Preparation

Definition : Algebraic and Transcendental

Let $K$ be a field, $L$ be an extension field of $K$, $\alpha$ be an element of $L$. If

$$\begin{eqnarray*} &&\exists a_n(\neq 0),a_{n-1},\cdots,a_1,a_0\in K,\\ &&a_n\alpha^n+a_{n-1} \alpha^{n-1}+\cdots a_1 \alpha+a_0=0, \end{eqnarray*}$$

then $\alpha$ is algebraic over $K$. And if $\alpha$ is not algebraic over $K$, then $\alpha$ is transcendental over $K$.

Let $\alpha$ be fixed and $K[X]$ be a 1-variable polynomial ring over $K$. Then the following mapping $\phi$ is a homomorphism by the substitution principle.

$$\begin{array}{ccccc} \phi &:&K[X] &\longrightarrow &L\\ &... ...tebox{90}{$\in$}\\ & &f(X) &\longmapsto&f(\alpha) \end{array}$$

Then the following propositions are equivalent to each other.

$\alpha$ is algebraic over $K$.	$\Leftrightarrow \ker\phi\neq (0)$.	$\Leftrightarrow \phi$ is not one-to-one.
$\alpha$ is transcendental over $K$.	$\Leftrightarrow \ker\phi=(0)$.	$\Leftrightarrow \phi$ is one-to-one.

By the homomorphism theorem of rings,

$$\begin{eqnarray*} &K[X]/\ker\phi \simeq \mathrm{Im}\,\phi \subset L,\\ &\text{\itshape where} \quad \mathrm{Im}\,\phi=\phi(K[X]). \end{eqnarray*}$$

As $K[X]/\ker\phi$ is isomorphic to an integral domain, $\ker\phi$ is a prime ideal of $K[X]$. As $K[X]$ is a principal ideal domain, if $\alpha$ is algebraic over $K$, there exists a monic irreducible polynomial $f_0(X)$ which generates $\ker\phi$. $f_0(X)$ is called a minimal polynomial of $\alpha$, and $\deg f_0$ is an algebraic degree of $\alpha$ over $K$. For example, if $K=\mathbb{Q}$, a rational number is an algebraic number of degree 1, $\sqrt{2}$ or $i\,(=\sqrt{-1})$ are algebraic numbers of degree 2.

The set of algebraic numbers over $\mathbb{Q}$ is often denoted by the symbol $\overline{\mathbb{Q}}$. It is known that $\overline{\mathbb{Q}}$ is a countable field (see 4). Especially it took more than 2300 years to prove that $\pi\notin\overline{\mathbb{Q}}$.

Lemma 1 : Factorial v.s. Power

$$\forall M,C>1,\;\exists n_0\in \mathbb{N},\; n>n_0 \Longrightarrow n!>MC^n.$$ (1)

[ proof ]
Set $n>2MC^2$, then

$$\begin{eqnarray*} &n!&=n(n\!-\!1)\cdots(n\!-\!\Bigl\lfloor\frac{n}{2}\Bigr\rflo... ...ft(\frac{n}{2}\right)^{\frac{n}{2}} >(MC^2)^{\frac{n}{2}}>MC^n. \end{eqnarray*}$$

[ Q.E.D. ]

Or Stirling's formula estimates the factorial more precisely.

$$n!\sim \sqrt{2\pi}\,n^{n+\frac{1}{2}}e^{-n}.$$

cf.

$$n!=\Gamma(n+1)=\int_0^\infty e^{-x}x^n dx.$$

Lemma 2 : Equality and Inequality

Let

$$\begin{eqnarray*} &f(x)&\stackrel{\mathrm{def}}{=}\tsum_{j=0}^m c_j x^j,\\ &I(t)&\stackrel{\mathrm{def}}{=}\int_0^t e^{t-u}f(u)du, \end{eqnarray*}$$

where $c_j\!\in\!\mathbb{R}$, $t\!\in\!\mathbb{C}$ and the path of integral is the line from 0 to $t$, then

$$I(t) =e^t\tsum_{j=0}^m f^{(j)}(0)-\tsum_{j=0}^m f^{(j)}(t),$$ (2)

$$\vert I(t)\vert \leqq \vert t\vert e^{\vert t\vert}\overline{f}(\vert t\vert),$$ (3)

where $$f^{(j)}=\frac{d^j f}{dx^j}$$, $$\overline{f}(x)=\tsum_{j=0}^m\vert c_j\vert x^j$$.

