"The steady state of mathematical research is to be completely stuck."

The steady state of mathematical research is to be completely stuck. It is a process that Charles Fefferman of Princeton, himself a onetime math prodigy turned Fields medalist, likens to "playing chess with the devil." The rules of the devil’s game are special, though: The devil is vastly superior at chess, but, Fefferman explained, you may take back as many moves as you like, and the devil may not. You play a first game, and, of course, "he crushes you." So you take back moves and try something different, and he crushes you again, "in much the same way." If you are sufficiently wily, you will eventually discover a move that forces the devil to shift strategy; you still lose, but — aha! — you have your first clue.

Source: http://www.nytimes.com/2015/07/26/magazine/the-singular-mind-of-terry-tao.html

Completions of Fields

We proved earlier that every valuation $\abs{\cdot}$ is equivalent to one that satisfies the triangle inequality. Thus $\abs{\cdot}$ gives $K$ the structure of a metric space by letting $\rho(x, y) = \abs{x - y}^r$.

Theorem Two valuations $\abs{\cdot}_1$ and $\abs{\cdot}_2$ induce the same topology on $K$ iff they are equivalent.

Proof From the definition of the topology, it is clear that if $\abs{x}_1^r = \abs{x}_2$, then $\abs{\cdot}_1$ and $\abs{\cdot}_2$ induce the same topology on $K$.

Now suppose $\abs{\cdot}_1$ and $\abs{\cdot}_2$ are nontrivial and induce the same topology on $K$ There exists an $x_0$ such that $\abs{x_0}_1 > 1$, and let $r = \frac{\log\abs{x_0}_2}{\log\abs{x_0}_1}$. I claim that this is the desired $r$. Since $\abs{x_0^{-n}}_1\to 0$, $x_0^{-n}\to 0$, so $\abs{x_0^{-n}}_2 \to 0$, which means that $\abs{x_0}_2 > 1$ as well. Thus $r > 0$.

Fix another $x\in K^\times$, and let $\epsilon = \log\abs{x}_2 - r\log\abs{x}_1$. Suppose for the sake of contradiction that $\epsilon \neq 0$. Since $\abs{\cdot}_1$ and $\abs{\cdot}_2$ are equivalent,

\[     \abs{x_0^{e_i} x^{f_i}}_1\to 0 \iff \abs{x_0^{e_i} x^{f_i}}_2\to 0, \] or equivalently, \[     e_i\log\abs{x_0}_1 + f_i\log \abs{x}_1\to -\infty \iff r(e_i\log\abs{x_0}_1 + f_i \log\abs{x}_1) + \epsilon f_i\to -\infty.\] Let $f_n = -n\cdot \operatorname{sgn}(\epsilon)$, $e_n = \left\lceil \frac{-f_n\log\abs{x}_1}{\log \abs{x_0}_1} \right\rceil$. Then

\[     0 \le e_i\log\abs{x_0}_1 + f_i\log \abs{x}_1 \le \abs{x_0}_1.   \] However, \[    \epsilon f_i \le r(e_i\log\abs{x_0}_1 + f_i \log\abs{x}_1) + \epsilon f_i \le \epsilon f_i + r\abs{x_0}_1, \] so the first series does not converge to $-\infty$ while the second does, contradiction. Thus for all $x\in K^\times$, $\log\abs{x}_2 - r\log\abs{x}_1 = 0$. $\boxed{}$

Given a valuation $v$ on $K$, we can complete $K$ with respect to the topology induced by $v$ to get a new metric space $K_v$.

Proposition There is a unique way to give $K_v$ the structure of a field such that addition and multiplication are continuous with respect to the valuation $v$.

Proof Note that multiplication and addition are clearly continuous on $K$, so they extend uniquely to multiplication and addition on $K$. Thus $K_v$ already has a ring structure. It suffices to show that $K_v$ has inverses. This is also not hard to show and is left as an exercise to the reader.$\boxed{}$

Proposition Suppose $v$ is a ((non-)Archimedian) valuation on $K$. Then $v$ extends uniquely to a ((non-)Archimedian) valuation on $K_v$.

