Let \(f\) be an irreducible polynomial in \(F[x]\).
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 \[D(u) = (u_1 - u_2)^2 (u_1 - u_3)^2 \cdots (u_{n-1} - u_n)^2 = \prod_{i<j} (u_i - u_j)^2.\]
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
Lemma. Let \(f\) be a polynomial over \(F\).
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.
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 \(F\).
Proposition. Let \(f\) be a polynomial over \(F\).
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 \(\beta_1\).
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.
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:
If \(K\) is a splitting field of \(f\) over \(F\), we may refer to \(G(K/F)\) as the Galois group of \(f\).
Corollary.
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 \(\beta_1,\cdots,\beta_r\).
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.
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
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 \(L\).
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 \(G = G(K/F)\).
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,
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:
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
The Galois extensions and Galois groups are listed in the graph below.
Now we prove that \(H\cong G = A_5\).
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;