Recall

Throughout, let $K$ be a field of characteristic $p\ne 0$ and $E/K$ a cyclic extension of order $p^{m-1}$ with $m >1$. The algebraic closure $\overline K^\mathrm{a}$, the separable algebraic closure $\overline K^{\mathrm{s}}$ are always fixed. We use $\mathbf{F}_p$ to denote the finite field of $p$ elements.

For proposition 2 in the post, let $G$ be the Galois group of the extension of $\overline K^\mathrm{s}/K$ (which is, the projective limit of $\mathrm{Gal}(K’/K))$, with $K’$ running over all finite and separable extension of $K$; see this post for the definition of projective limit). The reader is expected to know how to induce a long exact sequence from a short exact sequence, for example from this post.

In this post (the reader is urged to make sure that he or she has understood the concept of characters and more importantly Hilbert’s theorem 90), we have shown that if $[E:K]=p$, then $E=K(x)$ where $x$ is the zero of a polynomial of the form $X^p-X-\alpha$ where $\alpha \in K$. In this belated post, we want to show that, whenever it comes to an extension of order $p^{m-1}$, we are running into the a polynomial of the form $X^p-X-\alpha$. The theory behind is called Artin-Schreier theory, which has its own (highly non-trivial) nature.

Artin-Schreier extensions

Definition 1. An Artin-Schreier polynomial $A_\alpha(X) \in K[X]$ is of the form

An immediate property of Artin-Schreier polynomials that one should notice is the equation

To see this, one should notice that for $x,y \in K$ we have $(x+y)^p=x^p+y^p$.

With this equation we can easily show that

Proposition 1. If $A_\alpha(X)$ has a root in $K$, then all roots of $A_\alpha(X)$ is in $K$. Otherwise, $A_\alpha(X)$ is irreducible over $K$. In this case, let $x$ be a root of $A_\alpha(X)$, then $K(x)/K$ is a cyclic extension of degree $p$.

Proof. We suppose that $x \in K$ is a root of $A_\alpha(X)$. Then

Therefore, by induction, we see easily that $x, x+1, \cdots, x+p-1$ are roots of $A_\alpha(X)$, all of which are in $K$.

Now we suppose that $A_\alpha(X)$ has no root in $K$. Let $x \in \overline K$ be a root of $A_\alpha(X)$. Then in $\overline K[X]$, the polynomial will be written in the form

because, again due to the equation $A_\alpha(X+Y)=A_\alpha(X)+A_\alpha(Y)-A_\alpha(0)$, we can see that $x,x+1,\dots,x+p-1$ are roots of $A_\alpha$.

By contradiction we suppose that $A_\alpha$ is reducible, say $A_\alpha(X)=f(X)g(X)$ where $1 \le d=\deg f < p$ and $f,g \in K[X]$. It follows that

where $\{n_1,\dots,n_d\} \subset \{1,2,\cdots,p\}$. If we expand the polynomial above, we see

Therefore $\left(\sum_{j=1}^{d}n_j-dx\right) \in K$ which is absurd because we then have $x \in K$. Therefore we see that $A_\alpha$ is irreducible.

To see that $K(x)/K$ is Galois, we first notice that this extension is normal : $K(x)$ contains all roots of $A_\alpha(X)$. This extension is separable because all roots of $A_\alpha(X)$, namely $x,x+1,\dots,x+p-1$, are pairwise distinct, i.e. $A_\alpha(X)$ has no multiple roots.

Finally, to see why the Galois group of $K(x)/K$ is cyclic, we notice the action of the Galois group $G$ over the roots of $A_\alpha(X)$. Since $A_\alpha(X)$ is irreducible, there exists $\sigma \in G$ such that $\sigma(x)=x+1$. We see easily that $\sigma^j(x)=x+j$ so $\sigma$ generates $G$ which has period $p$. $\square$

The correspondence between extensions of degree $p$ and polynomials of the form $X^p-X-\alpha$ inspires us to consider them in a distinguished manner.

Definition 2. The field extension $E/K$ is called an Artin-Schreier extension if $E=K(x)$ for some $\alpha \in L \setminus K$ such that $x^p-x\in K$.

Consider the map $\wp:\overline K^\mathrm{s} \to \overline K^\mathrm{s}$ defined by $u \mapsto u^p-u$. We certainly want to find the deep relation between Artin-Schreier extensions of a given field $K$ and the map $\wp$. One of the key information can be found through the following correspondence.

Proposition 2. There is an isomorphism $\operatorname{Hom}(G,\mathbf{F}_p) \cong K/\wp(K)$.

Proof. We first notice that $\wp$ is a $G$-homomorphism, that is, it commutes with the action of $G$ on $\overline K^\mathrm{s}$. Indeed, for any $x \in \overline K^\mathrm{s}$ and $g \in G$, we have

On the other hand, $\wp$ is surjective. Indeed, for any $a \in \overline{K}^\mathrm{s}$, the equation $X^p-X=a$ always has a solution in $\overline K^\mathrm{s}$ because the polynomial $X^p-X-a$ is separable.

We can also see that the kernel of $\wp$ is $\mathbf{F}_p$. This is because the splitting field of $X^p-X$ is the field of $p^1$ elements, which has to be $\mathbf{F}_p$ itself. Therefore we have obtained a short exact sequence

where $\iota$ is the embedding. Taking the long exact sequence of cohomology, noticing that, by Hilbert’s Theorem 90, $H^1(G,\overline{K}^\mathrm{s})=0$, we have another exact sequence

where the first arrow is induced by $\wp$ and the second by $\iota$. Therefore we have $\operatorname{Hom}(G,\mathbf{F}_p) \cong K/\wp(K)$. One can explicitly show that there is a surjective map $K \to \operatorname{Hom}(G,\mathbf{F}_q)$ with kernel $\wp(K)$ that defines the isomorphism. For $c \in K$, one solves $x^p-x=c$, then $\varphi_c:g\mapsto g(x)-x$ is the desired map. The key ingredient of the verification involves the (infinite) Galois correspondence, but otherwise the verification is very tedious. We remark that for any $\varphi \in \operatorname{Hom}(G,\mathbf{F}_p)\setminus\{0\}$ and put $H=\ker\varphi$. Then $K^H/K$ is an Artin-Schreier extension with Galois group $G/H$ and on the other hand $H=\mathrm{Gal}(\overline K^\mathrm{s}/K^H)$. $\square$

“Artin-Schreier of higher order”

We conclude this post by showing that, under a certain condition, one can find an Artin-Schreier extension $L/E$ such that $L/K$ is cyclic of order $p^m$.

Lemma 1. Let $\beta \in E$ be an element such that $\operatorname{Tr}_K^E(\beta)=1$, then there exists $\alpha \in K$ such that $\sigma(\alpha)-\alpha = \beta^p-\beta$, where $\sigma$ is the generator of $\operatorname{Gal}(E/K)$.

Proof. Notice that $\operatorname{Tr}_K^E(\beta^p)=\operatorname{Tr}_K^E(\beta)^p=1$, which implies that $\operatorname{Tr}_K^E(\beta^p-\beta)=0$. By Hilbert’s theorem 90, such $\alpha$ exists. $\square$

Lemma 2. The polynomial $f(X)=X^p-X-\alpha$ is irreducible over $E$; that is, let $\theta$ be a root of $f$, then $E(\theta)$ is an Artin-Schreier extension of $E$.

Proof. By contradiction, we suppose that $\theta \in E$. By Artin-Schreier, all roots of $f$ lie in $E$. In particular, $\sigma(\theta)$ is a root of $f$. Therefore

which implies that

It follows that $\sigma\theta-\theta-\beta$ is a root of $g(X)=X^p-X$. This implies that $\sigma\theta-\theta-\beta\in\mathbf{F}_p \subset K$ and therefore

However, by assumption and Artin-Schreier, $\sigma\theta-\theta \in \mathbf{F}_p \subset K$ we therefore have $\operatorname{Tr}_K^E(\sigma\theta-\theta)=0$ and finally

which is absurd. $\square$

Proposition 3. The field extension $K(\theta)/K$ is Galois, cyclic of degree $p^m$ of $f$, whose Galois group is generated by an extension $\sigma^\ast$ of $\sigma$ such that

Proof. First of all we show that $K(\theta)=E(\theta)$. Indeed, since $K \subset E$, we have $K(\theta) \subset E(\theta)$. However, since $\theta \not \in E$, we must have $K \subset E \subsetneq K(\theta)$. Therefore $p=[E(\theta):K(\theta)][K(\theta):E]$, which forces $E(\theta)$ to be exactly $K(\theta)$.

Let $h(X)$ be the minimal polynomial of $\theta$ over $K$ of degree $p^m$. Then we give an explicit expression of $h$. Notice that since $f(X)$ is the polynomial of $\theta$ over $E$ of degree $p$, we must have $f(X)|h(X)$. For any $k$, we see that $f^{\sigma^k}(X)|g^{\sigma^k}(X)$ too. However, since $\sigma$ fixes $K$, we must have $g^{\sigma^k}(X)=g(X)$, from which it follows that $f^{\sigma^k}(X)|g(X)$ for all $0 \le k \le p^{m-1}-1$. Since the degree of each $f^{\sigma^k}(X)$ is $p$, we obtain

Knowing that $\theta$ is a root of $g$, we see that $\theta+\beta$ is a root of $g(X)$ too because

and by induction we see that for $0 \le k \le p^{m-1}-1$, $f^{\sigma^k}(X)$ has a root in the form

By Artin-Schreier, all roots of $f^{\sigma^k}(X)$ lie in $E(\theta)$ and therefore $h(X)$ splits in $E(\theta)$. Since $E(\theta)/E$ is separable, $E/K$ is separable, we see also $E(\theta)/K$ is separable, which means that $E(\theta)=K(\theta)$ is Galois over $K$.