Proof Obvious by the continuity of $v$. The details are left as an exercise to the reader. $\boxed{}$.

Theorem (Big Ostrowski, Part II) Suppose $v$ is an Archimedian valuation on the number field $K$. Then there exist an embedding $\sigma: K\to \BC$ and a real number $r > 0$ such that $\abs{x}_v = \abs{\sigma(x)}_{\BC}^r$.

Proof Let $K_v$ be the completion of $K$ with respect to $K$. It suffices to show that there exists a valuation-preserving isomorphism from either $K_v\to \BR$ or $K_v\to \BC$.

Note that since we have an embedding $\BQ\to K$ and the restriction of $v$ onto $\BQ$ must be the absolute value norm on $\BQ$, there exists an embedding $\BQ\to K_v$ that extends uniquely to an embedding $\BR\simeq \BQ_v \to K_v$. Thus it suffices to show that every element of $K_v$ has degree at most 2 over $\BR$.

Let $x\in K_v$, and define the function $f_x: \BC\to \BR$ as $z\mapsto \abs{x^2 - (z + \bar{z})x + z\cdot \bar{z}}_{\BC}$. Note that $x$ having degree at most 2 over $\BR$ is equivalent to $f_x(z) = 0$ for some $z\in \BC$, so suppose for the sake of contradiction that $f_x$ never hits 0. Note that the region $\{z: \abs{f_x(z)}_{\BC} < R\}$ is bounded for any $R$. Thus $f_x$ attains a minimum $r>0$ at $z_0$. Since $f_x^{-1}(r)$ is closed and bounded, it is compact. Thus we can pick $z_0$ such that $\abs{z_0}_{\BC}$ is maximal.

Let $0 < \epsilon < r$ be a real number, and let $z_1$ be a root of $z^2 - (z_0 + \bar{z}_0)z+ z_0\cdot \bar{z}_0 + \epsilon = 0$. Since this is also a real quadratic, we have $\abs{z_1}_{\BC}^2 = z_1\cdot \bar{z_1} = z_0\cdot \bar{z}_0 + \epsilon > \abs{z_0}_{\BC}^2$, so $f_x(z_1) > r$.

Now consider $g_n(t) = (t^2 - (z_0 + \bar{z}_0) t+ z_0\cdot \bar{z}_0)^n - (-\epsilon)^n\in \BR[t]$. Note that $g_n(t)$ can be factored uniquely into (real) linear terms and irreducible quadratic terms, so if $t_1, \ldots, t_{2n}\in \BC$ are the roots of $g_n$, then \[ g_n(x)^2 = g_n(t)\cdot \overline{g_n(t)} = \prod_{i = 1}^{2n} (t^2 - (t_i + \bar{t_i})t + t_i\cdot \bar{t_i}) \] Furthermore, $z_1$ is always a root of $g_n(t)$, so without loss of generality $z_1 = t_1$. Thus \[ \abs{g_n(x)}^2_v = \prod_{i = 1}^{2n} \abs{x^2 - (t_i + \bar{t_i})x + t_i\cdot \bar{t_i}}_v\ge f_x(z_1)\cdot r^{2n - 1} \] However, $g_n(x) = r^n - (-\epsilon)^n$. Thus $f_x(z_1)\le \frac{(r^n + (-\epsilon)^n)^2}{r^{2n - 1}}$, so taking $n\to\infty$ gives $f_x(z_1) \le r$, contradiction.$\boxed{}$

Archimedian Valuations on Number Fields

Definition A valuation $\abs{\cdot}$ is non-Archimedian if $\abs{x + y} \le \max(\abs{x}, \abs{y})$. Otherwise, we say that the valuation is Archimedian.

Remark It's not hard to see that if $\abs{\cdot}$ is Archimedian, then so is $\abs{\cdot}^r$ for any real number $r$.

