Ze Chen## Galois Theory (I)

#### Review: Field Theory

##### Roots of Polynomials

##### Primitive Element Theorem

#### Symmetries of Polynomials

##### Symmetric Polynomials

##### Discriminant

#### Fields and Polynomials

##### Splitting Field

##### Isomorphism of Field Extension

##### Fixed Field

#### Fundamentals of Galois Theory

##### Galois Extension

##### The Fundamental Theorem

#### Applications

##### Cubic Equation

##### Quartic Equation

##### Quintic Equation

##### Remark

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 \(f(x)\).

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 \(\alpha\).

\(S_n\) 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.

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 \[D = (u_1 - u_2)^2 = s_1^2 - 4s_2.\]

We assume that \(F\) has characteristic \(0\) hereinafter.

\(f\) **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.

- 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 \(K/F\).

- An \(F\)-
- 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.

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 \(\mathbb{C}(u)\).

**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.** 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, \(G(L/F) \cong G/H\).

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, \([K:F] = 3\).

**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 | \(G=A_4\) | \(G=S_4\) |

**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 \[G'\cong C_2,\quad H \cong C_2.\]

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;
```