To see why $K(\theta)/K$ is cyclic, we consider an homomorphism $\sigma^\ast$ of $K(\theta)$ such that $\sigma^{\ast}|_E=\sigma$ and that $\sigma^\ast(\theta)=\theta+\beta$. It follows that $\sigma^\ast \in \operatorname{Gal}(K(\theta)/K)$ because its restriction on $K$, which is the restriction of $\sigma$ on $K$, is the identity. We see then for all $0 \le n \le p^{m}$, one has

In particular,

from which it follows that $(\sigma^\ast)^{p^{m-1}}$ has order $p$, which implies that $\sigma^\ast$ has order $p^m$, thus the Galois group is generated by $\sigma^\ast$. $\square$

References

• Jean-Pierre Serre, Local Fields (chapter X) (link).
• Serge Lang, Algebra, chapter VI (link)

线性代数笔记。

基础知识

行列式

二阶行列式

$\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}$

三阶行列式

对于更高阶的行列式，一般将行列式转为三角形，这样只用计算对角线的乘积即可。

余子式

$n$阶行列式，把第$a_{ij}$所在的行列删除，留下的$n-1$阶行列式称为余子式，记为$M_{ij}$。

代数余子式

代数余子式的$A_{ij} = -1^{i+j}M_{ij}$。

伴随矩阵

代数余子式的转置称为伴随矩阵，只有方阵才有伴随矩阵，记为$A^*$。

伴随矩阵的性质：

$AA^* = A^*A = |A|E$

$A^{-1} = \frac{1}{|A|}A^*（存在A^{-1}）$

$(A^*)^{-1}=(A^{-1})^*=\frac{1}{|A|}A$

$|A^*|=|A|^{n-1}$

$(kA)^*=k^{n-1}A^*$

矩阵的逆

$A^{-1} = \frac{1}{|A|}A^*（存在A^{-1}）$

$A^{-1} = \frac{1}{|A|}A^{*}=\frac{1}{ad-bc}\begin{bmatrix}d &-b\ -c & a\end{bmatrix}$

例：

线性代数的本质

b站视频链接

什么是向量

向量对于不同的学科有不一样的定义。

物理中的向量有长度和方向决定，长度和方向不变可以随意移动，它们表示的是同一个向量。

计算机中的向量更多的是对数据的抽象，可以根据面积和价格定义一个房子$\begin{bmatrix}100m^2\\700000￥\end{bmatrix}$。

数学中的向量可以是任意东西，只要保证两个向量的相加$\vec v + \vec w$以及数字和向量相乘$2\vec v$是有意义的即可。

线性代数中的向量可以理解为一个空间中的箭头，这个箭头起点落在原点。如果空间中有许多的向量，可以点表示一个向量，即向量头的坐标。

向量的基本运算

向量的加法：可以理解为在坐标中两个向量的移动。

数字和向量相乘：可以理解为向量的缩放。

线性组合、张成空间、基

线性组合

两个数乘向量称为两个向量的线性组合$a\vec v+ b\vec w$。

两个不共线的向量通过不同的线性组合可以得到二维平面中的所有向量。

两个共线的向量通过线程组合只能得到一个直线的所有向量。

如果两个向量都是零向量那么它只能在原点。

张成空间

所有可以表示给定向量线性组合的向量的集合称为给定向量的张成空间（span）。

一般来说两个向量张成空间可以是直线、平面。

三个向量张成空间可以是平面、空间。

如果多个向量，并且可以移除其中一个而不减小张成空间，那么它们是线性相关的，也可以说一个向量可以表示为其他向量的线性组合$\vec u = a \vec v + b\vec w$。

如果所有的向量都给张成的空间增加了新的维度，它们就成为线性无关的$\vec u \neq a \vec v + b\vec w$。

基

向量空间的一组及是张成该空间的一个线性无关向量集。

矩阵与线性变换

严格意义上来说，线性变换是将向量作为输入和输出的一类函数。

变化可以多种多样，线性变化将变化限制在一个特殊类型的变换上，可以简单的理解为网格线保持平行且等距分布。

线性变化满足一下两个性质：

• 线性变化前后直线依旧是直线不能弯曲。
• 原点必须保持固定。

可以使用基向量来描述线性变化：

通过记录两个基向量$\hat{i}$,$\hat{j}$的变换，就可以得到其他变化后的向量。

已知向量$\vec v = \begin{bmatrix}-1\\2\end{bmatrix}$

变换之前的$\hat i$和$\hat j$：

Introduction

Polynomial is of great interest in various fields, such as analysis, geometry and algebra. Given a polynomial, we try to extract as many information as possible. For example, given a polynomial, we certainly want to find its roots. However this is not very realistic. Abel-Ruffini theorem states that it is impossible to solve polynomials of degree $\ge 5$ in general. For example, one can always solve the polynomial $x^n-1=0$ for arbitrary $n$, but trying to solve $x^5-x-1=0$ over $\mathbb{Q}$ is not possible. Galois showed that the flux of solvability lies in the structure of the Galois group, depending on whether it is solvable group-theoretically.

In this post, we will explore the theory of solvability in the modern sense, considering extensions of arbitrary characteristic rather than solely number fields over $\mathbb{Q}$.

Solvable Extensions

Definition 1. Let $E/k$ be a separable and finite field extension, and $K$ the smallest Galois extension of $k$ containing $E$. We say $E/k$ is solvable if $G(K/k)$ (the Galois group of $K$ over $k$) is solvable.

Throughout we will deal with separable extensions because without this assumption one will be dealing with normal extensions instead of Galois extensions. Although we will arrive at a similar result.

Proposition 1. Let $E/k$ be a separable extension. Then $E/k$ is solvable if and only if there exists a solvable Galois extension $L/k$ such that $k \subset E \subset L$.

Proof. If $E/k$ is solvable, it suffices to take $L$ to be the smallest Galois extension of $k$ containing $E$. Conversely, Suppose $L/k$ is a solvable and Galois such that $k \subset E \subset L$. Let $K$ be the smallest Galois extension of $k$ containing $E$, i.e. we have $k \subset E \subset K \subset L$. We see $G(K/k) \cong G(L/k)/G(L/K)$ is a homomorphism image of $G(L/k)$ and it has to be solvable. $\square$

Next we introduce an important concept concerning field extensions.

Definition 2. Let $\mathcal{C}$ be a certain class of extension fields $F \subset E$. We say that $\mathcal{C}$ is distinguished if it satisfies the following conditions:

1. Let $k \subset F \subset E$ be a tower of fields. The extension $k \subset E$ is in $\mathcal{C}$ if and only if $k \subset F$ is in $\mathcal{C}$ and $F \subset E$ is in $\mathcal{C}$.
2. If $k \subset E$ is in $\mathcal{C}$ and if $F$ is any given extension of $k$, and $E,F$ are both contained in some field, then $F \subset EF$ is in $\mathcal{C}$ too. Here $EF$ is the compositum of $E$ and $F$, i.e. the smallest field that contains both $E$ and $F$.
3. If $k \subset F$ and $k \subset E$ are in $\mathcal{C}$ and $F,E$ are subfields of a common field, then $k \subset FE$ is in $\mathcal{C}$.

When dealing with several extensions at the same time, it can be a great idea to consider the class of extensions they are in. For example, Galois extension is not distinguished because normal extension does not satisfy 1. That’s why we need to have the fundamental theorem of Galois theory, a.k.a. Galois correspondence, because not all intermediate subfields are Galois. Separable extension is distinguished however. We introduce this concept because:

Proposition 2. Solvable extensions form a distinguished class of extensions. (N.B. these extensions are finite and separable by default.)

Proof. We verify all three conditions mentioned in definition 2. To make our proof easier however, we first verify 2.

Step 1. Let $E/k$ be solvable. Let $F$ be a field containing $k$ and assume $E, F$ are subfields of some algebraically closed field. We need to show that $EF/F$ is solvable. By proposition 1, there is a Galois solvable extension $K/k$ such that $K \supset E \supset k$. Then $KF$ is Galois over $F$ and $G(KF/F)$ is a subgroup of $G(K/k)$. Therefore $KF/F$ is a Galois solvable extension and we have $KF \supset EF \supset F$, which implies that $EF/F$ is solvable.

Step 2. Consider a tower of extensions $E \supset F \supset k$. Assume now $E/k$ is solvable. Then there exists a Galois solvable extension $K$ containing $E$, which implies that $F/k$ is solvable because $K \supset F$. We see $E/F$ is also solvable because $EF=E$ and we are back to step 1.

Conversely, assume that $E/F$ is solvable and $F/k$ is solvable. We will find a solvable extension $M/k$ containing $E$. Let $K/k$ be a Galois solvable extension such that $K \supset F$, then $EK/K$ is solvable by step 1. Let $L$ be a Galois solvable extension of $K$ containing $EK$. If $\sigma$ is any embedding of $L$ over $k$ in a given algebraic closure, then $\sigma K = K$ and hence $\sigma L$ is a solvable extension of $K$. [This sentence deserves some explanation. Notice that $L/k$ is not necessarily Galois, therefore $\sigma$ is not necessarily an automorphism of $L$ and $\sigma L \ne L$ in general . However, since $K/k$ is Galois, the restriction of $\sigma$ on $K$ is an automorphism so therefore $\sigma K = K$. The extension $\sigma L / \sigma K$ is solvable because $\sigma L$ is isomorphic to $L$ and $\sigma K = K$.]

We let $M$ be the compositum of all extensions $\sigma L$ for all embeddings $\sigma$ of $L$ over $k$. Then $M/k$ is Galois and so is $M/K$ [note: this is the property of normal extension; besides, $M/k$ is finite]. We have $G(M/K) \subset \prod_{\sigma}G(\sigma L/K)$ which is a product of solvable groups. Therefore $G(M/K)$ is solvable, meaning $M/K$ is a solvable extension. We have a surjective homomorphism $G(M/k) \to G(K/k)$ (given by $\sigma \mapsto \sigma|_K$) and therefore $G(M/k)$ has a normal subgroup whose factor group is solvable, meaning $G(M/k)$ is solvable. Since $E \subset M$, we are done.

