A note in Galois Theory

Theorem 1: Let $$\mathbb{G} = \{ \sigma_1,\cdots,\sigma_n\}$$ be a finite group, $$x_1,\cdots,x_n$$ be indeterminates and let $$\rho$$ be the regular representation of $$\mathbb{G}$$. Let $$x = (x_1,\cdots,x_n)^T$$ and $$G = (\rho(\sigma_1)x,\cdots,\rho(\sigma_n)x)$$ , $$A' = diag(x_1,\cdots,x_n)$$. Set $$A=GA'G^{-1}$$. Suppose there exists $$\alpha:=(\alpha_1,\cdots,\alpha_n) \in \mathbb{C}^n$$ such that: 1) $$det(G(\alpha)) \neq 0$$ 2) $$A = A(\alpha)$$ has rational coefficients 3) the characteristic polynomial $$\chi_A(t)$$ is irreducible over $$\mathbb{Q}$$. Then $$Gal(\chi_A(t)/\mathbb{Q}) = \mathbb{G}$$. Proof: On the one hand we have \[ GA' = AG = GA' = (x_1\rho(\sigma_1)x,\cdots,x_n\rho(\sigma_n)x) \] On the other hand we have \[ GA' = AG = A (\rho(\sigma_1)x,\cdots,\rho(\sigma_n)x) = (A\rho(\sigma_1)x,\cdots,A\rho(\sigma_n)x) \] and it follows that \[ (A\rho(\sigma_1)x,\cdots,A\rho(\sigma_n)x) = GA' = (x_1\rho(\sigma_1)x,\cdots,x_n\rho(\sigma_n)x) \] and so we get \begin{equation}\label{eq:1} A\rho(\sigma_i)x = x_i \rho(\sigma_i)x \text{ for } i=1,\cdots,n \end{equation} Let $$K:=\mathbb{Q}(\alpha_1,\cdots,\alpha_n)$$. From $$A\rho(\sigma_i) \alpha = \alpha_i \rho(\sigma_i)\alpha$$ follows for $$i=1$$, $$\rho(\sigma_1) = 1$$ that $$A\alpha = \alpha_1 \alpha$$. Application of $$\tau \in Gal(K/\mathbb{Q})$$ at the coefficients of the last equation gives us with $$\tau(A) = A$$ (since $$A$$ has rational coefficients) that \begin{equation}\label{eq:2} A\tau(\alpha) = \tau(\alpha_1)\tau(\alpha) = \alpha_{\tau} \tau(\alpha) \end{equation} where we have set $$\alpha_{\tau} = \tau(\alpha_1)$$. This means that $$\tau(\alpha)$$ is an eigenvector of $$A$$ to the eigenvalue $$\alpha_{\tau}$$. But because of equation $$(1)$$ we have: \begin{equation}\label{eq:3} A\rho(\sigma_{\tau}) \alpha = \alpha_{\tau} \rho(\sigma_{\tau}) \alpha \end{equation} Since the eigenvectors $$\rho(\sigma_1)\alpha,\cdots,\rho(\sigma_n)\alpha$$ of $$A$$ build a basis of $$K^n$$, it means that every eigenspace has dimension $$1$$. Especially there exists $$a_{\tau} \in K$$ such that \begin{equation}\label{eq:4} \tau(\alpha) = a_{\tau} \rho(\sigma_{\tau}) \alpha \end{equation} If we build the polynomials with roots from the coefficients of the left and right side of $$(4)$$, then they must be because of $$(4)$$ be equal: \begin{equation}\label{eq:5} \chi_A(t) = \prod_{i=1}^n (t -\alpha_i) =^{(4)} \prod_{j=1}^n (t - a_{\tau} \alpha_j) \end{equation} This means that the coefficients of both polynomials must be equal, especially we get \begin{equation}\label{eq:6} \sum_{i=1}^n \alpha_i = a_{\tau} \sum_{i=1}^n \alpha_i \end{equation} By a theorem of Frobenius concerning the factorization of the group determinant, we have \begin{equation}\label{eq:7} det( G(x_1,\cdots,x_n)) = (x_1+\cdots+x_n) p(x_1,\cdots,x_n) \end{equation} for some polynomial $$p$$. From $$(7)$$ follows by assumption \[ 0 \neq