[ proof ]
While $(e^x)^{(j)}=e^x$ for all non-negative integer $j$, if $j>m$ then $f^{(j)}(x)=0$. So using definite integral by parts repeatedly

$$\begin{eqnarray*} &I(t)&=\Big[-e^{t-u}f(u)\Big]_0^t +\displaystyle \int_0^t e^{... ...ts\\ & &=e^t\tsum_{j=0}^m f^{(j)}(0)-\tsum_{j=0}^m f^{(j)}(t). \end{eqnarray*}$$

Next if $u$ is a point on the line from 0 to $t$, then

$$\begin{eqnarray*} &&\vert u\vert\leqq \vert t\vert,\;\vert t-u\vert\leqq \vert ... ...leqq \overline{f}(\vert u\vert)\leqq \overline{f}(\vert t\vert). \end{eqnarray*}$$

So

$$\begin{eqnarray*} &\vert I(t)\vert&\leqq \biggl\vert\displaystyle \int_0^t \ver... ...vert = \vert t\vert e^{\vert t\vert}\overline{f}(\vert t\vert). \end{eqnarray*}$$

[ Q.E.D. ]

Fundamental theorem of symmetric functions

Let $f\in K[X_1,\cdots,X_n]$. If

$$\forall \sigma\in S_n, \; f(X_{\sigma(1)},\cdots,X_{\sigma(n)}) =f(X_1,\cdots,X_n),$$

that is, $f$ is a symmetric polynomial on $n$ variables $\{X_1,\cdots,X_n\}$ over $K$, then $f$ is a polynomial on $\{\sigma_1,\cdots,\sigma_n\}$ over $K$, where

$$\sigma_1\!\stackrel{\mathrm{def}}{=}\!\sum_{j=1}^n X_j,\; ... ...ots,\; \sigma_n\!\stackrel{\mathrm{def}}{=}\!X_1X_2\cdots X_n$$

which are called the $n$ elementary symmetric polynomials.

For instance

$$\tsum_{j=1}^n X_j=\sigma_1,\; \tsum_{j=1}^n X_j^2=\sigma_1... ...^n X_j^3=\sigma_1^3-3\sigma_1 \sigma_2+3\sigma_3,\; \cdots\;.$$

The proof of this theorem is not easy. Please refer to some textbooks of algebra.

Lemma 3 : Differential of a certain polynomial

Leibniz' formula

$$(f_1 f_2)^{(j)}=\sum_{j_1=0}^{j} \binom{j}{j_1}f_1^{(j_1)}f_2^{(j-j_1)}$$

is proved by the mathematical induction on $j$ and can be extended to a product of $n$ functions.

$$(f_1\cdots f_n)^{(j)} =\sum_{j_1+\cdots j_n=j} \frac{j!}{j_1!\cdots j_n!} f_1^{(j_1)}\cdots f_n^{(j_n)}.$$

In this article, the differential of the polynomial

$$f(x)=x^{p-1}(x-\theta_1)^p\cdots (x-\theta_r)^p$$ (4)

where $p$ is a positive integer, is important.

$$\begin{eqnarray*} &&f^{(j)}(x)=\displaystyle\sum_{j_0+\cdots j_r=j} \frac{j!}{j... ...1}(x-\theta_1)^{p-j_1}\cdots \binom{p}{j_r}(x-\theta_r)^{p-j_r} \end{eqnarray*}$$

First, if $\theta_1,\cdots,\theta_r\in\mathbb{Z}$, then $f^{(j)}(x)$

$$\begin{array}{l\vert c\vert c\vert l} x\diagdown\,j&0\cdo... ...ts&\vdots\\ \theta_r &=0 &=0 &\equiv 0 \pmod{p!} \end{array}$$

Second, if $g(x)\stackrel{\mathrm{def}}{=} (x-\theta_1)\cdots (x-\theta_r)\in\mathbb{Z}[x]$, then according to the fundamental theorem of symmetric functions, as $f^{(j)}(0)$ and $$\tsum_{k=1}^r f^{(j)}(\theta_k)$$ are symmetric polynomials on $\{\theta_1,\cdots,\theta_r\}$ over $\mathbb{Z}$, they are polynomials on the coefficients of $g$, that is, they are integers. So $f^{(j)}(x)$

$$\begin{array}{l\vert c\vert c\vert l} x\diagdown\,j&0\cdo... ...!\pmod{p!}\right)\\ \theta_r &=0 &=0 &\text{---} \end{array}$$

e is transcendental e and pi are transcendental e and pi are transcendental