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\begin{cnabstract}

本文主要分为两个部分，前半部分主要研究了六次循环域的结构。首先将代数数论的一般理论、二次域和循环三次域的主要理论和近期文献中有关三次、六次循环域的结果, 进行了整理综合。 在此基础上, 我们给出了一般六次循环域的整基，素分解的方法，具体给出了多个例子。例如，我们解决了最简单的复六次循环域——7次分圆域的判别式，整基，素分解。同时，利用一般的类数和单位的理论，计算了它的类数和单位群。其次，我们给出了一个实六次循环域，在解决该例子时先利用线性预解子的方法判定出所给多项式$f(x)=x^6-x^5-6x^4+6x^3+8x^2-8x+1$的伽罗瓦群为$C_6$，即得到其对应的分裂域为六次循环域。之后，通过六次循环域的结果给出该例子的二次子域和三次循环子域，从而进一步得到它的整基、素分解，并且我们还计算了这个域的类数。

六次循环域整基和素分解的一般结果是在利用解析数论方法得到判别式的前提下，利用其子域结构得出来的。

第二部分研究的是函数域上的伽罗瓦扩张。在经典的代数和代数几何理论中，L\"{u}roth定理揭示了一元函数域$K(x)$和基域$K$的中间域$E$是基域的单扩张（中间域不等于基域时为非代数扩张）。进一步地，本文假定$K(x)/E$为伽罗瓦扩张，不利用L\"{u}roth定理，证明了$E=K(u)$,其中$u\in K(x)$，可以被$\operatorname{Gal}(K(x)/E)$的初等对称多项式所确定。我们在其后也给出了一个经典$\operatorname{Gal}(K(x)/E)=D_3$的例子。

\keywords{伽罗瓦理论,六次循环域,整基,素分解,L\"{u}roth定理}

\end{cnabstract}

\begin{enabstract}

We have two parts in this thesis. In the first part, we study the structure of cyclic sextic field with many details on its subfields: quadratic field and cyclic cubic field. We reorganize main theory of algebraic number theory and some recent references on cyclic cubic field and cyclic sextic field. Based on these results, we solve the integral basis of the cyclic sextic field as well as find the prime decomposition algorithm. Precisely, we give some examples applying our results. For example, we solve discriminant, integral basis, prime decomposition of 7-cyclotomic field. We also compute its unit group and class number using general theory. We also give an example for real cyclic sextic field. Using linear resolvent method, we first determine that the minimal polynomial has Galois group $C_6$. Then we also find the quadratic subfield and cyclic cubic subfield in order to get the integral basis and prime decomposition. We compute the class number as well.

The general results for integral basis and prime decomposition of cyclic sextic field are given by structures of its subfields based on its discriminant which is computed through analytic number theory's method. 

In the second part, we discuss Galois extensions of a function field. In the classical theory of algebra and algebraic geometry, L\"{u}roth's theorem reveals that any intermediate field $E/K$ of $K(x)/K$ (where $x$ is transcendental extension over $K$) is a simple extension. More precisely, assuming that $K(x)/E$ is Galois, without using L\"{u}roth's theorem, we prove that $E=K(u)$, where $u\in K(x)$ can be determined by the elementary symmetric polynomials. We then give a classical example for $\operatorname{Gal}(K(x)/E)=D_3$.   

\enkeywords{Galois Theory,Cyclic Sextic Fields, Integral Basis, Prime Decomposition, L\"{u}roth's Theorem}

\end{enabstract}

\chapter{The Legendre-Jacobi-Kronecker Symbol}

\label{chap:appA}

All the symbols with the form $\left(\frac{a}{b}\right)$ I used in the paper are Kronecker Symbols. But first, let us review the definition and some properties of Legendre symbol.

\section{The Legendre Symbol}

In number theory, the Legendre symbol is a multiplicative function with values {1,-1,0} that is a quadratic character modulo an \textbf{odd} prime number $p$: its value on a (nonzero) quadratic residue mod p is 1 and on a non-quadratic residue (non-residue) is −1. Its value on zero is 0. 

