Ultra filter

DEFINITION 4.1   A set $\mathfrak{F}$ of subsets of a set $X$ is called a filter on $X$ if the following conditions are satisfied.

1. $\mathfrak{F} \ni A,B \qquad \implies \qquad A \cap B \in \mathfrak{F}$ .
2. $\mathfrak{F} \ni A, \quad A \subset A_1\subset X \quad \implies \qquad A_1 \subset \mathfrak{F}$ .

DEFINITION 4.2   A maximal filter on a set $X$ is called an ultra filter on $X$ .

Those readers who are not familiar with the arguments are invited to read for example [15] or the book of Bourbaki [2].

LEMMA 4.3   Let $\mathcal U$ be a filter on a set $X$ The following statements are equivalent.

1. $\mathcal U$ is an ultrafilter. That means, a maximal filter.
2. for any subset $S \subset X$ , we have either $S \in \mathcal U$ or $\complement S\in \mathcal U$

DEFINITION 4.4   A principal filter on a set $X$ is an ultra filter of the form $\mathcal F_a=\{S\subset X\vert a\in S\}$ where $a$ is an element of $X$ . A ultrafilter which is not principal filter is called non-principal.

LEMMA 4.5   For any ultrafilter $\mathcal U$ , the following statements are equivalent.

1. $\mathcal U$ is principal.
2. $\mathcal U$ is not free. That means, $\cap_{U\in \mathcal U} U \neq \emptyset$ .
3. There exists a member $E$ of $\mathcal U$ which is a finite set ( $\char93 E<\infty$ ).
4. There exists a co-finite subset $Y$ of $X$ (that means, $\char93 (X \setminus Y) <\infty$ ,) such that $Y\notin \mathcal U$ .

In particular, if $\mathcal U$ is a non-principal ultrafilter on a set $X$ , then any co-finite subset $Y$ of $X$ of is a member of $\mathcal U$ .

An ultrafilter $\mathcal U$ on a set $X$ may be identified with a point of Stone-Cech compactification of ($X$ with discrete topology). A non principal ultrafilter is identified with a boundary point.

DEFINITION 4.6   Let $\mathfrak{K}$ be a number field with the ring of integers $\mathfrak{O}$ . Let $\mathcal U=\{U_\lambda\}$ be a non-principal ultrafilter on the set $\operatorname{Spm}(\mathfrak{O} )$ of all primes of $\mathfrak{O}$ of height 1.

Let $\mathfrak{I}_{\mathcal U}$ be an ideal of $\prod_{\mathfrak{p}\in \operatorname{Spm}(\mathfrak{O})} \mathfrak{O}/\mathfrak{p}$ defined as follows:

$$\mathfrak{I}_{\mathcal U}= \left\{ \left . (f_\mathfrak{p})_{\mat... ... that } f_\mathfrak{p}=0 \quad\text{ for } \forall \mathfrak{p}\in U \right\}$$

Then we define a ring $\mathfrak{K}_{\mathcal U}$ as follows:

$$\mathfrak{K}_{\mathcal U}= \left( \prod_{\mathfrak{p}\in \operato... ...}(\mathfrak{O})} (\mathfrak{O}/\mathfrak{p}) \right)/\mathfrak{I}_{\mathcal U}$$

We denote by $\pi_{\mathcal U}$ the canonical projection from $\prod (\mathfrak{O}/\mathfrak{p})$ to $\mathfrak{K}_{\mathcal U}$ .

LEMMA 4.7   $\mathfrak{K}_{\mathcal U}$ is a field of characteristic 0 .

PROOF.. Indeed, let $f=\pi_{\mathcal U}((f_\mathfrak{p}))$ be a non zero element in $\mathfrak{K}_{\mathcal U}$ . Let $E_1=\{\mathfrak{p}\in \operatorname{Spm}(\mathfrak{O}); f_\mathfrak{p}\neq 0\}$ . Then for any $E\in \mathcal U$ , intersection $E\cap E_1$ is non empty. Maximality of $\mathcal U$ now implies that $E_1$ itself is a member of $\mathcal U$ . The inverse $g=(g_\mathfrak{p})$ of $f$ in $\mathfrak{K}_{\mathcal U}$ is given by the following formula.

$$g_\mathfrak{p}= \begin{cases} f^{-1} & \text{if } \mathfrak{p}\in E_1 \\ 0 & \text{otherwise} \end{cases}$$

If $n=0$ in $\mathfrak{K}_{\mathcal U}$ for a positive integer $n$ , then there exists $E_0 \in \mathcal U$ such that $n\in \cap_{\mathfrak{p}\in E_0}\mathfrak{p}$ . On the other hand, as we have mentioned in Lemma 4.5 above, being a member of a non-principal filter $\mathcal U$ , $E_0$ cannot be a finite set. This is a contradiction, since non-zero member $n$ in $\mathfrak{O}$ has only finite ``zeros'' on the ``arithmetic curve'' $\operatorname{Spm}(\mathfrak{O} )$ . Thus the characteristic of $\mathfrak{K}_{\mathcal U}$ is zero.

$\qedsymbol$

The definition above is partly inspired by works of Kirchberg (See [12] for example.) We would like to give a little explanation on $\pi_{\mathcal U}$ . We regard it as a kind of `limit'. If we are given a member $U$ of $\mathcal U$ and we have an element, say, $h_\mathfrak{p}$ of $\mathfrak{O}/\mathfrak{p}$ for each primes $\mathfrak{p}\in U$ , then, by assigning arbitrary element to `exceptional' primes (that means, primes which are not in $U$ ), we may interpolate $h$ and consider

$$\pi_{\mathcal U} ((h_{\mathfrak{p}})).$$

The element ('limit') does not actually depend on the interpolation. Thus we may refer to the element without specifying the interpolation. In particular, this applies to the case where we have $h_\mathfrak{p}$ for almost all primes $\mathfrak{p}$ . The same type of argument applies for polynomials. We summarize this in the following Lemma.

LEMMA 4.8   Suppose we have a co-finite subset $Y$ of $\operatorname{Spm}(\mathfrak{O} )$ and a collection $\{F_\mathfrak{p}\}_{\mathfrak{p}\in Y} \in (\mathfrak{O}/\mathfrak{p})[T_1,T_2,\dots,T_n,U_1,U_2,\dots,U_n]$ of polynomials. Assume we have a bound $d$ for the degrees of the polynomials. That means,

$$\deg(F_\mathfrak{p})\leq d \qquad (\forall \mathfrak{p}\in Y).$$

Then we may define the `limit'

$$\pi_{\mathcal U}(( F_\mathfrak{p}))$$

by taking `limit' of each of the coefficients. The same arguments also applies for polynomial maps.

For any non-principal ultra filter $\mathcal U$ on $P=$(prime numbers) , We may consider the following ring.

$$\mathbb{Q}_{\mathcal {U}}^{(\infty)}=\prod_p \mathbb{F}_{p^\infty} /($$$U$ �� $0$$$)$$

It turns out that,

LEMMA 4.9  

1. $\mathbb{Q}_{\mathcal {U}}^{(\infty)}$ is an algebraically closed field of characteristic 0 .
2. $\mathbb{Q}_{\mathcal {U}}^{(\infty)}$ has the same cardinality as $\mathbb{C}$ .

Thus we conclude that

PROPOSITION 4.10   As an abstract field,

$$\mathbb{Q}_{\mathcal {U}}^{(\infty)}\cong \mathbb{C}.$$