Step 3. If $F/k$ and $E/k$ are solvable and $E,F$ are subfields of a common field, we need to show that $EF$ is solvable over $k$. By step 1, $EF/F$ is solvable. By step 2, $EF/k$ is solvable. $\square$

Solvable By Radicals

Definition 2. Let $F/k$ be a finite and separable extension. We say $F/k$ is solvable by radicals if there exists a finite extension $E$ of $k$ containing $F$, and admitting a tower decomposition

such that each step $E_{i+1}/E_i$ is one of the following types:

1. It is obtained by adjoining a root of unity.
2. It is obtained by adjoining a root of a polynomial $X^n-a$ with $a_i \in E_i$ and $n$ prime to the characteristic.
3. It is obtained by adjoining a root of an equation $X^p-X-a$ with $a \in E_i$ if $p$ is the characteristic $>0$.

For example, $\mathbb{Q}(\sqrt{-2})/\mathbb{Q}$ is solvable by radicals. We consider the polynomial $f(x)=x^2-2x+3$. We know its roots are $x_1=-1-\sqrt{-2}$ and $x_2=-1+\sqrt{-2}$. However let’s see the question in the sense of field theory. Notice that

Therefore $f(x)=0$ is equivalent to $(x-1)^2=-2$. Then $x-1=\sqrt{-2}$ and $x-1=-\sqrt{-2}$ in $\mathbb{Q}(\sqrt{-2})$ are two equations that make perfect sense. Thus we obtain our desired roots. The field gives us the liberty of basic arithmetic, and the radical extension gives us the method to look for a radical root.

It is immediate that the class of extensions solvable by radicals is a distinguished class.

In general, we are adding “$n$-th root of something”. However, when the characteristic of the field is not zero, there are some complications. For example, talking about the $p$-th root of an element in a field of characteristic $p>0$ will not work. Therefore we need to take good care of that. The second and third types are nods to Kummer theory and Artin-Schreier theory respectively, which are deduced from Hilbert’s theorem 90’s additive and multiplicative form. We interrupt the post by introducing the respective theorems.

Let $K/k$ be a cyclic extension of degree $n$, that is, $K/k$ is Galois and $G(K/k)$ is cyclic. Suppose $G(K/k)$ is generated by $\sigma$. Then we have the celebrated “Theorem 90”:

Theorem 1 (Hilbert’s theorem 90, multiplicative form). Notation being above, let $\beta \in K$. The norm $N_{k}^{K}(\beta)=1$ if and only if there exists an element $\alpha \ne 0$ in $K$ such that $\beta = \alpha/\sigma\alpha$.

To prove this, we need Artin’s theorem of independent characters. With this, we see the second type of extension in definition 2 is cyclic.

Theorem 2. Let $k$ be a field, $n$ an integer $>0$ prime to the characteristic of $k$, and assume that there is a primitive $n$-th root of unity in $k$.

1. Let $K$ be a cyclic extension of degree $n$. Then there exists $\alpha \in K$ such that $K = k(\alpha)$ and $\alpha$ satisfies an equation $X^n-a=0$ for some $a \in k$.
2. Conversely, let $a \in k$. Let $\alpha$ be a root of $X^n-a$. Then $k(\alpha)$ is cyclic over $k$ of degree $d|n$, and $\alpha^d$ is an element of $k$.

All in all, theorem 2 states that a $n$-th root of $a$ yields a cyclic extension. However we don’t drop the assumption that $n$ is prime to the characteristic of $k$. When this is not the case, we will use Artin-Schreier theorem.

Theorem 3 (Hilbert’s theorem 90, additive form). Let $K/k$ be a cyclic extension of degree $n$. Let $\sigma$ be the generator of $G(K/k)$. Let $\beta \in K$. The trace $\mathrm{Tr}_k^K(\beta)=0$ if and only if there exists an element $\alpha \in K$ such that $\beta = \alpha-\sigma\alpha$.

This theorem requires another application of the independence of characters.

Theorem 4 (Artin-Schreier). Let $k$ be a field of characteristic $p$.

1. Let $K$ be a cyclic extension of $k$ of degree $p$. Then there exists $\alpha \in K$ such that $K=k(\alpha)$ and $\alpha$ satisfies an equation $X^p-X-a=0$ with some $a \in k$.
2. Conversely, given $a \in k$, the polynomial $f(X)=X^p-X-a$ either has one root in $k$, in which case all its roots are in $k$, or it is irreducible. In the latter case, if $\alpha$ is a root then $k(\alpha)$ is cyclic of degree $p$ over $k$.

In other words, instead of looking at the $p$-th root of unity in a field of characteristic $p$, we look at the root of $X^p-X-a$, which still yields a cyclic extension.

Now we are ready for the core theorem of this post.

Theorem 5. Let $E$ be a finite separable extension of $k$. Then $E$ is solvable by radicals if and only if $E/k$ is solvable.

Proof. First of all we assume that $E/k$ is solvable. Then there exists a finite Galois solvable extension of $k$ containing $E$ and we call it $K$. Let $m$ be the product of all primes $l$ such that $l \ne \operatorname{char}k$ but $l|[K:k]$. Let $F=k(\zeta)$ where $\zeta$ is a primitive $m$-th root of unity. Then $F/k$ is abelian and is solvable by radical by definition.

Since solvable extensions form a distinguished class, we see $KF/F$ is solvable. There is a tower of subfields between $F$ and $KF$ such that each step is cyclic of prime order, because every solvable group admits a tower of cyclic groups, and we can use Galois correspondence. By theorem 2 and 4, we see $KF/F$ is solvable by radical because extensions of prime order have been determined by these two theorems. It follows that $E/k$ is solvable by radicals: $KF/F$ is solvable by radicals, $F/k$ is solvable by radicals $\implies$ $KF/k$ is solvable by radicals $\implies$ $E/k$ is solvable by radicals because $KF \supset E \supset k$.

The elaboration of the “if” part is as follows. In order to prove $E/k$ is solvable by radicals, we show that there is a much bigger field $KF$ containing $E$ such that $KF/k$ is solvable by radical. First of all there exists a finite Galois solvable extension $K/k$ containing $E$. Next we define a cyclotomic extension $F/k$ with the following intentions

1. $F/k$ should be solvable by radicals.
2. $F$ contains enough primitive roots of unity, so that we can use theorem 2 freely.

To reach these two goals, we decide to put $F=k(\zeta)$ where $\zeta$ is a $m$-th root of unity and $m$ is the radical of $[K:k]$ divided by the characteristic of $k$ when necessary. This field $F$ certainly ensures that $F/k$ is solvable by radical. For the second goal, we need to take a look of the subfield between $F$ and $KF$. Let $k = K_0 \subset K_1 \subset \dots \subset K_n = K$ be a tower of field extensions such that every step $K_{i+1}/K_i$ is of prime degree [this is possible due to the solvability of $K/k$]. These prime numbers can only be factors of $[K:k]$ Then in the lifted field extension $F=K_0F \subset K_1F \subset \dots \subset K_nF=KF$ we do not introduce new prime numbers. Why do we consider prime factors of $[K:k]$? Let’s say $[K_{i+1}F:K_iF] = \ell$ is a prime number. If $\ell=\operatorname{char}k$ then we can use theorem 4. Otherwise we still have $\ell|[K:k]$ so we use theorem 2. However this theorem requires a primitive $\ell$-th root to be in $K_{i}F$. Our choice of $m$ and $\zeta$ guaranteed this to happen because $\ell|m$ and therefore a primitive $\ell$-th root of unity exists in $F$. We can make $m$ bigger but there is no necessity. The “only if” part does nearly the same thing, with an alternation of logic chain.

Conversely, assume that $E/k$ is solvable by radicals. For any embedding $\sigma$ of $E$ in $E^{\mathrm{a}}$ over $k$, the extension $\sigma E/k$ is also solvable by radicals. Hence the smallest Galois extension $K$ of $E$ containing $k$, which is a composite of $E$ and its conjugates is solvable by radicals. Let $m$ be the product of all primes unequal to the characteristic dividing the degree $[K:k]$ and again let $F=k(\zeta)$ where $\zeta$ is a primitive $m$-th root of unity. It will suffice to prove that $KF$ is solvable over $F$, because it follows that $KF$ is solvable by $k$ and hence $G(K/k)$ is solvable because it is a homomorphic image of $G(KF/k)$. But $KF/F$ can be decomposed into a tower of extensions such that each step is prime degree and of the type described in theorem 2 and theorem 4. The corresponding root of unity is in the field $F$. Hence $KF/F$ is solvable, proving the theorem. $\square$

Introduction

There are a lot of important linear algebraic groups that are widely used in mathematics, physics and industry. Some of them have nice visualisations. For example, it is widely known that $SU(2) \cong S^3$ and $SO(3) \cong \mathbb{RP}^3$. The group $SL(2,\mathbb{R})$ is not less important than them but the visualisation concerning this group is not very easy to be found. In this post we show that

where $D$ is the open unit disk. In other words, $SL(2,\mathbb{R})$ can be considered as a donut, not the shell of it ($S^1 \times S^1$) but the “content” or “flesh” of it. More formally, the inside of a solid torus.

The related core theory can be found in Iwasawa decomposition, but to access it we need Lie group and Lie algebra theories, which involves differential geometry and certainly goes beyond the scope of this post. Interested readers can refer to Lie Groups Beyond an Introduction chapter 6 for Iwasawa decomposition theory.

Immediate topological consequences

Before we establish the homeomorphism

we first see what we can derive from it.

• Is $SL(2,\mathbb{R})$ compact?

No. Since $D$ is not compact, $S^1 \times D$ cannot be compact.

• What is the fundamental group of $SL(2,\mathbb{R})$?

