4 out of 5 dentists recommend this WordPress.com site

## Month: December, 2016

### 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.