Example(s) If $K = \BQ$, for any prime $p$, $\abs{\cdot}_p$ is non-Archimedian, while $\abs{\cdot}_\infty$ is Archimedian.

Example Let $K$ be a number field, and suppose $\mf{p}$ is a prime ideal of $K$. For each $x$, let $v_{\mf{p}}(x)$ be the greatest integer $n$ such that $x\in \mf{p}^n$. Then the function given by $\abs{x} = \mc{N}(\mf{p})^{-v_{\mf{p}}(x)}$ is a non-Archimedian norm. (This is called the $\mf{p}$-adic valuation on $K$ and is denoted $\abs{\cdot}_{\mf{p}}$.)

Proposition A valuation $\abs{\cdot}$ is non-Archimedian if and only if there exists a real number $M$ such that $\abs{m} < M$ for all $m\in \BZ$. (i.e. bounded on the rational integers)

Proof Note that the proposition is unchanged by replacing $\abs{\cdot}$ with $\abs{\cdot}^r$ for any real number $r > 0$, so we may assume that $\abs{\cdot}$ satisfies the triangle inequality.

For one direction, if $\abs{\cdot}$ is non-Archimedian, then $\abs{m}\le \abs{1}$, so the boundedness condition clearly holds.

Now we prove the other direction. Suppose $\abs{m} < M$ for all $m\in \BZ$. Then
\[ \begin{align*} \abs{x + y}^n &= \abs{(x + y)^n} \\ &= \abs{\sum_{i = 0}^n \binom{n}{i} x^k y^{n-i}} \\ &\le \sum_{i = 0}^n \abs{\binom{n}{i}}\cdot \abs{x^k y^{n-i}}\\ &\le M(n + 1)\max(\abs{x}^n, \abs{y}^n) \end{align*} \] Thus $\abs{x + y}\le (M(n + 1))^{1 / n}\max(\abs{x}, \abs{y})$ for all positive integers $n$, so taking $n\to\infty$ gives $\abs{x + y}\le \max(\abs{x}, \abs{y})$. $\boxed{}$.

For the remainder of this post, suppose $K$ is a number field (i.e. a finite extension of $\BQ$). Let $\mathcal{O}_K$ denote the ring of integers in $K$.

Lemma Suppose $\abs{\cdot}$ is a non-Archimedian valuation on $K$. Then for all $x\in \mc{O}_K$, $\abs{x} \le 1$.

Proof Let the minimal polynomial of $x$ be $x^n + a_{n-1}x^{n-1} + \ldots + a_0$, where $a_0, \ldots, a_{n-1}\in \BZ$. Then $\abs{a_i}\le i$ for all $i$, so \[ \begin{align*} \abs{x}^n = \abs{x^n} &= \abs{\sum_{i = 0}^{n - 1}a_i x^i}\\ &\le \max_{0\le i\le n - 1} \abs{a_i x^i}\\ &\le \max_{0\le i\le n - 1} \abs{x}^i\\ &= \max(1, \abs{x}^{n-1}) \end{align*} \] This is only possible if $\abs{x} \le 1$. $\boxed{}$.

Theorem (Big Ostrowski, Part I) Suppose $\abs{\cdot}$ is a (nontrivial) non-Archimedian norm on $K$. Then $\abs{\cdot}$ is equivalent to $\abs{\cdot}_{\mf{p}}$ for some prime ideal $\mf{p}\subseteq \mc{O}_K$.

Proof The proof of this theorem very closely resembles the proof of the non-Archimedian case of Little Ostrowski's Theorem. However, instead of rational primes, we have prime ideals, and we have to make a little tweak with the class group. In any case....

