Galois Theory (I)

Galois Theory (I)

Review: Field Theory

Roots of Polynomials

Let $f$ be an irreducible polynomial in $F[x]$.

$f$ has no multiple roots in any extension of $F$, unless $f'=0$.
If $F$ has characteristic $0$, then $f$ has no multiple roots in any extension of $F$.

Heuristically, we write a polynomial with multiple roots as \[f(x) = g(x)(x-\alpha)^2.\] Then \[f'(x) = g'(x)(x-\alpha)^2 + 2g(x)(x-\alpha)\] has a common factor $(x-\alpha)$ with $f(x)$.

Primitive Element Theorem

Let $K$ be an extension of $F$. If the extension $K/F$ is generated by a single element $\alpha$, then $\alpha$ is called the primitive element of $K/F$. We have the following theorem:

Primitive Element Theorem. If $F$ has characteristic $0$, then any extension $K/F$ has a primitive element.

The extension $K=\mathbb{Q}\qty(\sqrt{2},\sqrt{3})$ is of degree $4$. We could easily prove that $K=\mathbb{Q}\qty(\sqrt{2}+\sqrt{3})$, i.e. the extension has a primitive element $\alpha = \sqrt{2} + \sqrt{3}{}$.

We may verify that \[\sqrt{2} = \frac{1}{2}\qty(\alpha^3-9\alpha)\] and that \[\sqrt{3} = -\frac{1}{2}\qty(\alpha^3-11\alpha).\] Such decomposition exists because $\alpha^i$ could always be written as linear combinations of $\sqrt{2}{}$, $\sqrt{3}{}$ and $\sqrt{6}{}$. By solving a linear equation we retrieve $\sqrt{2}{}$ and $\sqrt{3}{}$ from powers of $\alpha$.

Symmetries of Polynomials

Symmetric Polynomials

$S_n$ acts on $R[u]$ by \[f=f(u_1,\cdots,u_n)\mapsto f(u_{\sigma 1},\cdots,u_{\sigma n}) = \sigma f.\]

This action is straightforward for someone who has experience with quantum mechanics. By regarding $f$ as the wavefunction of a many-body system, $\sigma f$ becomes the wavefunction after permutation of particles.

A polynomial $g$ is symmetric if $g$ is invariant under any permutation, i.e. if all monomials in the same orbit have the same coefficient.

Theorem. Any symmetric polynomial could be written (uniquely) as a linear combination of elementary symmetric polynomials.

We define the elementary symmetric polynomials by \begin{align*} s_1 &= u_1 + u_2 + \cdots + u_n, \\ s_2 &= u_1u_2 + u_1u_3 + \cdots, \\ s_3 &= u_1u_2u_3 + \cdots, \\ s_n &= u1 u_2 \cdots u_n. \end{align*}

We have \[u_1^2 + \cdots + u_n^2 = s_1^2 - 2s_2.\]

Corollary. If $f(x) = x^n - a_1x^{n-1} + \cdots + a_n$ has coefficients in $F$ and splits over $K$ with roots $\alpha_1$, $\cdots$, $\alpha_n$. Let $g\qty(u_1,\cdots,u_n)$ be a symmetric polynomial with coefficients in $F$, then $g(\alpha_1,\cdots,\alpha_n)$ is in $F$.

From the Vieta's formulae we know that the roots $x_1$ and $x_2$ of \[x^2 - bx + c = 0\] satisfies \[x_1 + x_2 = b,\quad x_1x_2 = c\] even if the roots are in $\mathbb{R}\backslash \mathbb{Q}$ while the coefficients are in $\mathbb{Q}$. Moreover, symmetric polynomials like \[x_1^2 + x_2^2 = (x_1+x_2)^2 - 2x_1x_2\] evaluate to values in $\mathbb{Q}$ as well.

Proposition. Let $h$ be a symmetric polynomial and $\qty{p_1,\cdots,p_k}$ be the orbit of $p_1$ under $S_n$, then \[h(p_1,\cdots,p_k)\] is a symmetric polynomial.

For the simplest case we take \[h = p_1 + \cdots + p_k,\] i.e. the sum of every element in the orbit.

Discriminant

Let the roots of \[P(x) = x^n - s_1 x^{n-1} + s_2 x^{n-2} - \cdots \pm s_n\] be $\qty{u_1,\cdots,u_n}$. The discriminant of $P$ is defined by \[D(u) = (u_1 - u_2)^2 (u_1 - u_3)^2 \cdots (u_{n-1} - u_n)^2 = \prod_{i<j} (u_i - u_j)^2.\]

$D(u)$ is a symmetric polynomial with integer coefficients.
- In particular, if the coefficients are in $F$, then $D(u)$ evaluates to an element in $F$.
If $\qty{\alpha_i}$ are elements in a field, then $D(u) = 0$ if and only if $P$ has a multiple root.

We are able to write $D$ as a sum of elementary symmetric polynomials.

In the case of degree 2, we have \[D = (u_1 - u_2)^2 = s_1^2 - 4s_2.\]

Fields and Polynomials

We assume that $F$ has characteristic $0$ hereinafter.

Splitting Field

$f$ splits over $K$ if \[f(x) = (x-\alpha_1) \cdots (x-\alpha_n)\] in $K$.

Let $f$ be a polynomial over $F$. The splitting field of $f$ is an extension $K/F$ such that

$f$ spilits in $K$; and
$K$ is generated by the roots, i.e. \[K = F\qty(\alpha_1,\cdots,\alpha_n).\]

Lemma. Let $f$ be a polynomial over $F$.

Let $F\subset L\subset K$ be a field, and $K$ be the splitting field of $f$, then $K$ is also the splitting field of $f$ as a polynomial over $L$.
Every $f$ over $F$ has a splitting field.
The splitting field is a finite extension of $F$. Every finite extension of $F$ is contained in a splitting field.