Notice there is a (strong) deformation retract between $S^1 \times D$ and $S^1$. Therefore $\pi_1(SL(2,\mathbb{R})) = \pi_1(S^1)=\mathbb{Z}$.

• Connectedness of $SL(2,\mathbb{R})$?

It is connected because $S^1$ and $D$ are connected. It is not simply connected because the fundamental group is not trivial.

• What is the dimension of $SL(2,\mathbb{R})$ as a manifold?

The dimension is $3$.

The Iwasawa decomposition

If we directly jump to the conclusion without mentioning Lie theory, one will see the decomposition comes from nowhere. Instead of defining $K$, $A$ and $N$ that will appear later and show that there is no discrepancy, we deduce the decomposition without the usage of Lie theory. Instead, we consider the action of $SL(2,\mathbb{R})$ on the upper half plane, because group action is likely to expose more information of the group.

Consider the group action of $SL(2,\mathbb{R})$ on the upper half plane

given by

Up to an explosion of calculation, one can indeed verify that this is a group action and in particular

As one may guess, it is not wise to continue without investigating the action first, or we will be lost in calculation. We first show that this action is transitive by showing that for any $z=x+yi \in \mathfrak{H}$, there is some $\sigma \in SL(2,\mathbb{R})$ such that $\sigma(z)=i$:

Let’s play around the last linear equation system:

We can put $c=0$ and $a=\frac{1}{\sqrt{y}}$ so that $b=-\frac{x}{\sqrt{y}}$ and $d=\sqrt{y}$. That is,

We have therefore proved:

The action of $SL(2,\mathbb{R})$ on $\mathfrak{H}$ is transitive.

Proof. For any $z,z’ \in \mathfrak{H}$, there exists $\sigma$ and $\sigma’$ such that $\sigma(z)=i$ and $\sigma’(z’)=i$. Then $\sigma’^{-1}(\sigma(z))=z’$, i.e. $\sigma’^{-1}\sigma$ sends $z$ to $z’$. $\square$

By working around $i$ on $\mathfrak{H}$ we can save ourselves from a lot of troubles. It is then desirable to find the stabiliser of $i$.

The stabiliser of $i \in \mathfrak{H}$ is $SO(2) \cong S^1$.

Proof. Suppose $\sigma=\begin{pmatrix} a & b \\c & d \end{pmatrix}$ stabilises $i$. Then first of all we have

Then

It follows that

Therefore $\sigma \in O(2) \cap SL(2) = SO(2)$ as expected. $\square$

With these being said, the action of $SL(2,\mathbb{R})$ on $i$ consists of $SO(2)$ that moves nothing and the rest that actually move things. In other words, $SL(2,\mathbb{R})/SO(2) \cong \mathfrak{H}$ as a $2$-manifold. In particular, the action is a isometry. We will find the effective part of the group action out. For $\sigma \in SL(2,\mathbb{R})$, we assume that $\sigma(i)=x+iy$. Then

Let $B$ be the upper triangular matrices in $SL(2,\mathbb{R})$ with positive diagonal elements. Then it is elements in $B$ that actually move things. According to this classification, we have obtained a decomposition

The matrix multiplication map $B \times SO(2) \to SL(2,\mathbb{R})$ is surjective.

Proof. Notice that every element of $B$ can be written in the form

For any $\sigma \in SL(2,\mathbb{R})$, suppose $\sigma(i)=x+iy$, then $\sigma(i)=\lambda_{x,y}(i)$, therefore $\lambda_{x,y}^{-1}\sigma(i)=i$, i.e. $\lambda_{x,y}^{-1}\sigma \in SO(2)$, i.e. $\lambda_{x,y}^{-1}\sigma$ is a stabiliser of $i$. The product $\sigma = \lambda_{x,y}(\lambda_{x,y}^{-1}\sigma)$ always lies in the image of $B \times SO(2)$. $\square$

We can decompose $B$ further:

Let $N$ be the group of upper triangular matrices in $SL(2,\mathbb{R})$ with $1$ on the diagonal line and let $A$ be the group of diagonal matrices with non-negative entries. Then $B=NA$. Let $K=SO(2) \subset SL(2,\mathbb{R})$, then we have obtained the so-called Iwasawa decomposition:

There is a diffeomorphism onto

Proof. It only remains to show injectivity. Suppose $n_1a_1k_1=n_2a_2k_2$. Applying both sides onto $i$ we obtain $n_1a_1(i)=n_2a_2(i)$. Suppose

Then we have $n_1a_1(i)=x_1+y_1i=n_2a_2(i)=x_2+y_2i$. It follows that $x_1=x_2$ and $y_1=y_2$, i.e. $n_1=n_2$ and $a_1=a_2$ and therefore $k_1=k_2$. $\square$

By investigating $N$ and $A$ further we obtain

The group $SL(2,\mathbb{R})$ is homeomorphic to $S^1 \times D$.

Proof. Notice that $N$ is homeomorphic to $\mathbb{R}$ and $A$ is homeomorphic to $\mathbb{R}_{>0}\cong \mathbb{R}$. $\square$

Notice the order of $N,A,K$ does not matter very much: $NAK,KAN,ANK,KNA$ are the same thing. This is because $AN=NA$ and for $nak \in SL(2,\mathbb{R})$, we have $(nak)^{-1}=k^{-1}a^{-1}n^{-1}$ which lies in the preimage of $K \times A \times N$ under matrix multiplication.

Immediate group-theoretical consequences

With the full Iwasawa decomposition in mind, we can scratch the surface of the rather complicated $SL(2,\mathbb{R})$.

The only continuous homomorphism of $SL(2,\mathbb{R})$ to $\mathbb{R}$ is trivial.

Proof. Let $f:SL(2,\mathbb{R}) \to \mathbb{R}$ be such a map. We have $f(kan)=f(k)+f(a)+f(n)$. We need to show that $f(k)=f(a)=f(n)=0$.

First of all, since $K$ is a compact subgroup of $SL(2,\mathbb{R})$, its image on $\mathbb{R}$ has to be a compact subgroup. On the other hand, $f$ on $A$ and $N$ can be constructed more explicitly. For $A$, we see $\begin{pmatrix}r & 0 \\ 0 & \frac{1}{r} \end{pmatrix} \mapsto r \mapsto \log{r}$ yields an isomorphism of $A$ and $\mathbb{R}$, in both algebraical and topological sense. For $N$ on the other hand, we immediately have an isomorphism $\begin{pmatrix}1 & x \\ 0 & 1\end{pmatrix} \mapsto x$. Therefore the image of $f$ on $A$ and $N$ can be realised as as $u\log{r}$ and $vx$ for some $u,v \in \mathbb{R}$. We use the fact that $AN=NA$ to determine $u$ and $v$. Notice that

applying $f$ on both sides, we have

For $u$, we consider the conjugate relation

Applying $f$ on both sides we obtain

This proves the triviality of $f$. $\square$

Let $f:SL(2,\mathbb{R}) \to GL(n,\mathbb{R})$ be a continuous homomorphism, then $f(SL(2,\mathbb{R})) \subset SL(n,\mathbb{R})$.

Proof. Consider the sequence of group homomorphisms

Since $SL(2,\mathbb{R})$ is connected, we see $\det\circ f(SL(2,\mathbb{R}))$ is connected, thus lying in $\mathbb{R}_{>0}$. We can then modify the sequence a little bit:

The map $\log \circ \det \circ f$ is a continuous homomorphism sending $SL(2,\mathbb{R})$ to $\mathbb{R}$, which is trivial, and therefore

This proves our assertion. $\square$

There are still a lot we can do without much Lie theory but Haar measure theory. The reader is advised to try this exercise set to see, for example, that the “volume” of $SL(2,\mathbb{R})/SL(2,\mathbb{Z})$ is $\zeta(2)$. In the references / further reading section the reader will also find a way to show that $SL(2,\mathbb{Z})\backslash SL(2,\mathbb{R})/SO(2,\mathbb{R})$ has volume $\frac{\pi}{3}$.

References / Further Reading

• Keith Conrad, Decomposing $SL(2,\mathbb{R})$. https://kconrad.math.uconn.edu/blurbs/grouptheory/SL(2,R).pdf
• Serge Lang, $SL(2,\mathbb{R})$. (He wrote this book for self-study because he didn’t know much about Lie theory. In chapter 3 one can see how to do calculus with $SL(2,\mathbb{R})$.)
• Anthony W. Knapp, Lie Groups Beyond an Introduction.
• Merrick Cai, The Volume of $SL(2,\mathbb{Z})\backslash SL(2,\mathbb{R})/SO(2,\mathbb{R})$. https://math.mit.edu/~mqt/math/teaching/mit/18-704/final-papers/cai-m_final.pdf (A more concrete example of doing calculus on this group.)

Heights

Definition. For a polynomial with coefficients in a number field $K$

the height of $f$ is defined to be

where

is the Gauss norm for any place $v$.

Here, $M_K$ refers to the canonical set of non-equivalent places on $K$. See first four pages of this document for a reference.

As one can expect, this can tell us about some complexity of a polynomial, just like how the height of an algebraic number tells us its complexity. Let us compute some examples.

Computing Heights

Let us consider the simplest one

first. Since $|x^2-1|_v=1$ for all places $v$, the height of $f$ is a sum of $0$, which is still $0$.

Next, we take care of a polynomial that involves prime numbers

We see $|g(x)|_\infty=2$, $|g(x)|_2=2^{-(-2)}=4$, $|g(x)|_3=3^{-(-1)}=3$, and the Gauss norm is $1$ for all other primes. Therefore

Put $u(x,y)=\sqrt{2}x^2 + 3\sqrt{2}xy+5y^2+7 \in \mathbb{Q}(\sqrt{2})[x,y]$, we can compute its height carefully. Notice that $|\sqrt{2}|_v=\sqrt{|2|_v}$ for all places $v$ and we therefore have

Height and Products

If $f \in K[s_1,\dots,s_n]$ and $g \in K[t_1,\dots,t_m]$ are two polynomials in different variables, then as a polynomial in $K[s_1,\dots,s_n;t_1,\dots,t_m]$, $fg$ has height $h(f)+h(g)$. This is immediately realised once we notice that the height of a polynomial is equal to the height of the vector of coefficients in appropriate projective space. The identity $h(fg)=h(f)+h(g)$ follows from the Segre embedding.