Furthermore, one can easily show that this symbol has the following properties:

\begin{proposition}

\begin{enumerate}

\item The Legendre symbol is periodic, if $a\equiv b(\text{mod }p)$, then $$\left(\frac{a}{p}\right)=\left(\frac{b}{p}\right)$$

\item The Legendre symbol is multiplicative, i.e. $$\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)=\left(\frac{a b}{p}\right)$$

\item We have the congruence $a^{(p-1)/2}\equiv \left(\frac{a}{p}\right)(\text{mod }p).$

\item There are as many quadratic residues as non-residues mod $p$, say $(p-1)/2$.

\item Let $p$ be an odd prime, then $$\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}, \left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$$

\item Let $p,q$ be two different odd primes, then we have reciprocity law: $$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{(p-1)(q-1)/4}$$

\end{enumerate}

\end{proposition}

\section{The Kronecker Symbol}

Now we extend the definition of the Legendre symbol \citep{cohen1993course}.

\begin{definition}

we define the Kronecker (or Kronecker-Jacobi) symbol $\left(\frac{a}{b}\right)$ for any $a$ and $b$ in $\mathbb{Z}$ as follows:

\begin{enumerate}

\item If $b=0$, then $\left(\frac{a}{0}\right)=1$ if $a=\pm1$, and is equal to 0 otherwise.

\item For $b\neq0$, firstly $\left(\frac{a}{1}\right)=1$. For other case write $b=\prod p$, where $p$ are not necessarily distinct primes (including $p=2$), or $p=-1$ to take care of sign. The we set $$\left(\frac{a}{b}\right)=\prod\left(\frac{a}{p}\right),$$ where $\left(\frac{a}{p}\right)$ is the Legendre symbol defined above for $p>2$, and where $p=2$ we define: $$\left(\frac{a}{2}\right)=\left\{\begin{array}{cc} 0, & \text{if } a \text{ is even}\\

(-1)^{(a^2-1)/8}, &  \text{if } a \text{ is odd.}

\end{array}\right.$$

and also $$\left(\frac{a}{-1}\right)=\left\{\begin{array}{cc} 1, & \text{if } a\geq0\\

-1, &  \text{if } a<0.

\end{array}\right.$$

\end{enumerate}

\end{definition}

Also the Kronecker symbol has the following simple properties:

\begin{proposition}

\begin{enumerate}

\item $\left(\frac{a}{b}\right)=0$ iff $(a,b)\neq1$

\item for all $a,b,c$, if $b c\neq0$, we have $$\left(\frac{ab}{c}\right)=\left(\frac{a}{c}\right)\left(\frac{b}{c}\right), \left(\frac{a}{bc}\right)=\left(\frac{a}{b}\right)\left(\frac{a}{c}\right)$$

\end{enumerate}

\end{proposition}

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\chapter{Resultant and Discriminant of a Polynomial}

\label{chap:appB}

We begin with some useful definitions regarding polynomials. The reason why we refer resultant is that we can calculate the discriminant of the polynomial through this tool, especially when the degree of polynomial is large. First of all, we give a definition of resultant\citep{healy2002resultants}:

\begin{definition}

Let $R$ be an integral domain, given two polynomials $f(x),g(x)\in R[x]$ with roots $\alpha_1,\dots,\alpha_m$ and $\beta_1,\dots,\beta_n$ respectively, then the resultant $\operatorname{Res}(f,g)$ of $f,g$ is defined to be $$\operatorname{Res}(f,g)=l(f)^nl(g)^m\prod_{i,j}(\alpha_i-\beta_j)$$ which is equivalent to both $$\operatorname{Res}(f,g)=l(f)^n\prod_{i}^m g(\alpha_i)$$ and $$\operatorname{Res}(f,g)=(-1)^{n m}l(g)^m\prod_{i}^n f(\beta_i)$$

\end{definition}

From this definition, on can easily see that $f,g$ have a common root in some if and only if $\operatorname{Res}(f,g)=0$.

