $Var(A)$ be the variety corresponding to the coordinate ring $A$. Then a map $Var(A)\to Var(B)$ defines a map $B\to A$, and vice-versa. Maybe $\text{spec }A$ is a formal representation of $Var(A)$. It is at least easy to remember which way the arrow goes this way.

How do we recover points from coordinate rings? Hilbert’s Nullstellensatz tells us that we can recover them using maximal ideals. Hence, our aim right now is to take an affine morphism, and construct a morphism between varieties. Hence, if the affine morphism is $B\to A$, we want to construct a map $Var(A)\to Var(B)$.

Given a ring homomorphism $\phi:R\to S$, for any prime ideal $p\in S$, $\phi^{-1}(p)$ is also prime. This is an elementary exercise in ring theory. It is however, not true in general that the inverse image of a maximal ideal is also maximal. For example, consider the map $\psi:\Bbb{Z}[x]\to\Bbb{Q}[x]$ defined by inclusion. Then the only maximal ideal of $\Bbb{Q}[x]$ is $(x)$, the inverse of which is also just $(x)$. It is easy to see that $(x)$ is not a maximal ideal in $\Bbb{Z}[x]$. For instance, $2\notin (x)$, and $2+zx\neq 1$ for any $z\in\Bbb{Z}[x]$.

We define the points of the affine scheme to be prime ideals. Why? Let us work this through. We have a scheme morphsim $\phi:B\to A$, where both $B$ and $A$ are coordinate rings. Now let us take a prime ideal in $A$. From the discussion above, we know that $\phi^{-1}(p)$ is a prime ideal in $B$. Hence, if prime ideals were points, we have taken a point in $A$, and mapped it to $B$. In a way, we have constructed a map from $Var(A)$ to $Var(B)$.

However, this is a little weird. Points correspond to maximal ideals, and not prime ideals. All maximal ideals are prime, but the converse is not true. Do we really have a map from $Var(A)\to Var(B)$? No. At least not in the traditional sense. What we have is a map from some “stuff” in $A$, which includes points, to “stuff” in $B$, which too includes points (possibly not all). Hence, something that’s not a point in $A$ may map to a point in $B$, and a point in $A$ may map to something that is not a point in $B$. We’re gonna call this “stuff” generic points. Hence, generic points in $A$ go to generic points in $B$. This is a classic example of formulating new definitions to suit our world-view.

Now that we have the concept of “generic” points, we also need a name for “actual points”. This name is “classical points”. Hence, we’ll refer to maximal ideals in $A$ as classical points.

So what exactly is a scheme? A scheme is a coordinate ring, whose prime ideals are its points. Simple. It generalizes the notion of a variety. How? A variety has a set of points and an associated coordinate ring. A scheme has a larger set of points, and an associated coordinate ring. Hence the generalization is in the set of points; at least in this instance. Also, as discussed before, although $k[x,y]/(y)$ and $k[x,y]/(y^2)$ correspond to different coordinate rings but the same variety, they correspond to different schemes. Why? If a scheme was the data of its “points” (read generic points), then the points of $k[x,y]/(y^2)$ are different from those of $k[x,y]/(y)$ (the cosets look different, for starters). Hence, we now allow for distinguishing between multiplicities.