But if variables coincide, things get different. For example, $h(x+1)=0$ but $h((x+1)^2)=2$. This is because we do not have $|fg|_\infty=|f|_\infty|g|_\infty$. Nevertheless, for non-Archimedean places, things are easier.

Gauss’s lemma. If $v$ is not Archimedean, then $|fg|_v=|f|_v|g|_v$.

Proof. First of all, it suffices to prove it for univariable cases. If $f$ and $g$ have multiple variables $x_1,\dots,x_n$, let $d$ be an integer greater than the degree of $fg$. Then the Kronecker substitution

reduces our study into $K[t]$. This is because, with such a $d$, this substitution gives a univariable polynomial with the same set of coefficients.

Therefore we only need to show that $|f(t)g(t)|_v=|f(t)|_v|g(t)|_v$. Without loss of generality we assume that $|f(t)|_v=|g(t)|_v=1$. Write $f(t)=\sum a_k t^k$ and $g(t)=\sum b_k t^k$, we have $f(t)g(t)=\sum c_jt^j$ where $c_j=\sum_{j=k+l}a_kb_l$.

We suppose that $|fg|_v<1$, i.e., $|c_j|_v<1$ for all $j$, and see what contradiction we will get. If $|a_j|=1$ for all $j$, then $|c_j|_v<1$ implies that $|b_k|_v<1$ for all $k$ and therefore $|g|_v<1$, a contradiction. Therefore we may assume that, without loss of generality, $|a_0|_v<1$ but $|a_1|_v=1$. Then, since

we have $|a_1b_{j-1}|_v=|b_{j-1}|_v<1$ for all $j \ge 1$. It follows that $|g(t)|_v<1$, still a contradiction. $\square$

So much for non-Archimedean case. For Archimedean case things are more complicated so we do not have enough space to cover that. Nevertheless, we have

Gelfond’s lemma. Let $f_1,\dots,f_m$ be complex polynomials in $n$ variables an set $f=f_1\cdots f_n$, then

where $d$ is the sum of the partial degrees of $f$, and $\ell_\infty(f)=\max_j|a_j|=|f|_\infty$.

Combining Gelfond’s lemma and Gauss’s lemma, we obtain

Mahler Measure

Is not actually given by Mahler initially. It was named after Mahler because he successfully extended it to multivariable cases in an elegant way. We will cover the original motivation anyway.

Original Version and Lehmer’s Conjecture

Say we want to find prime numbers large enough. Pierce came up with an idea. Consider $p(x) \in \mathbb{Z}[x]$, which is factored into

Consider $\Delta_n=\prod_i(\alpha^n_i-1)$. Then by some Galois theory, this is indeed an integer. So perhaps we may find some interesting integers in the factors of $\Delta_n$. Also, we expect it to grow slowly. Lehmer studied $\frac{\Delta_{n+1}}{\Delta_n}$ and observed that

So it makes sense to compare all roots of $p(x)$ with $1$. He therefore suggested the following function related to $p(x)$:

This number appears if we consider $\lim_{n \to \infty}\Delta_{n+1}/\Delta_n$.

He also asked the following question, which is now understood as Lehmer conjecture, although in his paper he addressed it as a problem instead of a conjecture:

Is there a constant $c$ such that, $M(p)>1 \implies M(p)>c$?

It remains open but we can mention some key bounds.

• Lehmer himself found that

and actually this is the finest result that has ever been discovered. It was because of this discovery that he gave his problem.

This polynomial has also led to the discovery of a large prime number $\sqrt{\Delta_{379}}=1, 794, 327, 140, 357$, although by studying $x^3-x-1$, we have found a bigger prime number $\Delta_{127}=3, 233, 514, 251, 032, 733$.

• Breusch (and later Smyth) discovered that if $p$ is monic, irreducible and nonreciprocal, i.e. it does not satisfy $p(x)=\pm x^{\deg p}f(1/x)$, then

• E. Dobrowlolski found that, t if $p(x)$ is monic, irreducible and noncyclotomic, and
  has degree $d$ then

for some $c>0$.

The General Version and Jensen’s Formula

Definition. For $f \in \mathbb{C}[x_1,\dots,x_n]$, the Mahler measure is defined to be

where $d\mu_i=\frac{1}{2\pi}d\theta_i$, i.e., $d\mu_1\dots d\mu_n$ corresponds to the (completion of) Harr measure on $\mathbb{T}^n$ with total measure $1$.

We see through Jensen’s formula that when $n=1$ this coincides with what we have defined before. Observe first that $M(fg)=M(f)M(g)$. Consider $f(t)=a\prod_{i=1}^{d}(t-\alpha_i)$, then

On the other hand, as an exercise in complex analysis, one can show that

Combining them, we see

Taking the logarithm we also obtain Jensen’s formula

We first give a reasonable and useful estimation of $M(f)$, which will be used to prove the Northcott’s theorem.

Definition. For $f(t)=a_dt^d+\dots+a_0$, the $\ell_p$-norm of $f$ is naturally defined to be

For $p=\infty$, we have $\ell_\infty(f)=\max_j|a_j|$.

Lemma 1. Notation being above, $M(f) \le \ell_1(f)$ and

Proof. To begin with, we observe those obvious ones. First of all,

Therefore

Next, by Jensen’s inequality

However, by Parseval’s formula, the last term equals

For the remaining inequality, we use Vieta’s formula

and therefore

for all $0 \le r \le d$. Replacing $|a_{d-r}|$ with $\ell_\infty(f)$, we have finished the proof. $\square$

Before proving Northcott’s theorem, we show the connection between Mahler measure and heights.

Proposition 1. Let $\alpha \in \overline{\mathbb{Q}}$ and let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Z}$. Then

and

Proof. Put $d=\deg(\alpha)$ and write

Choose a number field $K$ that contains $\alpha$ and is a Galois extension of $\mathbb{Q}$, with Galois group $G$. Then $(\sigma\alpha:\sigma \in G)$ contains every conjugate of $\alpha$ exactly $[K:\mathbb{Q}]/d$ times. Since $a_0,\dots,a_d$ are coprime, for any non-Archimedean absolute value $v \in M_K$, we must have $\max_i|a_i|_v=|f|_v=1$. Combining with Gauss’s lemma and Galois theory, we see

Now we are ready to compute the height of $\alpha$ to rediscover the Mahler’s measure. Notice that

We therefore obtain

The last term corresponds to what we have computed above about non-Archimedean absolute values so we break it down a little bit:

for some $u \mid \infty$, according to the product formula. On the other hand, for $v \mid \infty$,

All in all,

The second assertion follows immediately because

Northcott’s Theorem

The set of non-zero algebraic integers of height $0$ lies on the unit circle, and they are actually roots of unit, by Kronecker’s theorem. However keep in mind that algebraic integers on the unit circle are not necessarily roots of units. See this short paper.

When it comes to algebraic integers of small heights, things may get complicated, but Northcott’s theorem assures that we will be studying a finite set.

Northcott’s Theorem. Given an integer $N>0$ and a real number $H \ge1$, there are only a finite number of algebraic integers $\alpha$ satisfying $\deg(\alpha) \le N$ and $h(\alpha) \le \log H$.

Proof. Let $\alpha$ be a algebraic integer of degree $d<N$ and height $h(\alpha) \le \log H$. Suppose $f(t)=a_dt^d+\dots+a_0 \in \mathbb{Z}[t]$ is the minimal polynomial of $\alpha$. Then lemma 1 shows us that

On the other hand, by proposition 1,

we have actually

This gives rise to no more than $(2\lfloor (2H)^d \rfloor+1)^{d+1}$ distinct polynomials $f$, which produces at most $d(2\lfloor (2H)^d \rfloor+1)^{d+1}<\infty$ algebraic integers. Ranging through all $d \le N$ we get what we want. $\square$

We also have the Northcott property, where we do not care about degrees. A set $L$ of algebraic integers is said to satisfy Northcott property if, for every $T>0$, the set

is finite. Such a set $L$ is said to satisfy Bogomolov property if, there exists $T>0$ such that the set

is empty. As a matter of elementary topology, Northcott property implies Bogomolov property. It would be quite interesting if $L$ is a field. This paper can be quite interesting.

References / Further Reading

• Erico Bombieri, Walter Gubler, Heights in Diophantine Geometry.

• Michel Waldschmidt, Diophantine Approximation on Linear Algebraic Groups, Transcendence Properties of the Exponential Function in Several Variables.

• Chris Smyth, THE MAHLER MEASURE OF ALGEBRAIC NUMBERS: A SURVEY.

Introduction

Let $F$ be a non-Archimedean local field, meaning that $F$ is complete under the metric induced by a non-Archimedean absolute value $|\cdot|$. Consider the ring of integers

and its unique prime (hence maximal) ideal

The residue field $k=\mathfrak{o}_F/\mathfrak{p}$ is finite because it is compact and discrete. For compactness notice that $\mathfrak{o}_F$ is compact, and the canonical projection $\mathfrak{o}_F \to k$ is open. For discreteness, notice that $\mathfrak{p}$ is open, connected and contains the unit.

Let $f \in \mathfrak{o}_F[x]$ be a polynomial. Hensel’s lemma states that, if $\overline{f} \in k[x]$, the reduction of $f$, has a simple root $a$ in $k$, then the root can be lifted to a root of $f$ in $\mathfrak{o}_F$ and hence $F$. This blog post is intended to offer a well-organised proof of this lemma.