Splitting Field Theorem. Let $K$ be the splitting field of $f$ as a polynomial over $F$. Let $g$ be an irreducible polynomial over $F$. If $g$ has a root in $K$, then $g$ split over $K$.

Outline of Proof. Let $K$ be generated by $\alpha_1$,$\cdots$,$\alpha_n$, and $\beta_1$ be a root of $G$ in $K$. Let \[p_1(\alpha) = \beta_1,\] and $\qty{p_1,\cdots,p_k}$ be the orbit of $p_1$ under $S_n$ and \[\beta_k = p_k(\alpha).\] We may prove that the coefficients of \[h(x) = (x-\beta_1)\cdots (x-\beta_k)\] are in $F$. Therefore, $g$ divides $h$ since $h$ has $\beta_1$ as a root and $g$ splits since $h$ splits.

Isomorphism of Field Extension

Let $K$ and $K'$ be extensions of $F$. An $F$-isomorphism is a field isomorphism between $K$ and $K'$ that fixes every element in $F$.
- An $F$-automorphism of $K$ is an $F$-isomorphism from $K$ to itself, i.e. the symmetries of the extension $K/F$.
The $F$-automorphisms of a finite extensions $K/F$ form a group $G(K/F)$, called the Galois group.
A finite extension $K/F$ is called a Galois extension if \[\abs{G(K/F)} = \qty[K:F].\]

For any quadratic extension $K/F$, e.g. $\mathbb{C}/\mathbb{R}$ and $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$, the Galois group has two elements: the trivial one, and the conjugation, i.e. \[a+bi \mapsto a-bi\] or \[a+b\sqrt{2} \mapsto a-b\sqrt{2}.\]

Let $F=\mathbb{Q}$ and $K=F[\sqrt[3]{2}]$. Since $x^3-2=0$ has two more roots which are complex and do not belong to $K$, we have \[G(K/F) = \qty{e}\] while $[K:F]=3$. Therefore, the extension is not Galois.

Lemma. Let $K$ and $K'$ be extensions of $F$.

Let $f(x)$ be a polynomial with coefficients in $F$, and $\sigma$ be an $F$-isomorphism from $K$ to $K'$. If $\alpha$ is a root of $f$ in $K$, then $\sigma(\alpha)$ is a root of $f$ in $K'$.
Let $K=F(\alpha_1,\cdots,\alpha_n)$, and $\sigma$ and $\sigma'$ be $F$-isomorphisms from $K$ to $K'$. If for $i=1,\cdots,n$ we have $\sigma(\alpha_i) = \sigma'(\alpha_i)$, then $\sigma = \sigma'$. If $\sigma$ fixes every $\alpha_i$, then $\sigma$ is the identity mapping.
If $f$ is an irreducible polynomial over $F$, and $\alpha$ and $\alpha'$ are roots of $f$ in $K$ and $K'$ respectively, then there exists a unique $F$-isomorphism $\sigma$ that maps $\alpha$ to $\alpha'$. If $F(\alpha) = F(\alpha')$ then $F$ is the identity mapping.

Proposition. Let $f$ be a polynomial over $F$.

An extension $L/F$ contains at most one splitting field of $f$ over $F$.
Splitting fields of $f$ over $F$ are all isomorphic to each other.

Fixed Field

Let $H$ be an automorphism group of $K$. The fixed field of $H$ consists of all the elements invariant under every elements of $H$, \[K^H = \Set{\alpha\in K|\sigma(\alpha)=\alpha,\ \text{for all } \sigma \in H}.\]

Theorem. Let $H$ be a finite automorphism group of $K$, and $F$ be the fixed field $K^H$. Let $\beta_1$ be an element of $K$ and $\qty{\beta_1,\cdots,\beta_r}$ be the $H$-orbit of $\beta_1$.

The irreducible polynomial of $\beta_1$ over $F$ is given by \[g(x) = (x-\beta_1)\cdots (x-\beta_r).\]
$\beta_1$ is algebraic over $F$, and the degree thereof is the cardinality of the $H$-orbit thereof.
- In particular, degree of $\beta_1$ divides the order of $H$.

Let $K=\mathbb{C}$ and $H = \qty{e, c}$ where $c$ denotes conjugation. Then $F = \mathbb{R}$ and the irreducible polynomial of $\beta \in \mathbb{C}$ is given by

$g(x) = x-\beta$ if $\beta \in \mathbb{R}$, or

$g(x) = (x-\beta)(x-\overline{\beta})$ if $\beta \in \mathbb{C}\backslash \mathbb{R}$.

Lemma. Let $K$ be an algebraic extension of $F$ and not be a finite extension. Then there exists elements in $K$ of arbitrary large degrees over $F$.

Fixed Field Theorem. Let $H$ be a finite automorphism group of $K$ and $F=K^H$ be the fixed field thereof. Then $K$ is a finite extension over $F$, the degree $[K:H]$ of which equals $\abs{H}$. \[[K:K^H] = \abs{H}.\]

Let $K=\mathbb{C}(t)$, and $\sigma$ and $\tau$ be $\mathbb{C}$-automorphisms of $K$, such that \[\sigma(t) = it\] and \[\tau(t) = t^{-1}.\] We have \[\sigma^4 = 1,\quad \tau^2 = 1,\quad \tau\sigma = \sigma^{-1}\tau.\] Therefore, $\sigma$ and $\tau$ generate the $D_4$ group.

We may easily verify that $u = t^4 + t^{-4}$ is transcendental over $\mathbb{C}$. Now we prove that $\mathbb{C}(u) = K^H$.

Since $u$ is fixed by $\sigma$ and $\tau$ we have $\mathbb{C}(u)\in K^H$. The $H$-orbits of $t$ is given by \[\qty{t,it,-t,-it,t^{-1},-it^{-1},-t^{-1},it^{-1}}.\] From the theorem above we obtain the irreducible polynomial of $t$ over $\mathbb{C}(u)$: \[x^8 - ux^4 + 1.\]

Therefore, $[K:\mathbb{C}(u)] \le 8$.