An important proposition shows the relationship of this definition to the \textbf{Sylvester's matrix} (Some books take that as definition): Let $S$ be the Sylvester's matrix of polynomials $f(x)$ and $g(x)$, then $\operatorname{Res}(f,g)=\det(S)$.

Also for convenience, I choose a definition of normalized discriminant \citep{janson2007resultant} of polynomial as follows in this paper.

\begin{definition}

Let $f$ be a polynomial of degree $n\geq1$ with coefficients in a field $F$. Let $F_1$ be an extension of $F$ where $f$ splits, and let $r_1,\dots,r_n$ be the roots of $f$ in $F_1$. Then the discriminant of $f$ is $$\operatorname{Disc}(f):=\prod_{1\leq i<j\leq n}(r_i-r_j)^2$$

\end{definition}

Note further that $\operatorname{Disc}(cf)=\operatorname{Disc}(f)$ for any nonzero constant. The following theorem give the relation between the discriminant and resultant.

\begin{proposition}\label{prop:disctri}

Let $f=a_nx^n+\cdots+a_0$ be polynomial of degree $n\geq1$ with coefficients in a field $F$. The the discriminant of $f$ is given by 

$$\operatorname{Disc}(f)=(-1)^{n(n-1)/2}a^{-(2n-1)}_n \operatorname{Res}(f,f')$$

\end{proposition}

\begin{example}\label{for:nxpxq}

Let $f(x)=x^n+px+q$ for $n\geq2$, then $f'(x)=nx^{n-1}+p$, then from the Proposition \ref{prop:disctri}, we have $$\operatorname{Disc}(f)=(-1)^{(n-1)(n-2)/2}(n-1)^{n-1p^n}+(-1)^{n(n-1/2)}n^nq^{n-1}$$

\end{example}

\chapter{Table of Orbit-length Partition for Small Degree Polynomials}

\label{chap:appD}

This table can be found in L. Soicher and J. McKay's \citep{soicher1985computing} articles.

\begin{table}[htbp]

\centering

\begin{tabular}{cccc}

\hline

$G$ & $x_1+x_2$ & $x_1+x_2+x_3$ & $x_1-x_2$\\

\hline

\textbf{Degree 3} &  &  &  \\

+$A_3$ &  & & $3^2$\\

$S_3$ &  & & 6\\

\textbf{Degree 4} &  &  &  \\

$Z_4$ & 2,4  & & $4^3$\\

+$V_4$ & $2^3$  & & $4^3$\\

$D_4$ & 2,4  & & 4,8\\

+$A_4$ &6  & & 12\\

$S_4$ &6  & & 12\\

\textbf{Degree 6} &  &  &  \\

$Z_6$ & 3,$6^2$  &2,$6^3$ & $6^5$\\

$S_3$ & $3^3$,6  &2,$6^3$ & $6^5$\\

$D_6$ & 3,$6^2$  &2,$6,12$ & 6,$12^2$\\

+$A_4$ &3,12  &$4^2,6^2$ & 6,$12^2$\\

$G_{18}$ &6,9  &2,18 & $6^2$,18\\

$G_{24}$ & 3,12 &$6^2$,8 &6,$12^2$\\

+$S_4/V_4$ & 3,12&$4^2$,12&6,24\\

$S_4/Z_4$ & 3,12 &8,12&6,24\\

$G_{36}^1$ & 6,9 &2,18&12,18\\

+$G_{36}^2$ & 6,9 &2,18&12,18\\

$G_{48}$& 3,12 &8,12 &6,24\\

+$\text{PSL}_2(5)$ & 15 & $10^2$ & 30\\

$G_{72}$ & 6,9 &2,18&12,18\\

$\text{PGL}_2(5)$ & 15&20&30\\

+$A_6$ & 15&20&30\\

$S_6$ & 15&20&30\\

\hline

\end{tabular}

\caption{Orbit-length Partition of Polynomial with Degree 3,4,6}

\label{tab:allorbits}

\end{table}

where ``+" means that the discriminants of the corresponding polynomial is square. Note that we need a non-linear resolvent to determine $PGL_2(5)$ and $S_6$ etc.

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