To do this, we need to use Newton’s method of approximating roots of $f(x)=0$, something like

We know that $a_n \to \zeta$ where $f(\zeta)=0$ at a $A^{2^n}$ speed for some constant $A$, in calculus (do Walter Rudin’s exercise 5.25 of Principles of Mathematical Analysis if you are not familiar with it, I heartily recommend.). Now we will steal Newton’s method into number theory to find roots in a non-Archimedean field, which is violently different from $\mathbb{R}$, the playground of elementary calculus.

We will also use induction, in the form of which I would like to call “double induction”. Instead of claiming that $P(n)$ is true for all $n$, we claim that $P(n)$ and $Q(n)$ are true for all $n$. When proving $P(n+1)$, we may use $Q(n)$, and vice versa.

This method is inspired by this lecture note, where actually a “quadra induction” is used, and everything is proved altogether. Nevertheless, I would like to argue that, the quadra induction is too dense to expose the motivation and intuition of this proof. Therefore, we reduce the induction into two arguments and derive the rest with more reasonings.

Hensel’s Lemma

Hensel’s Lemma. Let $F$ be a non-Archimedean local field with ring of integers $\mathfrak{o}_F=\{\alpha \in F:|\alpha| \le 1\}$ and prime ideal $\mathfrak{p}=\{\alpha \in F:|\alpha|<1\}$. Let $f \in \mathfrak{o}_F[x]$ be a polynomial whose reduction $\overline{f} \in k[x]$ has a simple root $a \in k$, then $a$ can be lifted to $\alpha \equiv a \mod \mathfrak{p}$, such that $f(\alpha)=0$.

By simple root we mean $\overline{f}(a)=0$ but $\overline{f}’(a) \ne 0$. Before we prove this lemma, we see some examples.

Examples and Applications

Square Root of 2 in 7-adic Numbers

Put $F=\mathbb{Q}_7$. Then $\mathfrak{o}_F=\mathbb{Z}_7$, $\mathfrak{p}=7\mathbb{Z}_7$ and $k=\mathbb{F}_7$. We show that square roots of $2$ are in $F$. Note $\overline{f}(x)=x^2-2=(x-3)(x+3) \in k[x]$, we therefore two simple roots of $\overline{f}$, namely $3$ and $-3$. Lifting to $\mathfrak{o}_F$, we have two roots $\alpha_1 \equiv 3 \mod 7\mathbb{Z}_7$ and $\alpha_2 \equiv -3 \mod 7\mathbb{Z}_7$, of $f$. For $\alpha_1$, we have

Hence we can put $\alpha=\sqrt{2}=3+7+2\cdot 7^2+6\cdot 7^3\cdots\in\mathbb{Z}_7 \subset \mathbb{Q}_7$. Likewise $\alpha_2$ can be understood as $-\sqrt{2}$. This expansion is totally different from our understanding in $\mathbb{Q}$ or $\mathbb{R}$.

Roots of Unity

Since $k$ is a finite field, we see $k^\times$ is a cyclic group of order $q-1$ where $q=p^n=|k|$ for some prime $p$. It follows that $x^{q-1}=1$ for all $x \in k^\times$. Therefore $f(x)=x^{q-1}-1$ has $q-1$ distinct roots in $k$. By Hensel’s lemma, $F$ contains all $(q-1)$st roots of unity. It does not matter whether $F$ is isomorphic to $\mathbb{Q}_p$ or $\mathbb{F}_q((t))$.

Proof of Hensel’s Lemma (with Explanation)

Pick any $a_0 \in \mathfrak{o}_F$ that is a lift of $a\mod\mathfrak{p}$. Define

then we claim that $a_n$ converges to the root we are looking for.

Step 1 - Establishing A Sequence by Newton’s Method

First of all, we need to show that $a_n \in \mathfrak{o}_F$, i.e., $|a_n| \le 1$ for all $n$. It suffices to show that $|f(a_{n-1})/f’(a_{n-1})| \le 1$. We firstly observe the case when $n=1$.

Since $\overline{f}(a)=0$ but $\overline{f}’(a) \ne 0$, we have $f(a_0) \in \mathfrak{p}$ but $f’(a_0)\not\in\mathfrak{p}$. As a result, $|f(a_0)|<1$ but $|f’(a_0)|=1$. As a result, $|f(a_0)/f’(a_0)|<1$, which implies that $f(a_0)/f’(a_0) \in \mathfrak{o}_F$ and therefore $a_1 \in \mathfrak{o}_F$.

By Taylor’s theorem.

for some $g_n \in \mathfrak{o}_F[x]$. When $n=1$, we see $g_1(a_1) \in \mathfrak{o}_F$ and as a result $|g_1(a_1)| \le 1$. Therefore

Since $a_1 \in \mathfrak{o}_F$, we also see that $f(a_1) \in \mathfrak{o}_F$ hence its absolute value is not greater than $1$. As a result $|f(a_1)/f’(a_1)| \le 1$, which implies that $a_2 \in \mathfrak{o}_F$.

This inspires us to claim the following two statements:

(a) $|f(a_n)| < 1$ for all $n \ge 0$.

(b) $|f’(a_n)|=|f’(a_0)|=1$ for all $n \ge 0$.

We have verified (a) and (b) for $n=0$ and $n=1$. Now assume that (a) and (b) are true for $n-1$, then, for $n$, we will verify as follows.

First of all, by (a) and (b) for $n-1$, we see $a_n \in \mathfrak{o}_F$.

Consider the Taylor’s expansion

where $h_n \in \mathfrak{o}_F[x]$. It follows that $|h_n(a_n)| \le 1$. Since $|f’(a_{n-1})|=1$, by (b) we actually have

To prove (b) for $n$, we consider the Taylor’s expansion

Notice that since $a_n \in \mathfrak{o}_F$, we have $f’’(a_{n-1}),g_n(a_n) \in \mathfrak{o}_F$. By (a) and (b) for $n-1$, we see

Hence

bearing in mind that for a non-Archimedean absolute value, $|x+y|=\max\{|x|,|y|\}$ iff $|x| \ne |y|$. Through this process we have also proved (b).

Step 2 - Validating the Convergence

We need to show that $\{a_n\}$ is a Cauchy sequence. To do this, it suffices to show that $|f(a_n)| \to 0$ sufficiently quick. Recall in the proof of (a) we have shown that $|f(a_n)| \le |f(a_{n-1})|^2$ for all $n$. By applying this relation inductively, we see $|f(a_n)| \le |f(a_0)|^{2^n}$. Since $|f(a_0)|<1$, it follows that $|f(a_n)| \to 0$ as $n \to \infty$.

For any $\varepsilon>0$, there exists $N>0$ such that $|f(a_n)| <\varepsilon$ for all $n \ge N$. As a result, for all $m>n>N$, we have

Therefore $\{a_n\}$ is Cauchy. Since $F$ is complete, $a_n$ converges to some $\alpha \in \mathfrak{o}_F \subset F$ such that $f(\alpha)=\lim_{n \to \infty}f(a_n)=0$.

Step 3 - Validating the Congruence

In local fields, congruence is determined by inequality. In fact, we only need to show that $|\alpha-a_0|<1$, which means that $\alpha-a_0 \in \mathfrak{p}$, and therefore $\alpha \equiv a \mod \mathfrak{p}$ as expected. To do this, we show by induction that $|a_n-a_0|<1$. For $n=1$ we see $|a_1-a_0|=|f_0|<1$.

Suppose $|a_{n-1}-a_0|<1$ then

Therefore $|\alpha-a_0|=\lim_{n \to \infty}|a_n-a_0|<1$, from which the result follows. $\square$

Stronger Version

In fact we have not explicitly used the fact that $a$ is a simple root. We only used the fact that $|f(a_0)|<1$ but $|f’(a_0)|=1$. Moreover, what really matters here is that $|f(a_n)|$ converges to $0$ quick enough. Therefore $1$ may be replaced by a smaller constant. For this reason we introduce a stronger version of Hensel’s lemma.

Hensel’s lemma, stronger version. Let $F$ be a non-Archimedean local field with ring of integers $\mathfrak{o}_F$. Suppose there exists $a \in \mathfrak{o}_F$ such that $|f(a)|<|f’(a)|^2$, then there exists some $b \in \mathfrak{o}_F$ such that $f(b)=0$ and $|b-a|<|f’(a)|$.

Instead of asserting $|f’(a_n)|=1$ for all $n$, we claim that $|f’(a_n)|=|f’(a_0)|$ (as it should be!). Instead of asserting $|f(a_n)|<1$, we claim that $|f(a_n)| \le \lambda^{2^n}|f’(a_0)|$ where $\lambda=|f(a_0)|/|f’(a_0)|^2$. The proof will be nearly the same.

For example, we can find a square root of $257$ in $\mathbb{Z}_2 \subset \mathbb{Q}_2$. The polynomial $f(x)=x^2-257$ is reduced to $\overline{f}(x)=x^2-1=(x-1)^2$ in $\mathbb{F}_2[x]$, where $1$ is not a simple root. Therefore we cannot apply the original version of Hensel’s lemma to this polynomial. Nevertheless, we see $f(1)=-256$ and $f’(1)=2$. Therefore $|f(1)|=\frac{1}{2^8}$ while $|f’(1)|=\frac{1}{2}$. We can apply Newton’s method here to find a square root of $257$ without worrying about repeated roots.

Ending Remarks

There are a lot of variants of Hensel’s lemma, for example you can do exercise 10.9 of Atiyah-MacDonald. In fact, we later even have Henselian ring and Henselisation of a ring.

There are some other proofs of Hensel’s lemma in this post, for example, since Newton’s method can also be understood as a contraction mapping, we can also prove it using properties of contraction mapping (see K. Conrad’s note).