Let $\abs{\cdot}$ be a non-Archimedian norm on $K$, let $I = \{x\in \mc{O}_K : \abs{x} < 1\}$. By the previous lemma, $I$ has to be a proper ideal of $\mc{O}_K$. Since $\abs{\cdot}$ is nontrivial, $I$ is also nonzero. Finally, if $xy\in I$, then either $x\in I$ or $y\in I$, so $I = \mf{p}$ for some prime ideal $\mf{p}$. It remains to show that $\abs{x} = c^{v_\mf{p}(x)}$ for some real number $0 < c < 1$.

Let $n = \abs{\mc{Cl}_K}$ be the size of the class group of $K$. Then for each prime ideal $\mf{q}$, there exists an $x_{\mf{q}}\in \mc{O}_K$ such that $\mf{q}^n = (x_{\mf{q}})$. In addition, we have \[\abs{x_{\mf{q}}} < 1\iff x_{\mf{q}}\in \mf{p} \iff \mf{q} = \mf{p}.\]Given an $x\in K$, we have that  $(x) = \prod_{\mf{q}} \mf{q}^{v_{\mf{q}}(x)}$ is the unique factorization of the fractional ideal $K$ into prime ideals. Then \[(x^n) = \prod_{\mf{q}} \mf{q}^{nv_{\mf{q}}(x)} = \prod_{\mf{q}} (x_{\mf{q}})^{v_{\mf{q}}(x)}\implies x^n = \prod_{\mf{q}} x_{\mf{q}}^{v_{\mf{q}}(x)}.\] Thus \[\abs{x} = \abs{x^n}^{1/n} = \prod_{\mf{q}} \abs{x_{\mf{q}}}^{v_{\mf{q}}(x)} = \abs{x_{\mf{p}}}^{v_{\mf{q}}(x)/n},\] as desired. $\boxed{}$

Introduction to Valuations

Here we will introduce the notion of a valuation on a field

Definition Let $K$ be a field. A valuation $\abs{\cdot}$ is a map $K\to \BR^{\ge 0}$ such that the following constraints hold.

1. $\abs{x} = 0$ if and only if $x = 0$.
2. $\abs{xy} = \abs{x}\cdot \abs{y}$.
3. There exists a real number $c$ such that for all $x,y\in K$, $\abs{x + y}\le c\max(\abs{x}, \abs{y})$.

Example Let $K = \BQ$. The function $\abs{\cdot}$ given by embedding $\BQ$ into $\BR$ and then taking absolute value is a valuation (henceforth we will denote this $\abs{\cdot}_\infty$. This notation is a little funky, but should make a more sense after seeing Ostrowski's theorem below).

Example / Exercise Let $K = \BQ$ and let $p$ be a rational prime. Let $v_p(x)$ be the exponent of the power of $p$ dividing $x$. Then the function given by $\abs{x} = p^{-v_p(x)}$ is a valuation on $\BQ$. (This is the standard $p$-adic valuation on $\BQ$, and is denoted $\abs{\cdot}_p$.)

Definition / Example The trivial valuation $\abs{\cdot}_0$ is the one that takes on the value of 1 on all of $K^\times$.

Proposition Suppose $\abs{\cdot}$ is a valuation on $K$, and let $r$ positive real number. Then $\abs{\cdot}^r$ is also a valuation on $K$.

Proof Obvious. $\boxed{}$

Definition Two nontrivial valuations $\abs{\cdot}_1$ and $\abs{\cdot}_2$ are equivalent if there exists a real number $r > 0$ such that $\abs{x}_1^r = \abs{x}_2$ for all $x\in K$.

Proposition Let $\abs{\cdot}$ be a valuation on $K$ such that property (3) holds with $c = 2$. Then $\abs{\cdot}$ satisfies the triangle inequality; i.e. $\abs{x + y} \le \abs{x} + \abs{y}$ for all $x, y\in K$.