From the fixed field theorem we know $[K:K^H] = 8$.

We have already proved that $\mathbb{C}(u) \subset K^H$.

Thus we conclude that $\mathbb{C}(u) = K^H$.

The example above illustrates the Lüroth theorem: Let $F \subset \mathbb{C}(t)$ be a field that contains $\mathbb{C}$ and not be $\mathbb{C}$ itself. Then $F$ is isomorphic to $\mathbb{C}(u)$.

Fundamentals of Galois Theory

Galois Extension

Lemma.

Let $K/F$ be a finite extension. Then $G(K/F)$ is a fintie group and \[\abs{G(K/F)}\mid [K:F].\]
Let $H$ be a finite automorphism group of $K$. Then $K$ is a galois extension of $K^H$, and $H=G(K/K^H)$.

Lemma. Let $\gamma_1$ be the primitive element of a finite extension $K/F$ and $f(x)$ be the irreducible polynomial of $\gamma_1$ over $F$. Let $\gamma_1,\cdots,\gamma_r$ be roots of $f$ in $K$. Then there exists a unique $F$-automorphism $\sigma_i$ of $K$ such that $\sigma_i(\gamma_1) = \gamma_i$. These are all the $F$-automorphisms. Therefore, $\abs{G(K/F)}=r$.

Characteristics of Galois Extension. Let $K/F$ be a finite extension and $G$ be the Galois group thereof.Then the following statements are equivalent:

$K/F$ is a Galois extension, i.e. $\abs{G} = [K:F]$.
$K^G=F$.
$K$ is a splitting field over $F$.

If $K$ is a splitting field of $f$ over $F$, we may refer to $G(K/F)$ as the Galois group of $f$.

Corollary.

Every finite extension $K/F$ is contained in a Galois extension.
If $K/F$ is a Galois extension, and $L$ is an intermediate field, then $K/L$ is also a Galois extension, and $G(K/L)$ is a subgroup of $G(K/F)$.

Theorem. Let $K/F$ be a Galois extension and $G$ be its Galois group, and $g$ be a polynomial over $F$ that splits in $K$ with roots $\beta_1,\cdots,\beta_r$.

$G$ acts on $\qty{\beta_i}$.
If $K$ is the splitting field of $G$ over $F$, then the permutation of $G$ on $\qty{\beta_i}$ is faithful and $G$ is a subgroup of $S_r$.
If $g$ is irreducible over $F$, then the action of $G$ on $\qty{\beta_i}$ is transitive.
If $K$ is the splitting field of $g$ over $F$ and that $g$ is irreducible over $F$, then $G$ is a transitive subgroup of $S_r$.

The Fundamental Theorem

The Fundamental Theorem. Let $K$ be a Galois extension over $F$, and $G$ be the Galois group thereof. Then there is a one-to-one correspondence between \[\qty{\text{subgroup } H \text{ of } G}\] and \[\qty{\text{intermediate field } L \text{ of } K/F}.\] The mapping \[H\mapsto K^H\] is the inverse of \[L\mapsto G(K/L).\]

Corollary.

If $L\subset L'$ are intermediate fields, and the corresponding subgroups are $H$ and $H'$, then $H\supset H'$.
The subgroup corresponding to $F$ is $G(K/F)$, while to $G$ is the trivial group $\qty{e}$.
If $L$ corresponds to $H$, then $[K:L]=H$ and $[L:F]=[G:H]$.

Corollary. A finite extension $K/F$ has finitely many intermediate fields $F\subset L\subset K$.

Let $F=\mathbb{Q}$, and $\alpha=\sqrt{2}$, $\beta=\sqrt{5}$, and $K = F(\alpha,\beta)$. Then $G=G(K/\mathbb{Q})$ is of order $4$. Since \[F(\alpha),\quad F(\beta),\quad F(\alpha\beta)\] are three distinct intermediate fields, $G$ has three nontrivial subgroups and therefore $G$ is the Klein group. Therefore, there are not other intermediate fields.

Theorem. Let $K/F$ is a Galois extension and $G$ be the Galois group thereof. Let $L=K^H$ where $H$ is a subgroup of $G$. Then

$L/F$ is Galois if and only if $H$ is a normal subgroup of $G$; and
in this case, $G(L/F) \cong G/H$.

Applications

Cubic Equation

Let \[f(x) = x^3 - a_1 x^2 + a_2 x - a_3\] be a irreducible polynomial over $F$ and $K$ be the splitting field thereof. Let $\qty{\alpha_1,\alpha_2,\alpha_3}$ denotes the roots of $f$. We have \[ F\subset F(\alpha_1) \subset F(\alpha_1,\alpha_2) = F(\alpha_1,\alpha_2,\alpha_3) = K \] since \[\alpha_3 = a_1 - \alpha_1 - \alpha_2.\] Let $L=F(\alpha_1)$, then \[ f(x) = (x-\alpha_1) q(x) \] over $L$.

If $q$ is irreducible over $L$, then $[K:L] = 2$ and $[K:F] = 6$.
If $q$ is reducible over $L$, then $L=K$ and $[K:F]=3$.

Let $f(x) = x^3 + 3x + 1$. $f(x)$ is irreducible over $\mathbb{Q}$ and has one real root and two complex roots. Therefore, $[K:F]=6$.

Let $f(x) = x^3 - 3x + 1$. $f(x)$ is irreducible over $\mathbb{Q}$. If $\alpha_1$ is one root then $\alpha_2 = \alpha_1^2 - 2$ is another. Therefore, $[K:F]=3$.

Let $f$ be a irreducible polynomial (having no multiple root therefore) over $F$. Then $G(K/F)$ is a transitive subgroup of $S_3$. We have either $G = S_3$ or $G = A_3$, corresponding to $[K:F] = 6$ and $[K:F] = 3$ respectively.

