algebraic number theory-part 3

This post is mainly based on the book, ‘Fermat’s dream’ by three Japanese mathematicians Kazuya Kato, Nobushige Kurokawa and Takeshi Saito. This post is concerned with the local and global problems in algebraic number theory.

We already know that in Dedekind rings, especially like algebraic integer rings, the most important object is the non-zero prime ideal. We know thus that any non-zero ideal I in a Dedekind ring R, there is a decomposition I=\prod_{P,\text{prime ideal}}P^{e_P} with e_P\in\mathbb{Z}. Of course, this decomposition still works for fractional ideals in the fractional field K of R. So, in fact, if we fix a prime ideal P, then any fractional ideal I gives an integer e_P. Since now P is fixed, we can write this number as v_P(I)(when there is no confusion ,we will write it simply as v(I)). It is clear that for two fractional ideals I,J, we have that v_P(IJ)=v_P(I)+v_P(J). Moreover, it is not hard to show that v_P(I+J)\geq \min{v_P(I),v_P(J)}. In terms of commutative algebra, we say that v_P:K^{\times}\rightarrow\mathbb{Z},k\mapsto v_P(Rk) defines a discrete valuation on K(here we require that the map be surjactive). We can set v_P(0)=+\infty so that the two conditions above are satisfied on the whole K. We call R a discrete valuation ring and K a discrete valuation field.

The most basic example is perhaps the p-adic numbers. Suppose that R=\mathbb{Z}, K=\mathbb{Q}. Then any non-zero prime ideal correspond to an unique prime number p. And for any non-zero rational number a, we can write it as a=p^nA/B where A,B are two integers prime to p. Then this n is just v_p(a). It is easy to verify that the valuation thus defined is indeed a discrete valuation.

The really interesting point is that we can define some topology on K using this discrete valuation. Indeed, using the properties of the discrete valuation, we can define a distance d on K. For any k,k'\in K, we set d_v(k,k')=e^{-v(k-k')}(and we set d_v(k,k)=0,compatible with the above definition). This is indeed a distance thanks to the inequality above. Note we here choose e. In fact, we can change e for any real number greater than 1, and the resulting distance is equivalent to the one defined above in the sense that they define the same topology on K.

There is an important concept to introduce before going any further. We define A=\{k\in K|v(k)\geq0\} to be the valuation ring of K. Note that this is indeed a ring, closed under addition and subtraction and multiplication. Since the map v:A\rightarrow \mathbb{N} is surjective(this condition is essential), we can choose a\in A such that v(a)=1. Then for any ideal I\subset A, we set n=\inf_{k\in A}v(k). Then for any k\in I, since v(k/a^n)\geq0, we must have, according to the definition of A, k/a^n\in A, that is a^n divides k for any k\in I. Moreover, if k\in I with v(k)=n, then v(k/a^n)=0. It is easy to show that if an element k\in A has v(k)=0,then since v(1/k)=0, which implies that 1/k\in A, thus k is a unit in A. So, we have that k/a^n is a unit, which means that I=(a^n). So, for any ideal in A, it is always of the form (a^n) for some n\in\mathbb{N}. So the valuation ring is rather simple under the discrete valuation. In the above example, we have that A=\{k\in\mathbb{Q}|k=p^na/b,a,b\in\mathbb{Z},n\geq0\}. So in general, A\neq R.

Now return to the topology defined above. It is not hard to see that the algebraic operations on K is continuous under this topology, so this makes K into a topological field. And A a topological ring, too. K being a metric space, a natural question is, whether K is complete with respect to this distance. How to complete this space if it is not complete? Perhaps we should get inspired from the above example. One way to complete A is to consider the inverse limit of the system A/(a^n). So we define \hat{A}=\varprojlim_n A/(a^n), the completion of A(in fact, \hat{A}, being a closed subspace of \prod_n A/(a^n) which is compact due the Tychonoff’s theorem, thus is compact, and as a result is complete). And we set K_v to be the fractional field of \hat{A}. We can show that K_v is in fact locally compact, which arises from the observation that one neighborhood of the origin of K_v is just \hat{A}, which is compact and open at the same time due to the discrete valuation condition and the fact that \hat{A}=\{k\in K_v|v(k)\geq 0\}(note that, we can induce a discrete valuation, which we denote still as v, on A so that its restriction to A coincide with v). This is a very good signal, since this implies us that we can do integration on K after finding the Haar measure on K.

We call R/P the residual field of K( it is indeed a field, since P is a prime ideal, and is not zero, thus is maximal, so R/P is a field). Note that, when K is a number field(finite algebraic extension of \mathbb{Q}), We see that R is a \mathbb{Z}-module of rank [K:\mathbb{Q}], so is for all non-zero ideals, and thus we have that R/P is a finite set, thus is a finite field. So, all in all, given a prime ideal P\neq 0, we can define a valuation, thus a distance on K, with a residual field, R/P. We call a number field a global field, while a local field refers to a field with a discrete valuation such that it is complete with respect to the distance induced by the valuation, with a complementary condition that the residual field is a finite field. Note that in fact, we haven’t defined what a residual field is in general. Suppose that a field K with a discrete valuation v, and consider its valuation ring A=\{k\in K|v(k)\geq0\}. We saw that there is only one non-zero prime ideal of A, that is (a). And we set A/(a) to be the residual field of K. So, we say that a field is a local field if it is complete topological field and its residual field is finite. One question is, are there many local fields? Not so many, in fact we can show that any local field is the completion of some number field K with respect to some discrete valuation v induced by some prime ideal in \mathfrak{O}_K. Note that, using the projective-limit definition of \hat{A}, we can see that the residual field of K is the same as any of its completions K_v.  So, in fact, local fields appear in rather limited domains, just those number fields.

So, in this post, we have associated any non-zero prime ideal of an algebraic integer ring with a discrete valuation, and thus defined a distance, and at the end, completed the field to get a local field.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s