Introduction

In the previous post we are convinced that the Galois group of a separable irreducible polynomial $f$ can be realised as a subgroup of the symmetric group, the elements of which permute the roots of $f$. We worked on cubic polynomials over a field with characteristic not equal to $2$ and $3$, and this definitely works with $\mathbb{Q}$. In this post we go one step further.

Let $f \in \mathbb{Q}[X]$ be an irreducible polynomial of prime degree $p$. Since it is also separable (see lemma 9.12.1 on the stack project), we can safely work on its Galois group $G$. One immediately wants to question the position of $\mathfrak{S}_p$. Indeed we have $G \subset \mathfrak{S}_p$. The question is, when does the equality hold? It is not likely to have an immediate answer. However, we have some interesting sufficient conditions, which will be discussed in this post.

Generators of the Symmetric Group

We present some handy results in finite group theory that will be used in the main result. One may skip this section until needed. I will collapse the proof in case one wants to treat it as an exercise.

Lemma 1. Let $p$ be a prime number. The symmetric group $\mathfrak{S}_p$ is generated by $[12 \cdots p]$ and an arbitrary transposition $[rs]$.

Proof. We prove this by presenting several sets of generators of $\mathfrak{S}_n$ where $n$ is a positive integer.

1. It is generated by cycles. This is a really, really routine verification and sometimes this is assumed as a fact.

2. It is generated by transpositions, i.e., $2$-cycles. It suffices to show that a cycle is a product of transpositions. Indeed, for any cycle $[i_1\dots i_k]$ in $\mathfrak{S}_n$, we have $[i_1\cdots i_k]=[i_1i_2][i_2i_3]\cdots[i_{k-1}i_k]$. This proves our statement.

3. It is generated by translations of the form $[1k]$. It suffices to show that a transposition is generated as such. For any transposition $[rs]$, we have $[rs]=[1r][1s][1r]$.

4. It is generated by adjacent translations, i.e. the generators can be of the form $[k-1 ,k]$. This follows from the following identity:

1. It is generated by two elements: $\sigma=[12]$ and $\tau=[12\cdots n]$. This follows from the following identity:

Now, back to the case when $n=p$ is prime. Put $\sigma=[rs]$ and $\tau=[12\cdots p]$. If $s-r=1$ then it is already proved in 5 by several conjugations. Therefore we may assume that $d=s-r>1$. From now on integers may be a number in either $\mathbb{Z}$ or $\mathbf{F}_p=\mathbb{Z}/p\mathbb{Z}$, depending on the context. Recall that $\mathbf{F}_p$ is a field. Pick the integer $w$ such that $dw=1$ in $\mathbf{F}_p$. By conjugation we see $\tau$ and $\sigma$ generate

The product of elements above is $[1,1+wd]=[12]$. Therefore we are still back to 5. $\square$

Computing the Galois Group

We have many good reasons to study the Galois group of something. It would be great if the group can be written down explicitly. In this section we show that the group can be revealed by the number of nonreal roots.

The Simplest Case

Proposition 1. Let $f(X) \in \mathbb{Q}[X]$ be an irreducible polynomial of prime degree. If $f$ has precisely two nonreal roots, then the Galois group $G$ over $\mathbb{Q}$ is $\mathfrak{S}_p$.

Proof. Let $L$ be the splitting field of $f$. It suffices to show that $G$ contains a transposition and a $p$-cycle, which is $[12\cdots p]$. By the Sylow’s theorem, $G$ has a subgroup $H$ of order $p$, which can only be cyclic. Say $H=\langle \sigma \rangle$. Suppose $\sigma$ is of cycle type $(k_1,\dots,k_r)$. Then the period of $\sigma$, which equals $p$, is the least common multiple of $k_1,\dots,k_r$, where $k_1+\dots+k_r=p$. This can only happen when $r=1$ and $k_1=p$. Therefore $\sigma$ is a $p$-cycle.

In fact, $\sigma$ can be considered as $[12\dots p]$. Suppose the order of roots of $f$ is given, for which we have $\sigma=[i_1 i_2 \dots i_p]$. Then If we re-order these roots, by putting the $k$th root to be the original $i_k$th root, then we can write $\sigma=[12\dots p]$. (This re-ordering is, in fact, a conjugation.)

It remains to prove that $G$ contains a transposition. Let $\alpha$ and $\beta$ be two nonreal roots of $f$. Since $\overline{\alpha}$ is also a root of $f$ (because coefficients of $f$ are real; if $\sum_{n=0}^{p}a_n\alpha^n=0$, then $\sum_{n=0}^{p}a_n\overline{\alpha}^n=\sum_{n=0}^{p}\overline{a_n\alpha^n}=\overline{0}=0$) we see $\beta=\overline{\alpha}$. Therefore complex conjugation over $\mathbb{Q}(\alpha)$ extends to $L$ as an element of order $2$, which is a transposition in $G$. This proves our assertion. $\square$

For example, consider the polynomial

With calculus one can show that it has exactly three roots, hence it has two nonreal roots. Eisenstein’s criterion shows that $f$ is irreducible. Therefore we are allowed to use proposition 1. The Galois group of $f$ is $\mathfrak{S}_5$.

This also works fine when $p=2$ or $3$. The case when $p=2$ is nothing but working around a quadratic polynomial. When $f(X)$ is irreducible of degree $3$, and it has two nonreal roots, we also know that it has an irrational root. Let the roots be $a+bi,a-bi,c$ where $b \ne 0$ and $c$ is irrational. We see

Therefore the Galois group is $\mathfrak{S}_3$.

“Linear” Generalisation

It is way too ambitious to restrict ourselves in one single pair of roots. Also, it seems we have ignored the alternating group $\mathfrak{A}_p$ for no reason. Oz Ben-Shimol gave us a nice way to work around this (see arXiv:0709.2868). The whole paper is not easy but the result is pretty beautiful and generalised what we said above as $p \ge 5$.

Proposition 2. Let $f \in \mathbb{Q}[X]$ be an irreducible polynomial of prime degree $p \ge 5$. Suppose that $f$ has $k>0$ pairs of nonreal roots. If $p \ge 4k+1$, then the Galois group $G$ is isomorphic to $\mathfrak{A}_p$ or $\mathfrak{S}_p$. If $k$ is odd then $G \cong \mathfrak{S}_p$.

The proof is done by showing that $\mathfrak{A}_p \subset G \subset \mathfrak{S}_p$. As the index of $\mathfrak{A}_p$ is $2$, $G$ can only be one of them. The solvability of $G$ is also concerned here.

Indeed, what we have proved in “the simplest case” is nothing but $k=1$. When $p \ge 5$ we clearly have $p \ge 1+4 \times 1$. This refined the result of A. Bialostocki and T. Shaska (see arXiv:math/0601397), and the inequality used to be

When $k$ is big enough, we have $k(k\log{k}+2\log{k}+3) \ge 4k+1$. Oz Ben-Shimol’s result is a refinement because it is saying, $p$ does not need to that big. He also offered a refined algorithm to compute the Galois group, which we will present below. Also, computing $4k+1$ is much easier than computing $k^2\log{k}$ plus something.

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Input: An irreducible polynomial f(X) over Q with prime degree p >= 5
Output: The Galois group Gal(f/Q)
begin
    r:=NumberOfRealRoots(f(X))
    k:=(p-r)/2
    if k>0 and p>=4k+1 then
        if k is odd then
            Gal(f/Q)=S_p;
        else
            if ∆(f) is a complete square then
                Gal(f/Q)=A_p;
            else
                Gal(f/Q)=S_p;
            endif;
        endif;
    else
        ReductionMethod(f(X));
    endif;
end;

Here, $\Delta(f)$ is the discriminant of $f$. We have seen that whether $\Delta$ is a perfect square matters a lot. The discussion of ReductionMethod can be trailed in Oz Ben-Shimol’s paper.

【阅读时间】20min 4589 words
【内容简介】从【直观理解】线性代数的本质笔记出发，继续讨论几个线性代数中的概念，正交，正规，正定及转置的直观解释。旨在能帮助读者在看完后不会忘记什么是正交矩阵，什么是正规矩阵，转置部分进行了深入挖掘，希望找出一些几何直观的解释

在之前的【直观理解】线性代数的本质的笔记中，详细讨论了特征值与特征向量的几何直观意义

起初，研究线性代数，也是因为深入了解矩阵（变换）对机器学习中的很多优美公式的推导和理解有帮助。上篇笔记中，3B1B团队的讲解内容中没有涉及几个线性代数中的概念，且这些概念在做矩阵分解时会被用到。以上一篇笔记中的直观理解为基础（矩阵 = 变换）在这里做一个整理和记录

正交矩阵

可能很多人已经有一个概念：正交（Orthogonal） = 垂直。但我们知道，正交的一定垂直，垂直的不一定正交（比如空间中两个不相交直线垂直）。提及垂直，首先出现你脑海中的特点是什么呢？我想是勾股数 $a^2 + b^2 = c^2$ ， 还有 $\cos (\frac {\pi}{2}) = 1$

那什么是正交矩阵呢？在讲这个概念之前，变换中有一种特殊变换：旋转变换。这种变换除了原点外没有特征向量，特征值恒为1，不对网格进行伸缩。或者说，这个变换保证了新列空间内和原列空间内所有对应向量的长度不变

三维情况下，单位矩阵（对角线为1，其他为0，即基向量构成的矩阵）$\mathbf E = \left [ \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0&0&1 \end{smallmatrix} \right ]$ 如下图所示

根据基变换原理，易得旋转变换的矩阵表达式 $\mathbf R = \left [ \begin{smallmatrix} 0&-0.8 &-0.6 \\ 0.8&-0.36&0.48 \\ 0.6&0.48&-0.64 \end{smallmatrix} \right ]$ 计算得特征向量为 $(0,-0.75,1)$，发现这条向量即旋转轴！