Galois Theory of Cubic Equation. Let $K$ be the splitting field of a cubic polynomial $f$ over $F$, and $G = G(K/F)$.

If $D$ is a square over $F$, then $[K:F] = 3$ and $G = A_3$.
If $D$ is not a square over $F$, then $[K:F] = 6$ and $G = S_3$.

The discriminant of $f=x^2+3x+1$ is $-5\cdot 3^2$, which is not a square. Therefore, $[K:F] = 6$.

The discriminant of $f=x^2-3x+1$ is $3^4$, which is a square. Therefore, $[K:F] = 3$.

Quartic Equation

Galois Theory of Quartic Equation. Let $G$ be the Galois group of an irreducible quartic polynomial. Then the discriminant $D$ of $f$ is a square in $F$ if and only if $G$ does not contain odd permutations. Therefore,

If $D$ is a square in $F$, then $G = A_4$ or $D_2$.
If $D$ is not a square in $F$, then $G = S_4$, $D_4$ or $C_4$.

Denote the roots by $\alpha_i$ and let \[ \beta_1 = \alpha_1 \alpha2 + \alpha_3 \alpha_4,\quad \beta_2 = \alpha_1 \alpha_3 + \alpha_2 \alpha_4,\quad \beta_3 = \alpha_1 \alpha_4 + \alpha_2 \alpha_3, \] and \[ g(x) = (x-\beta_1)(x-\beta_2)(x-\beta_3). \]

	$D$ is a square	$D$ is not a square
$g$ is reducible	$G=D_2$	$G=D_4$ or $C_4$
$g$ is irreducible	$G=A_4$	$G=S_4$

Quintic Equation

Proposition. Let $F$ be a subfield of $\mathbb{C}$. Then $\alpha\in\mathbb{C}$ is solvable over $F$ if it satisfy either of the following two equivalent conditions:

There is a chain of subfields of $\mathbb{C}$ \[F = F_0 \subset F_1 \subset \cdots \subset F_r = K\] such that
- $\alpha\in K$; and
- $F_j = F_{j-1}(\beta_j)$, where a power of $\beta_j$ is in $F_{j-1}.
There is a chain of subfields of $\mathbb{C}$ \[F = F_0 \subset F_1 \subset \cdots \subset F_r = K\] such that
- $\alpha\in K$; and
- $F_{j+1}/F_j$ is a Galois extension of prime degree.

Theorem. Let $f$ be a irreducible quintic polynomial over $F$ and that the Galois group thereof is $A_5$ or $S_5$. Then the roots of $F$ are not solvable over $F$.

Proof. We assume that $G=A_5$, and let

$K$ be the splitting field of $f$,
$F'/F$ be a Galois extension of prime degree $p$,
$F'$ be the splitting field of $g$ with roots $\beta_1,\cdots,\beta_p$,
$\alpha_1,\cdots,\alpha_5$ be the roots of $f$, and
$K'$ denote the splitting field of $fg$.

The Galois extensions and Galois groups are listed in the graph below.

Now we prove that $H\cong G = A_5$.

$H\cap H'$ is the trivial group $\qty{e}$ since an element thereof fixes all the roots $\alpha_i$ and $\beta_j$.
We restrict the canonical mapping $\mathcal{G} \mapsto \mathcal{G}/H \cong G'$ to $H'$, i.e. permutation of $\beta$ in $K' = F(\alpha,\beta)$ to permutation of $\beta$ in $F' = F(\beta)$.
- The kernel is the trivial group $\qty{e}$, i.e. this is an injection.
- Either $H'$ is trivial, or $H'$ is the cyclic group of order $p$.
  - $H'$ can not be trivial since $G=A_5$ has no normal subgroups.
  - Therefore, $H'$ is the cylic group of order $p$, and $\abs{H} = \abs{G} = \abs{A_5}$. Now we restrict the canonical mapping $\mathcal{G} \mapsto \mathcal{G}/H' \cong G$ to $H$, i.e. permutation of $\alpha$ in $K' = F(\alpha,\beta)$ to permutation of $\alpha$ in $F' = F(\alpha)$. Again this is an interjection and we found that $H=A_5$.

Take $x^4 - 4x^2 + 2$ for example. The roots are given by \[\alpha = \pm\sqrt{2\pm\sqrt{2}}.\] Taking $F = \mathbb{Q}$, $K'=K = F(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ and $F' = F(\sqrt{2})$, we find that \[K'=K=F(\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2-\sqrt{2}}).\] $K'=K$ looks in $F'$ like \[K'=K=F'(\sqrt{2+\sqrt{2}})\] since \[(\sqrt{2}-1)\sqrt{2+\sqrt{2}} = \sqrt{2-\sqrt{2}}.\] We find that \[G'\cong C_2,\quad H \cong C_2.\]

Remark

To obtain the Galois group of a certain polynomial, one may go to the Magma online calculator and submit the following code:

P<x> := PolynomialAlgebra(Rationals());
f := x^4 - 2*x^2 - 1;
G := GaloisGroup(f);
print G;

2021/3/4 22:06:34

[title=Galois Theory (I)]

## Galois Theory (I)

[[toc]]

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#### Review: Field Theory

##### Roots of Polynomials

Let $f$ be an irreducible polynomial in $F[x]$.
- $f$ has no multiple roots in any extension of $F$, unless $f'=0$.
- If $F$ has characteristic $0$, then $f$ has no multiple roots in any extension of $F$.

> Heuristically, we write a polynomial with multiple roots as
> $$f(x) = g(x)(x-\alpha)^2.$$
> Then
> $$f'(x) = g'(x)(x-\alpha)^2 + 2g(x)(x-\alpha)$$
> has a common factor $(x-\alpha)$ with $f(x)$.