Proof We are given that $\abs{x + y}\le 2\max(\abs{x}, \abs{y})$ for all $x, y\in K$. By induction we see that \[\abs{\sum_{i = 1}^{2^s} x_i}\le 2^s \max_{1\le i\le 2^s}\abs{x_i}\] for all positive integers $s$. In particular, if $2^{s - 1}\le n< 2^s$, \[ \abs{\sum_{i = 1}^{n} x_i}\le 2^s \max_{1\le i \le n}\abs{x_i} < 2n \max_{1\le i \le n}\abs{x_i} \] One immediate consequence is that $\abs{n} < 2n$ for all positive integers $n$. Now for all $x, y\in K$ and for all positive integers $n$, we have \[ \begin{align*} \abs{x + y}^n = \abs{(x + y)^n} &= \abs{\sum_{i = 0}^n \binom{n}{i} x_i y^{n - i}} \\ &\le 2(n + 1)\max_{0\le i\le n} \abs{\binom{n}{i} x^i y^{n - i}} \\ &\le 2(n + 1)\sum_{i = 0}^n \abs{\binom{n}{i} x^i y^{n - i}} \\ &\le 4(n + 1)\sum_{i = 0}^n \binom{n}{i} \abs{x^i y^{n - i}} \\ &= 4(n + 1)(\abs{x} + \abs{y})^n. \end{align*} \] Taking $n\to \infty$ gives $\abs{x + y}\le \abs{x} + \abs{y}$, as desired. $\boxed{}$

Corollary Let $\abs{\cdot}$ be a valuation on $K$. Then $\abs{\cdot}$ is equivalent to a valuation that satisfies the triangle inequality.

Proof Pick $c$ such that property (3) holds, and pick $r > 0$ such $c^r \le 2$. Then $\abs{\cdot}^r$ satisfies the triangle inequality. $\boxed{}$

Theorem ([Little] Ostrowski's Theorem) When $K = \BQ$, all nontrivial valuations on $K$ are equivalent to either $\abs{\cdot}_\infty$ or $\abs{\cdot}_p$ for some prime $p$.

Proof First suppose $\abs{n}\le 1$ for all positive integers $n$. Set $I = \{n\in \BZ: \abs{n} < 1\}$; note that $I$ is a proper ideal of $\BZ$. Since $\abs{\cdot}$ is nontrivial, $I$ is nonempty, so $I = (n)$ for some positive integer $n > 1$. However, if $\abs{n} < 1$, then $\abs{p} < 1$ for some prime $p$ dividing $n$, which means that $p\in (n)$. The only way this is possible is if $n = p$, or $I = (p)$. Thus $\abs{q} = 1$ for all primes $q\neq p$, which means that $\abs{n} = \abs{p}^{v_p(n)}$ for all $n\in \BQ^\times$. As $\abs{p} < 1$, this is equivalent to the $p$-adic valuation.

Now suppose there exists an integer $m$ such that $\abs{m} > 1$. For every positive integer $n > 1$, we have that $m^k < n^{1 + \fl{\frac{k\log m}{\log n}}}$. Thus if we write out $m^k$ in base $n$ and use the triangle inequality we have \[ \abs{m}^k = \abs{m^k}\le (n-1)\left(1 + \fl{\frac{k\log m}{\log n}}\right)\max(1, \abs{n}^{ \fl{\frac{k\log m}{\log n}}}) \] As $k\to \infty$, $\abs{m}^k$ grows larger than $1 + \fl{\frac{k\log m}{\log n}}$, so we must have $\abs{n} > 1$ for all positive integers $n > 1$. In addition, taking $k\to \infty$ shows that for any pair of positive integers $m, n > 1$, \[ \abs{m}\le \abs{n}^{\displaystyle\frac{\log m}{\log n}}\implies \frac{\log\abs{m}}{\log m} = \frac{\log\abs{n}}{\log n} \] Thus there is a real number $r$ such that $\abs{n} = n^r$ for all positive integers $n$. Thus for all $x\in \BQ, \abs{x} = \abs{x}_\infty^r$. Furthermore, $\abs{n} > 1$ for all positive integers $n$, so $r > 1$. Thus $\abs{\cdot}$ is equivalent to $\abs{\cdot}_\infty$. $\boxed{}$