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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