##### Primitive Element Theorem

Let $K$ be an extension of $F$. If the extension $K/F$ is generated by a single element $\alpha$, then $\alpha$ is called the **primitive element** of $K/F$. We have the following theorem:

**Primitive Element Theorem.** If $F$ has characteristic $0$, then any extension $K/F$ has a primitive element.

> The extension $K=\mathbb{Q}\qty(\sqrt{2},\sqrt{3})$ is of degree $4$. We could easily prove that $K=\mathbb{Q}\qty(\sqrt{2}+\sqrt{3})$, i.e. the extension has a primitive element $\alpha = \sqrt{2} + \sqrt{3}{}$.
> :::
> :::
> We may verify that
> $$\sqrt{2} = \frac{1}{2}\qty(\alpha^3-9\alpha)$$
> and that
> $$\sqrt{3} = -\frac{1}{2}\qty(\alpha^3-11\alpha).$$
> Such decomposition exists because $\alpha^i$ could always be written as linear combinations of $\sqrt{2}{}$, $\sqrt{3}{}$ and $\sqrt{6}{}$. By solving a linear equation we retrieve $\sqrt{2}{}$ and $\sqrt{3}{}$ from powers of $\alpha$.

#### Symmetries of Polynomials

##### Symmetric Polynomials

$S_n$ acts on $R[u]$ by
$$f=f(u_1,\cdots,u_n)\mapsto f(u_{\sigma 1},\cdots,u_{\sigma n}) = \sigma f.$$

> This action is straightforward for someone who has experience with quantum mechanics. By regarding $f$ as the wavefunction of a many-body system, $\sigma f$ becomes the wavefunction after permutation of particles.

A polynomial $g$ is **symmetric** if $g$ is invariant under any permutation, i.e. if all monomials in the same orbit have the same coefficient.

**Theorem.** Any symmetric polynomial could be written (uniquely) as a linear combination of elementary symmetric polynomials.

We define the **elementary symmetric polynomials** by
\begin{align*}
s_1 &= u_1 + u_2 + \cdots + u_n, \\
s_2 &= u_1u_2 + u_1u_3 + \cdots, \\
s_3 &= u_1u_2u_3 + \cdots, \\
s_n &= u1 u_2 \cdots u_n.
\end{align*}{ }
> We have
> $$u_1^2 + \cdots + u_n^2 = s_1^2 - 2s_2.$$

**Corollary.** If $f(x) = x^n - a_1x^{n-1} + \cdots + a_n$ has coefficients in $F$ and splits over $K$ with roots $\alpha_1$, $\cdots$, $\alpha_n$. Let $g\qty(u_1,\cdots,u_n)$ be a symmetric polynomial with coefficients in $F$, then $g(\alpha_1,\cdots,\alpha_n)$ is in $F$.

> From the Vieta's formulae we know that the roots $x_1$ and $x_2$ of 
> $$x^2 - bx + c = 0$$
> satisfies
> $$x_1 + x_2 = b,\quad x_1x_2 = c$$
> even if the roots are in $\mathbb{R}\backslash \mathbb{Q}$ while the coefficients are in $\mathbb{Q}$. Moreover, symmetric polynomials like
> $$x_1^2 + x_2^2 = (x_1+x_2)^2 - 2x_1x_2$$
> evaluate to values in $\mathbb{Q}$ as well.

**Proposition.** Let $h$ be a symmetric polynomial and $\qty{p_1,\cdots,p_k}$ be the orbit of $p_1$ under $S_n$, then
$$h(p_1,\cdots,p_k)$$
is a symmetric polynomial.

> For the simplest case we take
> $$h = p_1 + \cdots + p_k,$$
> i.e. the sum of every element in the orbit.

##### Discriminant

Let the roots of
$$P(x) = x^n - s_1 x^{n-1} + s_2 x^{n-2} - \cdots \pm s_n$$
be $\qty{u_1,\cdots,u_n}$. The **discriminant** of $P$ is defined by
$$D(u) = (u_1 - u_2)^2 (u_1 - u_3)^2 \cdots (u_{n-1} - u_n)^2 = \prod_{i<j} (u_i - u_j)^2.$$

- $D(u)$ is a symmetric polynomial with integer coefficients.
  - In particular, if the coefficients are in $F$, then $D(u)$ evaluates to an element in $F$.
- If $\qty{\alpha_i}$ are elements in a field, then $D(u) = 0$ if and only if $P$ has a multiple root.

We are able to write $D$ as a sum of elementary symmetric polynomials.

> In the case of degree 2, we have
> $$D = (u_1 - u_2)^2 = s_1^2 - 4s_2.$$

#### Fields and Polynomials

::: {.alert .alert-info .mb-2 .pb-0}
We assume that $F$ has characteristic $0$ hereinafter.
:::

##### Splitting Field

$f$ **splits** over $K$ if
$$f(x) = (x-\alpha_1) \cdots (x-\alpha_n)$$
in $K$.

Let $f$ be a polynomial over $F$. The splitting field of $f$ is an extension $K/F$ such that
- $f$ spilits in $K$; and
- $K$ is generated by the roots, i.e.
$$K = F\qty(\alpha_1,\cdots,\alpha_n).$$

**Lemma.** Let $f$ be a polynomial over $F$.
- Let $F\subset L\subset K$ be a field, and $K$ be the splitting field of $f$, then $K$ is also the splitting field of $f$ as a polynomial over $L$.
- Every $f$ over $F$ has a splitting field.
- The splitting field is a finite extension of $F$. Every finite extension of $F$ is contained in a splitting field.

**Splitting Field Theorem.** Let $K$ be the splitting field of $f$ as a polynomial over $F$. Let $g$ be an irreducible polynomial over $F$. If $g$ has a root in $K$, then $g$ split over $K$.

*Outline of Proof.* Let $K$ be generated by $\alpha_1$,$\cdots$,$\alpha_n$, and $\beta_1$ be a root of $G$ in $K$. Let
$$p_1(\alpha) = \beta_1,$$
and $\qty{p_1,\cdots,p_k}$ be the orbit of $p_1$ under $S_n$ and
$$\beta_k = p_k(\alpha).$$
We may prove that the coefficients of
$$h(x) = (x-\beta_1)\cdots (x-\beta_k)$$
are in $F$. Therefore, $g$ divides $h$ since $h$ has $\beta_1$ as a root and $g$ splits since $h$ splits.