det(G(\alpha_1,\cdots,\alpha_n)) = (\alpha_1+\cdots+\alpha_n) p(\alpha_1,\cdots,\alpha_n) \] which means that $$0 \neq \alpha_1+\cdots+\alpha_n$$ and because of $$(6)$$ we get \begin{equation}\label{eq:8} a_{\tau} = 1 \end{equation} From $$(4)$$ and $$(8)$$ it follows that \begin{equation}\label{eq:9} P_{\tau} \alpha = \tau(\alpha) = \rho(\sigma_{\tau}) \alpha \end{equation} where $$P_{\tau}$$ is the permutation matrix which corresponds to $$\tau$$. Since \[ \chi_A(t) = \prod_{i=1}^n t-\alpha_i \] is separable, it follows that $$\alpha_k \neq \alpha_l$$ for $$k \neq l$$. This means that the two permutation matrices $$P_{\tau}$$ and $$\rho(\sigma_{\tau})$$ from $$(9)$$ must be equal, that is \begin{equation}\label{eq:10} P_{\tau} = \rho(\sigma_{\tau}) \end{equation} From this we get a mapping: \[ \phi: Gal(K/\mathbb{Q}) \rightarrow \mathbb{G}, \tau \mapsto \sigma_{\tau} \] $$\phi$$ is injective: Let $$\sigma_{\tau} = \phi(\tau) = \phi(\tau') = \sigma_{\tau'}$$. Then we get \[ P_{\tau} =^{(10)} \rho(\sigma_{\tau}) = \rho(\sigma_{\tau'}) = ^{(10)} P_{\tau'} \] and since the regular representation $$P$$ is faithful, it follows that $$\tau = \tau'$$. $$\phi$$ is surjective: Let $$m = |Gal(K/\mathbb{Q})|$$, $$n=|\mathbb{G}|=deg(\chi_A(t))$$. Since $$\chi_A(t)$$ is irreducible, $$deg(\chi_A(t)) = n$$ is a divisor of $$m = |Gal(K/\mathbb{Q})|$$ and it follows that $$n \le m$$. Since $$\phi$$ is injective it follows that $$m \le n$$, which means that $$m=n$$. Since the mapping $$\phi$$ is injective on two finite sets of equal size, this means that $$\phi$$ must also be surjective. $$\phi$$ is an antihomomorphism of groups: \[ \tau(\alpha) = \rho(\sigma_m)\alpha \rightarrow \phi(\tau) = \sigma_m\] \[ \tau'(\alpha) = \rho(\sigma_k)\alpha \rightarrow \phi(\tau') = \sigma_k\] \[ (\tau \tau')(\alpha) = \rho(\sigma_l)\alpha \rightarrow \phi(\tau \tau') = \sigma_l\] From this it follows that: \[ \rho(\sigma_l)\alpha = (\tau \tau')(\alpha) = \tau(\tau'(\alpha)) = \tau(\rho(\sigma_k) \alpha) = \rho(\sigma_k) \tau(\alpha) = \rho(\sigma_k) \rho(\sigma_m)\alpha \] Since $$\alpha_i$$ are separable, it follows that the permutation matrices $$\rho(\sigma_l)$$ and $$\rho(\sigma_k \sigma_m)$$ must be equal: \[ \rho(\sigma_l) = \rho(\sigma_k \sigma_m) \] and from this, since the regular representation is faithful, it follows that $$\sigma_l = \sigma_k \sigma_m$$. From this we get: \[ \phi(\tau \tau') = \sigma_l = \sigma_k \sigma_m = \phi(\tau') \phi(\tau) \] Since for two finite groups which are antiisomorph it follows that they are isomorph (if $$\phi(xy)= \phi(y)\phi(x)$$ is an antihomomorphism, then $$\Phi(x):=\phi(x)^{-1}$$ is a homomorphism), we conclude that \[ Gal(K/\mathbb{Q}) = \mathbb{G} \] Definition: Let $$\mathbb{G}$$ be a finite group with $$n$$ elements. Let $$\mathbb{Q}[x_1,\cdots,x_n]^{\mathbb{G}} = \mathbb{Q}[g_1,\cdots,g_m]$$. Then there exist polynomials $$s_j \in \mathbb{Q}[y_1,\cdots,y_m]$$ for $$j=1,\cdots,n$$ such that $$e_j(x_1,\cdots,x_n) = s_j(g_1(x_1,\cdots,x_n), \cdots, g_m(x_1,\cdots,x_n))$$ for $$j=1,\cdots,n$$ where $$e_j$$ is the $$j$$-elementary symmetric polynomial in $$x_k$$. Consider the polynomial $$p(t,y_1,\cdots,y_m) = t^n - s_1(y_1,\cdots,y_m) t^{n-1}+\cdots+(-1)^n s_n(y_1,\cdots,y_m)$$ which is a polynomial in $$\mathbb{Q}[t,y_1,\cdots,y_m]$$. We call $$p$$ the corresponding polynomial of the group $$\mathbb{G}$$ to the invariants $$g_1,\cdots,g_m$$. Let $$I=\{h \in \mathbb{Q}[y_1,\cdots,y_m] | h(g_1,\cdots,g_m) = 0 \}$$. Define the variety $$V_{\mathbb{G}} = \{ a \in \mathbb{Q}^m | h(a) = 0 \forall h \in I \}$$. We say that a polynomial is Hilbert irreducible over some variety, if it is irreducible and there exists a point in the variety such that the polynomial remains irreducible after specializing to the point of the variety and that other polynomials are not zero on this point. Lemma 1: The corresponding polynomial is irreducible in $$\mathbb{Q}[t,y_1,\cdots,y_m]$$. Proof: If the polynomial factored in a non-trivial way, then because of the $$t^n$$ term the factors must have degree less than $$n$$ in $$t$$ (consider the factorization in $$R=\mathbb{Q}(y_1,y_2,\ldots,y_n)[t]$$; note also that we can assume that the factors over $$R$$ are monic polynomials in $$t$$). Now specialise via $$y_i\mapsto g_i$$ and we get a non-trivial factorization of the specialised polynomial in $$\mathbb{Q}[g_1,\ldots,g_m][t]$$ and hence in $$\mathbb{Q}[x_1,\ldots,x_n][t]$$ and so in $$\mathbb{Q}(x_1,\ldots,x_n)[t]$$. But we know the complete factorization in this ring, it's just $$\prod(t-x_i)$$, so our given factorization must specialise into factors of the form $$\prod_{i\in I}(t-x_i)$$ for some subsets $$I$$ of $$\{1,2,\ldots,n\}$$ (with each $$I$$ not empty or the whole thing), and the constant term of each factor must be in $$\mathbb{Q}[g_1,\ldots,g_m]$$ and hence $$G$$-invariant. This is a contradiction. Lemma 2: Let $$f_1,\cdots,f_r \in \mathbb{Q}(x_1,\cdots,x_n)^{\mathbb{G}}$$ and suppose that the corresponding polynomial of $$\mathbb{G}$$ is Hilbert irreducible over the variety $$V_G$$. Then there exists $$\alpha = (\alpha_1,\cdots,\alpha_n) \in \mathbb{C}^n$$ such that: 1) $$f_1(\alpha),\cdots,f_r(\alpha)$$ are rational numbers. 2) $$\prod_{i=1}^n t - \alpha_i$$ is irreducible in $$\mathbb{Q}[t]$$. Proof: We can write \[ f_i = \frac{P_i(g_1,\cdots,g_m)}{Q_i(g_1,\cdots,g_m)} \text{ for } P_i,Q_i \in \mathbb{Q}[y_1,\cdots,y_m] \] The corresponding polynomial $$p(t,y_1,\cdots,y_m)$$ is irreducible in $$\mathbb{Q}[t,y_1,\cdots,y_m]$$ by Lemma 1. By assumption, we can find $$(a_1,\cdots,a_m)$$ in $$V_{\mathbb{G}}$$ such that: 1) $$p(t,a_1,\cdots,a_m)$$ is irreducible in $$\mathbb{Q}[t]$$. 