Ze Chen

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

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