##### Isomorphism of Field Extension

- Let $K$ and $K'$ be extensions of $F$. An $F$-**isomorphism** is a field isomorphism between $K$ and $K'$ that fixes every element in $F$.
  - An $F$-**automorphism** of $K$ is an $F$-**isomorphism** from $K$ to itself, i.e. the symmetries of the extension $K/F$.
- The $F$-automorphisms of a finite extensions $K/F$ form a group $G(K/F)$, called the **Galois group**.
- A finite extension $K/F$ is called a **Galois extension** if
$$\abs{G(K/F)} = \qty[K:F].$$

> For any quadratic extension $K/F$, e.g. $\mathbb{C}/\mathbb{R}$ and $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$, the Galois group has two elements: the trivial one, and the conjugation, i.e.
> $$a+bi \mapsto a-bi$$
> or
> $$a+b\sqrt{2} \mapsto a-b\sqrt{2}.$$

> Let $F=\mathbb{Q}$ and $K=F[\sqrt[3]{2}]$. Since $x^3-2=0$ has two more roots which are complex and do not belong to $K$, we have
> $$G(K/F) = \qty{e}$$
> while $[K:F]=3$. Therefore, the extension is not Galois.

**Lemma.** Let $K$ and $K'$ be extensions of $F$.
- Let $f(x)$ be a polynomial with coefficients in $F$, and $\sigma$ be an $F$-isomorphism from $K$ to $K'$. If $\alpha$ is a root of $f$ in $K$, then $\sigma(\alpha)$ is a root of $f$ in $K'$.
- Let $K=F(\alpha_1,\cdots,\alpha_n)$, and $\sigma$ and $\sigma'$ be $F$-isomorphisms from $K$ to $K'$. If for $i=1,\cdots,n$ we have $\sigma(\alpha_i) = \sigma'(\alpha_i)$, then $\sigma = \sigma'$. If $\sigma$ fixes every $\alpha_i$, then $\sigma$ is the identity mapping.
- If $f$ is an irreducible polynomial over $F$, and $\alpha$ and $\alpha'$ are roots of $f$ in $K$ and $K'$ respectively, then there exists a unique $F$-isomorphism $\sigma$ that maps $\alpha$ to $\alpha'$. If $F(\alpha) = F(\alpha')$ then $F$ is the identity mapping.

**Proposition.** Let $f$ be a polynomial over $F$. 
- An extension $L/F$ contains at most one splitting field of $f$ over $F$.
- Splitting fields of $f$ over $F$ are all isomorphic to each other.

##### Fixed Field

Let $H$ be an automorphism group of $K$. The **fixed field** of $H$ consists of all the elements invariant under every elements of $H$,
$$K^H = \Set{\alpha\in K|\sigma(\alpha)=\alpha,\ \text{for all } \sigma \in H}.$$

**Theorem.**{id="theorem-irreducible-polynomial-fixed-field"} Let $H$ be a finite automorphism group of $K$, and $F$ be the fixed field $K^H$. Let $\beta_1$ be an element of $K$ and $\qty{\beta_1,\cdots,\beta_r}$ be the $H$-orbit of $\beta_1$.
- The irreducible polynomial of $\beta_1$ over $F$ is given by
$$g(x) = (x-\beta_1)\cdots (x-\beta_r).$$
- $\beta_1$ is algebraic over $F$, and the degree thereof is the cardinality of the $H$-orbit thereof.
  - In particular, degree of $\beta_1$ divides the order of $H$.

> Let $K=\mathbb{C}$ and $H = \qty{e, c}$ where $c$ denotes  conjugation. Then $F = \mathbb{R}$ and the irreducible polynomial of $\beta \in \mathbb{C}$ is given by
> - $g(x) = x-\beta$ if $\beta \in \mathbb{R}$, or
> - $g(x) = (x-\beta)(x-\overline{\beta})$ if $\beta \in \mathbb{C}\backslash \mathbb{R}$.

**Lemma.** Let $K$ be an algebraic extension of $F$ and not be a finite extension. Then there exists elements in $K$ of arbitrary large degrees over $F$.

**Fixed Field Theorem.** Let $H$ be a finite automorphism group of $K$ and $F=K^H$ be the fixed field thereof. Then $K$ is a finite extension over $F$, the degree $[K:H]$ of which equals $\abs{H}$.
$$[K:K^H] = \abs{H}.$$

> Let $K=\mathbb{C}(t)$, and $\sigma$ and $\tau$ be $\mathbb{C}$-automorphisms of $K$, such that
> $$\sigma(t) = it$$
> and
> $$\tau(t) = t^{-1}.$$
> We have
> $$\sigma^4 = 1,\quad \tau^2 = 1,\quad \tau\sigma = \sigma^{-1}\tau.$$
> Therefore, $\sigma$ and $\tau$ generate the $D_4$ group.

> We may easily verify that $u = t^4 + t^{-4}$ is transcendental over $\mathbb{C}$. Now we prove that $\mathbb{C}(u) = K^H$.
> :::
> :::
> Since $u$ is fixed by $\sigma$ and $\tau$ we have $\mathbb{C}(u)\in K^H$. The $H$-orbits of $t$ is given by
> $$\qty{t,it,-t,-it,t^{-1},-it^{-1},-t^{-1},it^{-1}}.$$
> From the [theorem above](#theorem-irreducible-polynomial-fixed-field) we obtain the  irreducible polynomial of $t$ over $\mathbb{C}(u)$:
> $$x^8 - ux^4 + 1.$$
> - Therefore, $[K:\mathbb{C}(u)] \le 8$.
> - From the fixed field theorem we know $[K:K^H] = 8$.
> - We have already proved that $\mathbb{C}(u) \subset K^H$.
> - Thus we conclude that $\mathbb{C}(u) = K^H$.