2) $$Q_i(a_1,\cdots,a_m) \neq 0$$ for $$i=1,\cdots,r$$. By application of the Hilbert Nullstellensatz we can find $$\alpha =( \alpha_1,\cdots,\alpha_n) \in \mathbb{C}^n$$ such that $$g_i(\alpha) = a_i$$ for $$i=1,\cdots,m$$. Then \[ e_j(\alpha) = s_j(g_1(\alpha),\cdots,g_m(\alpha)) = s_j(a_1,\cdots, a_m) \] and the polynomial $$p(t,a_1,\cdots,a_m)$$ factors over $$\mathbb{C}$$ as $$p(t,a_1,\cdots,a_m) = \prod_{i=1}^n t - \alpha_i $$. Then we get \[ f_i(\alpha) = \frac{P_i(g_1(\alpha),\cdots,g_m(\alpha))}{Q_i(g_1(\alpha),\cdots,g_m(\alpha))} = \frac{P_i(a_1,\cdots,a_m)}{Q_i(a_1,\cdots,a_m)} \] is a rational number, which was to be shown. Lemma 3: Let the notation be as in Theorem 1. Then the entries $$a_{ij}$$ of the matrix $$A$$ are elements of $$\mathbb{Q}(x_1,\cdots,x_n)^{\mathbb{G}}$$. Proof: We have $$A(x) = G(x) A'(x) G(x)^{-1}$$ and want to show that $$A(\rho(\sigma)x) = A(x)$$ for all $$\sigma \in \mathbb{G}$$. For this, it is sufficient to show that \[ A'(\rho(\sigma)x) = \rho(\sigma)A'(x)\rho(\sigma)^{-1} \] and \[ G(\rho(\sigma)x) = G(x) \rho(\sigma)^{-1} \] as it follows that \[ A(\rho(\sigma) x) = G(\rho(\sigma)x) A'(\rho(\sigma) x ) G(\rho(\sigma) x) ^{-1} \] \[ = G(x) \rho(\sigma)^{-1} \rho(\sigma) A'(x) \rho(\sigma)^{-1} \rho(\sigma) G(x)^{-1} \] \[ = G(x) A'(x) G(x)^{-1} = A(x) \] Corollary 1: Let $$\mathbb{G}$$ be a finite group such that the correspoding polynomial is Hilbert irreducible over $$V_{\mathbb{G}}$$. Then $$\mathbb{G}$$ is realizable over $$\mathbb{Q}$$ as a Galois group. Proof: Consider the regular representation of $$\mathbb{G}$$ and let $$A = (a_{ij})$$ be defined as in Theorem 1. Then by Lemma 3 we have $$a_{ij} \in \mathbb{Q}(x_1,\cdots,x_n)^{\mathbb{G}}$$. Hence by Lemma 2 there exists $$\alpha \in \mathbb{C}^n$$ such that 1) $$a_{ij}(\alpha)$$ are rational numbers. 2) The characteristic polynomial $$\chi_A(t)=\prod_{i=1}^n t-\alpha_i$$ is irreducible in $$\mathbb{Q}[t]$$. Since $$a_{ij}(\alpha)$$ are rational numbers, the matrix $$A(\alpha) = G(\alpha) A'(\alpha) G(\alpha)^{-1}$$ is defined for $$\alpha$$, hence $$det(G(\alpha)) \neq 0$$. By Theorem 1 therefore it follows that $$Gal(\chi_A(t) / \mathbb{Q}) = \mathbb{G}$$. Theorem 2: The following are equivalent: 1) Hilbert Irreducibility can be applied to the corresponding polynomial $$p(t,y_1,\cdots,y_m)$$ over the variety $$V_\mathbb{G}$$ 2) Every finite group $$\mathbb{G}$$ is Galois over $$\mathbb{Q}$$. Proof: 1) $$\rightarrow$$ 2): Corollary 1. 2) $$\rightarrow$$ 1) Let $$\mathbb{G} = Gal(K/\mathbb{Q}) = \{ \sigma_1,\cdots,\sigma_n\}$$ be a Galois group. By the normal basis theorem there exists a $$\theta \in K$$ such that $$\sigma_1(\theta),\cdots,\sigma_n(\theta)$$ build a basis of $$K$$ over $$\mathbb{Q}$$. Let $$\mathbb{Q}[x_1,\cdots,x_n]^\mathbb{G} = \mathbb{Q}[g_1,\cdots,g_m]$$ where $$\mathbb{G}$$ acts through the regular representation. Let $$I=\{h \in \mathbb{Q}[y_1,\cdots,y_m] | h(g_1,\cdots,g_m)=0\}$$ and $$V_\mathbb{G} = \{ a \in \mathbb{Q}^m | h(a) = 0 \forall h \in I \}$$. Let now $$Q_1,\cdots,Q_r \in \mathbb{Q}[y_1,\cdots,y_m]$$ be given, such that $$Q_i \notin I$$ for $$i=1,\cdots,r$$. Let $$Q:=Q_1\cdots Q_r$$. Since $$I$$ is a prime ideal