此时我们考虑从 $\mathbf R$矩阵下变到 $\mathbf E$的变换矩阵是多少，即求 $\mathbf R$ 矩阵的逆

观察形式大家就可以发现一个有趣的特点 $\mathbf R^{-1} = \mathbf R^T$

正交矩阵有一个几何直观的特点，表示一个旋转变换，并且矩阵的逆和矩阵的转置相等

正定与半正定矩阵

根据特征值和特征向量这篇笔记中的内容，我们知道特征值是对一个变换（矩阵）特性的有力表征，公式 $\mathbf A \mathbf{\vec v} = \lambda \mathbf{\vec v}$ 表示了变换中被留在张成空间内的向量就是特征向量的符号表达，其中 $\vec v$ 是特征向量，$\lambda$ 即特征值

我们对上式进行一些数学恒等变换，左乘 $\vec v^T$，得到

此时我们会发现一些巧合，先来看看正定矩阵的正规定义：若一个 n×n的矩阵 $\mathbf M$ 是正定的，当且仅当队友所有的非零实系数的向量 $\vec v$，都有 $\vec v^T \mathbf M \vec v > 0$

我们暂时不考虑复数情况（在机器学习预见复数域的内容较少），结合上面的二公式，发现保证 $\vec v^T \mathbf M \vec v > 0$ 即使得 $\lambda \vec v^T \vec v>0$，其中 $\vec v^T \vec v$一定大于等于0（由于 $\vec v$ 是一个1×n的向量，转置进行矩阵相乘实际效果计算元素的平方和），所以可以推出即正定矩阵就是使得特征值大于0

再回到正定矩阵的定义公式 $\vec x^T \mathbf M \vec x > 0$，我们已经有深刻的理解 $\mathbf M \vec x$ 表示对向量 $\vec x$ 进行变换，记变换后的向量为 $\vec y = \mathbf M \vec x$ ，则我们可以把正定矩阵的公式写成

这个公式是不是很熟悉呢？它是两个向量的内积，对于内积，有公式：

$\Vert \vec x \Vert \; \Vert \vec y \Vert$ 表示 $\vec x$ 和 $\vec y$的长度，$\theta$ 是它们之间的夹角。根据2-2式，可以得到 $\cos(\theta) > 0$，即它们之间的夹角小于90度

总结：如果说一个矩阵正定，则表示，一个向量经过此矩阵变换后的向量与原向量夹角小于90度

当然，加一个【半】字，是指这个小于变成小于等于

正规矩阵

矩阵中还有一张形状特殊的矩阵，被称为正规矩阵，定义为：如果矩阵 $\mathbf A$ 满足 $\mathbf A^T \mathbf A = \mathbf A \mathbf A^T$

更多的，如果矩阵 $\mathbf U$ 满足 $\mathbf U^T \mathbf U = \mathbf U \mathbf U^T = \mathbf I$，其中 $\mathbf I$ 是单位矩阵，则称矩阵 $\mathbf U$ 为酉矩阵

从变换的角度来看正规矩阵，先做一个变换 $\mathbf A$ 再做一个变换 $\mathbf A^T$。并且交换两个矩阵的位置，最终结果相同

矩阵的转置

什么是转置

在前面的三个描绘矩阵不同矩阵的概念中，多次使用了转置的概念。从矩阵形态的角度看，转置是将 $\mathbf A$ 的所有元素关于一条从第1行第1列元素出发的向右下方45度的射线作镜面翻转（下面的动图更加直观）

那么，从矩阵是表示变换的集合角度如何理解转置呢？

为什么转置

试图从另一个角度来理解其实也是为了回答另一个问题：为什么要定义转置这种操作呢？你可能会说，这就是一个【对角线镜像对称交换的操作】，从形式上来理解对一般人已完全足够。

这里要深究的原因也只是为了克服在学习机器学习的过程中，公式里若出现转置符号，无法完全理解带来的生涩感（俗称强迫症），对博主来说，一个直接动机源于SVD算法

首先，考虑矩阵的列向量有具体的列空间的含义（对应 $\hat {\imath}$ 和 $\hat {\jmath}$ 的变换位置），若进行转置操作，列空间的性质会被完全破坏，或者说，转换成了一个新的列空间

非方阵

考虑矩阵转置的几何含义是无意义的，或者说，对出现过矩阵转置的公式的进一步理解是没有帮助的

特别的，如果是向量形式（1×n的矩阵），转置很多时候出现，是为了进行二次型运算（即平方运算），设 $\mathbf x = \{x_1, x_2,\ldots, x_n\}$ 是一个1×n的矩阵

很多机器学习的教材中这里会是列矩阵，因为要切合列空间的概念。对于机器学习来说，这里的 $x_1$ 代表的数据的特征维度

计算二次型： $\mathbf x \mathbf x^T = x_1^2 + x_2^2 + \ldots + x_n^2$ ，记过为一个数，表示的是距离

方阵

从列空间的概念，转置是一种非常特殊的旋转。这种旋转结合了横向镜面等特性，详细可以参看下图

从这幅图可以看出，如果想从几何变换的角度（类似3b1b的方法）来理解转置相当困难。此时，需要考虑换一种思考角度，性质和数学计算（这可能也是解决一个数学问题最常用的两种手段，性质寻找共性，并推广演绎，数学计算导出同样问题的不同表现形式，并总结规律）

性质找规律

首先先用一个简单的例子，二维可视化，并寻找规律，得到下面图像

观察得到，和旋转有一定的关系的，但是其实这个几何意义已经十分抽象了，为了追根之底，从数学的角度进行一些更有意思的探究

数学计算

为了发掘 $\mathbf A$ 和 $\mathbf A^T$ 之间的关系，我们可以设有转换矩阵 $\mathbf T$，其描述了从 $\mathbf A$ 到 $\mathbf A^T$ 的变换，写成公式为：
$$
\mathbf T \mathbf A = \mathbf A^{T} \tag 1
$$
同时我们假设 $\mathbf A$ 是一个2×2的矩阵，如下所示所示（列的不同为字母的不同a和b，行的不同为小标的1和2）

保证变换不会压缩维度，即 $det(\mathbf A) \neq 0$，利用（1）式，未知数，消元计算后，可以算出 $\mathbf T$，其中 $det(\mathbf A) = a_1b_2 - b_1a_2$

前面的 $\frac{1}{det(\mathbf A)}$ 作为一个常数，保证了 $det(\mathbf A^{T})=det(\mathbf A)$，将后面关键的变换矩阵写成

把 （4）式进行整理，也写成矩阵相乘的形式，得到下式

对于 $\mathbf A$ 的列空间来说，有 $\mathbf a = \left [ \begin{smallmatrix} a_1 \\ a_2 \end{smallmatrix} \right ]$ 和 $\mathbf b = \left [ \begin{smallmatrix} b_1 \\ b_2 \end{smallmatrix} \right ]$ 观察到 $\mathbf T$ 中包含列空间的两个项，把 $\mathbf T’$ 整理得

令 $\mathbf c = \begin{bmatrix}b_{2} \\ -a_{2} \\ \end{bmatrix} $ $\mathbf d = \begin{bmatrix}-b_{1} \\ a_{1} \\ \end{bmatrix} $ ，观察后发现规律：

• $\mathbf c$ 变换到 $\mathbf A^T_{\hat {\jmath}}$ 为逆时针旋转90°
• $\mathbf d$ 变换到 $\mathbf A^T_{\hat {\imath}}$ 为顺时针旋转90°

$\mathbf c$ 和 $\mathbf d$ 组成的列空间设为 $\mathbf C$，写成公式为

将（7）式左乘 $\mathbf A$ 得到下式

再将（8）式所有乘以 $\mathbf A$ 的逆矩阵 $\mathbf A^{-1}$ 得到
$$
\mathbf C = \mathbf A^{-1} det(\mathbf A) \tag{9}
$$
所以，构造的这个 $\mathbf C$ 矩阵和 $\mathbf A$ 矩阵的逆矩阵有关

另外，（4）式，即 $\mathbf T’$ 矩阵还可以被写成

或者可以写成

由 $\mathbf T \mathbf A = \mathbf A^T$ 可得

这时候，可以得出一个结论，矩阵的转置的过程和矩阵的逆是有关系的，是矩阵逆的一个更加复杂的表现形式

有了这个作为基础，考虑一下具有对称性，构造 $\mathbf A \mathbf A^T$，这个复合变换有着很好的对称性和分解性（接下来为了方便，默认非粗体表示的矩阵变换），因为

考虑任何不进行压缩维度变换的矩阵都可以进行特征分解，则有

之后左乘（这里补充一下，所有的左乘操作在几何意义上来说，就是附加了一个新的变换）$A^{-1}$，得到

根据之前（1）式的例子进行计算，发现 $R_{AA^{T}} = (R^{-1})_{AA^{T}}$，并且 $R_{AA^T}$ 首先将空间关于 y 轴对称，之后旋转 $\alpha$ 角度，所以假设定义

利用上面的推导，带入（15）式，得到：

用更加清晰的符号改写（17）式得到

$M_y$ 表示的是关于y轴对称
$R_\alpha$ 表示逆时针旋转 $\alpha$ 度
$\Lambda$ 表示一种伸缩变换

这里的（18式）结论和上面的作图规律，还有（12）式从某种程度上来说有相似的地方

其实应该从矩阵分解的部分来再次思索矩阵转置的意义，可详见什么是PCA,SVD

最终感觉关于转置还是研究的不够透彻，可能需要拔高到另一个层面去理解会更加直观，但是限于水平，只能至此（并不是数学专业的学生）

总结

• $\mathbf A \mathbf A^T$ 的对称特性非常有用
• 正交 = 旋转 = 垂直，并且逆等于转置
• 正规矩阵的定义和转置息息相关，但是从形式上来看，约束条件是逐步变弱的，其实这些特征都是描述了空间一个变换的一些变化模式，比如旋转，伸缩的特殊模式，只需要有一个基本的直观理解，在机器学习领域就已经足够用了