The example above illustrates the **L&uuml;roth theorem**: Let $F \subset \mathbb{C}(t)$ be a field that contains $\mathbb{C}$ and not be $\mathbb{C}$ itself. Then $F$ is isomorphic to $\mathbb{C}(u)$.

#### Fundamentals of Galois Theory

##### Galois Extension

**Lemma.**
- Let $K/F$ be a finite extension. Then $G(K/F)$ is a fintie group and
$$\abs{G(K/F)}\mid [K:F].$$
- Let $H$ be a finite automorphism group of $K$. Then $K$ is a galois extension of $K^H$, and $H=G(K/K^H)$.

**Lemma.** Let $\gamma_1$ be the primitive element of a finite extension $K/F$ and $f(x)$ be the irreducible polynomial of $\gamma_1$ over $F$. Let $\gamma_1,\cdots,\gamma_r$ be roots of $f$ in $K$. Then there exists a unique $F$-automorphism $\sigma_i$ of $K$  such that $\sigma_i(\gamma_1) = \gamma_i$. These are all the $F$-automorphisms. Therefore, $\abs{G(K/F)}=r$.

**Characteristics of Galois Extension.** Let $K/F$ be a finite extension and $G$ be the Galois group thereof.Then the following statements are equivalent:
- $K/F$ is a Galois extension, i.e. $\abs{G} = [K:F]$.
- $K^G=F$.
- $K$ is a splitting field over $F$.

If $K$ is a splitting field of $f$ over $F$, we may refer to $G(K/F)$ as the Galois group of $f$.

**Corollary.**
- Every finite extension $K/F$ is contained in a Galois extension.
- If $K/F$ is a Galois extension, and $L$ is an intermediate field, then $K/L$ is also a Galois extension, and $G(K/L)$ is a subgroup of $G(K/F)$.

**Theorem.** Let $K/F$ be a Galois extension and $G$ be its Galois group, and $g$ be a polynomial over $F$ that splits in $K$ with roots $\beta_1,\cdots,\beta_r$.
- $G$ acts on $\qty{\beta_i}$.
- If $K$ is the splitting field of $G$ over $F$, then the permutation of $G$ on $\qty{\beta_i}$ is faithful and $G$ is a subgroup of $S_r$.
- If $g$ is irreducible over $F$, then the action of $G$ on $\qty{\beta_i}$ is transitive.
- If $K$ is the splitting field of $g$ over $F$ and that $g$ is irreducible over $F$, then $G$ is a transitive subgroup of $S_r$.

##### The Fundamental Theorem

**The Fundamental Theorem.** Let $K$ be a Galois extension over $F$, and $G$ be the Galois group thereof. Then there is a one-to-one correspondence between
$$\qty{\text{subgroup } H \text{ of } G}$$
and
$$\qty{\text{intermediate field } L \text{ of } K/F}.$$
The mapping
$$H\mapsto K^H$$
is the inverse of
$$L\mapsto G(K/L).$$

**Corollary.**
- If $L\subset L'$ are intermediate fields, and the corresponding subgroups are $H$ and $H'$, then $H\supset H'$.
- The subgroup corresponding to $F$ is $G(K/F)$, while to $G$ is the trivial group $\qty{e}$.
- If $L$ corresponds to $H$, then $[K:L]=H$ and $[L:F]=[G:H]$.

**Corollary.** A finite extension $K/F$ has finitely many intermediate fields $F\subset L\subset K$.

> Let $F=\mathbb{Q}$, and $\alpha=\sqrt{2}$, $\beta=\sqrt{5}$, and $K = F(\alpha,\beta)$. Then $G=G(K/\mathbb{Q})$ is of order $4$. Since
> $$F(\alpha),\quad F(\beta),\quad F(\alpha\beta)$$
> are three distinct intermediate fields, $G$ has three nontrivial subgroups and therefore $G$ is the Klein group. Therefore, there are not other intermediate fields.

**Theorem.** Let $K/F$ is a Galois extension and $G$ be the Galois group thereof. Let $L=K^H$ where $H$ is a subgroup of $G$. Then 
- $L/F$ is Galois if and only if $H$ is a normal subgroup of $G$; and
- in this case, $G(L/F) \cong G/H$.

#### Applications

##### Cubic Equation

Let
$$f(x) = x^3 - a_1 x^2 + a_2 x - a_3$$
be a irreducible polynomial over $F$ and $K$ be the splitting field thereof. Let $\qty{\alpha_1,\alpha_2,\alpha_3}$ denotes the roots of $f$. We have
$$
F\subset F(\alpha_1) \subset F(\alpha_1,\alpha_2) = F(\alpha_1,\alpha_2,\alpha_3) = K
$$
since
$$\alpha_3 = a_1 - \alpha_1 - \alpha_2.$$
Let $L=F(\alpha_1)$, then
$$
f(x) = (x-\alpha_1) q(x)
$$
over $L$.
- If $q$ is irreducible over $L$, then $[K:L] = 2$ and $[K:F] = 6$.
- If $q$ is reducible over $L$, then $L=K$ and $[K:F]=3$.

> Let $f(x) = x^3 + 3x + 1$. $f(x)$ is irreducible over $\mathbb{Q}$ and has one real root and two complex roots. Therefore, $[K:F]=6$.

> Let $f(x) = x^3 - 3x + 1$. $f(x)$ is irreducible over $\mathbb{Q}$. If $\alpha_1$ is one root then $\alpha_2 = \alpha_1^2 - 2$ is another. Therefore, $[K:F]=3$.

Let $f$ be a irreducible polynomial (having no multiple root therefore) over $F$. Then $G(K/F)$ is a transitive subgroup of $S_3$. We have either $G = S_3$ or $G = A_3$, corresponding to $[K:F] = 6$ and $[K:F] = 3$ respectively.