and $$Q_i \notin I$$, we must have $$Q \notin I$$, which means that there exists $$b \in \mathbb{Q}^n$$ such that \[ Q(g_1(b),\cdots,g_m(b)) \neq 0 \] that is $$\hat{f}(x_1,\cdots,x_n) = Q(g_1(x_1,\cdots,x_n),\cdots,g_m(x_1,\cdots,x_n))$$ is a reduced polynomial ( since $$\mathbb{Q}$$ has infinite elements) (see Algebra, Lang, p. 177, p. 312). Since the $$\sigma_i$$ by Artin (p. 311, Th. 12.2 Algebra, Lang) are algebraically independent and $$\hat{f}$$ is reduced, there exists $$\alpha \in K$$, such that \[ 0 \neq Q(g_1(\sigma_1(\alpha),\cdots,\sigma_n(\alpha)),\cdots,g_m(\sigma_1(\alpha),\cdots,\sigma_n(\alpha))) \] Let $$\alpha = \sum_{i=1}^n{\alpha_i \sigma_i(\theta)}$$ with $$\alpha_i \in \mathbb{Q}$$ Then we have for $$i=1,\cdots,m$$ that $$g_i(\sigma_1(\alpha),\cdots,\sigma_n(\alpha)) = p_i(\alpha_1,\cdots,\alpha_n)$$ is a rational number, where $$p_i$$ is some polynomial in $$\mathbb{Q}[x_1,\cdots,x_n]$$. But then also \[ 0 \neq Q(g_1(\sigma_1(\alpha),\cdots,\sigma_n(\alpha)),\cdots,g_m(\sigma_1(\alpha),\cdots,\sigma_n(\alpha))) \] \[ = Q(p_1(\alpha_1,\cdots,\alpha_n),\cdots, p_m(\alpha_1,\cdots,\alpha_n)) \] \[ = q(\alpha_1,\cdots,\alpha_n) \] for some polynomial $$q \in \mathbb{Q}[x_1,\cdots,x_n]$$. We can choose the $$\alpha_i$$ such that $$\alpha_i \neq \alpha_j$$ for $$i \neq j$$, otherwise replace $$\hat{\alpha}_j = \alpha_i + u$$ for some rational number $$u \neq 0$$ and we get the polynomial in u: \[ q(u) := q(\alpha_1,\cdots,\alpha_{j-1}, \alpha_i + u , \alpha_{j+1}, \cdots, \alpha_n) \] Since $$q(0) \neq 0$$, $$q$$ is not the zero polynomial in $$\mathbb{Q}[u]$$. By choosing $$u \neq 0$$ as a rational number to avoid the finitely many roots of $$q$$, we get $$q(u) \neq 0$$ and $$\hat{\alpha}_j = \alpha_i + u \neq \alpha_i$$. But then also $$\sigma_i(\alpha) \neq \sigma_j(\alpha)$$ for $$i \neq j$$, since $$\alpha$$ has pairwise distinct coordinates $$\alpha_k \neq \alpha_l$$ to the given basis $$\sigma_1(\theta),\cdots,\sigma_n(\theta)$$. The rational and separable polynomial \[ \prod_{i=1}^n (x -\sigma_i(\alpha)) \] is irreducible in $$\mathbb{Q}[x]$$, since the Galois group operates transitively on the roots. Let $$a_i := g_i(\sigma_1(\alpha),\cdots,\sigma_n(\alpha))$$ for $$i=1,\cdots,m$$. Then we have $$a:=(a_1,\cdots,a_m)$$ is an element of $$V_\mathbb{G}$$ and also \[ p(t,a_1,\cdots,a_m) = \sum_{i=0}^n (-1)^i s_i(a_1,\cdots,a_m) t^{n-i} \] \[ = \sum_{i=0}^n (-1)^i s_i(g_1(\sigma_1(\alpha),\cdots,\sigma_n(\alpha)),\cdots,g_m(\sigma_1(\alpha),\cdots,\sigma_n(\alpha))) t^{n-i} \] \[ = \sum_{i=0}^n (-1)^i e_i(\sigma_1(\alpha),\cdots,\sigma_n(\alpha)) t^{n-i} \] \[ = \prod_{i=1}^n (t -\sigma_i(\alpha)) \] is an irreducible polynomial in $$\mathbb{Q}[t]$$. Furthermore we have $$Q(a_1,\cdots,a_m) \neq 0$$, which means that $$Q_j(a_1,\cdots,a_m) \neq 0$$ for all $$j=1,\cdots,r$$, which was to be shown. This note can be downloaded from here (A note in Galois Theory, Orges Leka) as a pdf file.