**Galois Theory of Cubic Equation.** Let $K$ be the splitting field of a cubic polynomial $f$ over $F$, and $G = G(K/F)$.
- If $D$ is a square over $F$, then $[K:F] = 3$ and $G = A_3$.
- If $D$ is not a square over $F$, then $[K:F] = 6$ and $G = S_3$.

> The discriminant of $f=x^2+3x+1$ is $-5\cdot 3^2$, which is not a square. Therefore, $[K:F] = 6$.

> The discriminant of $f=x^2-3x+1$ is $3^4$, which is a square. Therefore, $[K:F] = 3$.

##### Quartic Equation

**Galois Theory of Quartic Equation.** Let $G$ be the Galois group of an irreducible quartic polynomial. Then the discriminant $D$ of $f$ is a square in $F$ if and only if $G$ does not contain odd permutations. Therefore,
- If $D$ is a square in $F$, then $G = A_4$ or $D_2$.
- If $D$ is not a square in $F$, then $G = S_4$, $D_4$ or $C_4$.

Denote the roots by $\alpha_i$ and let
$$
\beta_1 = \alpha_1 \alpha2 + \alpha_3 \alpha_4,\quad \beta_2 = \alpha_1 \alpha_3 + \alpha_2 \alpha_4,\quad \beta_3 = \alpha_1 \alpha_4 + \alpha_2 \alpha_3,
$$
and
$$
g(x) = (x-\beta_1)(x-\beta_2)(x-\beta_3).
$$

:::{.mb-2}
|---|---|---|
|&nbsp;|$D$ is a square|$D$ is not a square|
|$g$ is reducible| $G=D_2$ | $G=D_4$ or $C_4$ |
|$g$ is irreducible| $G=A_4$ | $G=S_4$ |
{.table .table-striped .text-center .w-auto .m-auto }
:::

##### Quintic Equation

**Proposition.** Let $F$ be a subfield of $\mathbb{C}$. Then $\alpha\in\mathbb{C}$ is **solvable** over $F$ if it satisfy either of the following two equivalent conditions:
- There is a chain of subfields of $\mathbb{C}$
$$F = F_0 \subset F_1 \subset \cdots \subset F_r = K$$
such that
  - $\alpha\in K$; and
  - $F_j = F_{j-1}(\beta_j)$, where a power of $\beta_j$ is in $F_{j-1}.
- There is a chain of subfields of $\mathbb{C}$
$$F = F_0 \subset F_1 \subset \cdots \subset F_r = K$$
such that
  - $\alpha\in K$; and
  - $F_{j+1}/F_j$ is a Galois extension of prime degree.

**Theorem.** Let $f$ be a irreducible quintic polynomial over $F$ and that the Galois group thereof is $A_5$ or $S_5$. Then the roots of $F$ are not solvable over $F$.

*Proof.* We assume that $G=A_5$, and let
- $K$ be the splitting field of $f$,
- $F'/F$ be a Galois extension of prime degree $p$,
- $F'$ be the splitting field of $g$ with roots $\beta_1,\cdots,\beta_p$,
- $\alpha_1,\cdots,\alpha_5$ be the roots of $f$, and
- $K'$ denote the splitting field of $fg$.

The Galois extensions and Galois groups are listed in the graph below.
:::{.text-center}
![](/scribble-files/35a0aa8c-c4ce-40a8-ae54-6f76cbb1360e/FFKK.jpg){style="height:13.18em;"}
:::

Now we prove that $H\cong G = A_5$.

- $H\cap H'$ is the trivial group $\qty{e}$ since an element thereof fixes all the roots $\alpha_i$ and $\beta_j$.
- We restrict the canonical mapping $\mathcal{G} \mapsto \mathcal{G}/H \cong G'$ to $H'$, i.e. permutation of $\beta$ in $K' = F(\alpha,\beta)$ to permutation of $\beta$ in $F' = F(\beta)$. 
  - The kernel is the trivial group $\qty{e}$, i.e. this is an injection.
  - Either $H'$ is trivial, or $H'$ is the cyclic group of order $p$.
    - $H'$ can not be trivial since $G=A_5$ has no normal subgroups.
    - Therefore, $H'$ is the cylic group of order $p$, and $\abs{H} = \abs{G} = \abs{A_5}$. Now we restrict the canonical mapping $\mathcal{G} \mapsto \mathcal{G}/H' \cong G$ to $H$, i.e. permutation of $\alpha$ in $K' = F(\alpha,\beta)$ to permutation of $\alpha$ in $F' = F(\alpha)$. Again this is an interjection and we found that $H=A_5$.

> Take $x^4 - 4x^2 + 2$ for example. The roots are given by
> $$\alpha = \pm\sqrt{2\pm\sqrt{2}}.$$
> Taking $F = \mathbb{Q}$, $K'=K = F(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ and $F' = F(\sqrt{2})$, we find that
> $$K'=K=F(\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2-\sqrt{2}}).$$
> $K'=K$ looks in $F'$ like
> $$K'=K=F'(\sqrt{2+\sqrt{2}})$$
> since
> $$(\sqrt{2}-1)\sqrt{2+\sqrt{2}} = \sqrt{2-\sqrt{2}}.$$
> We find that
> $$G'\cong C_2,\quad H \cong C_2.$$

##### Remark

To obtain the Galois group of a certain polynomial, one may go to the [Magma online calculator](http://magma.maths.usyd.edu.au/calc/) and submit the following code:
```
P<x> := PolynomialAlgebra(Rationals());
f := x^4 - 2*x^2 - 1;
G := GaloisGroup(f);
print G;
```

Post anonymously

	\(D\) is a square	\(D\) is not a square
\(g\) is reducible	\(G=D_2\)	\(G=D_4\) or \(C_4\)
\(g\) is irreducible	\(G=A_4\)	\(G=S_4\)