Galois Theory
Definition. Let $K\supset F$ be a field extension. An automorphism of $\varphi: K \to K$ is an $F$-automorphism if it fixes elements of $F$ (i.e., $\varphi$ restricted to $F$ is the identity). The set of all $F$-automorphisms is denoted $G(K/F)$, and is called the Galois group of $K$ over $F$.
$G(K/F) = \{\varphi: K\to K \ |\ \varphi\ \text{is a field automorphism,}\ \varphi(x)=x,\ \forall x\in F \}$
Corollary. If $G$ is a finite group, the $G$ is the Galois group of some field extension.
Proof.
By Cayley's theorem, $G$ is a subgroup of $S_n$ for some $n\ge 1$. Let $K\subset F$ be a Galois extension with Galois group $S_n$ (which exists by last proposition), then by theorem 2, $K\supset K^{G}$ is Galois with $G(K/K^{G}) = G$.
Definition. Let $f\in F[x]$ be a polynomial, the Galois group of $f$ over $F$ is $G(K/F)$, where $K$ is the splitting field of $f$ over $F$.
Theorem. Let $f\in \mathbb{Q}[x]$ be irreducible with deg $f=p$, a prime, if $f$ has precisely two non-real roots in $\mathbb{C}$, then the Galois group of $f$ over $\mathbb{Q}$ is $S_p$. In particular, the splitting field of $f$ over $\mathbb{Q}$ has degree $p!$.
Proof.
Let $K$ be the splitting field of $f$ over $\mathbb{Q}$. If $\alpha$ is a root of $f$ in $K$, then $[\mathbb{Q}(\alpha):\mathbb{Q}]=p$. Therefore,
So $p$ divides $[K:\mathbb{Q}]=|G(K/\mathbb{Q})|$ by theorem 1. By Cauchy's theorem, there exists an element $g \in G$ of order $p$. Let $X$ be the set of $p$ roots of $f$ (Note that $f$ does not have multiple roots because $Char \mathbb{Q}=0$, see page 54 here), then we have seen that $G\hookrightarrow S_X = S_p$. Under this embedding, $g$ must map onto a $p$-cycle, because the order of its image must also be $p$.
Since $f$ has two non-real roots, they must be complex conjugates $\{z,\overline{z}\}$, let $\alpha_3, \dots, \alpha_p$ be the other real roots of $f$, then $K=\mathbb{Q}(z,\overline{z},\alpha_3,\dots,\alpha_p)$. The conjugation map $\sigma: z\mapsto \overline{z}$ is an element of $G(K/\mathbb{Q})$ of order 2, and must map onto a transposition $(ij)$ in $S_p$ under the embedding because it only permutes two roots.
Now, $G(K/S)$ as a subgroup of $S_p$ contains a transposition $(ij)$ and a $p$-cycle $(a_1 \dots a_p)$, so they generate the whole group $S_p$ (by homework 8).
Proposition. Let $G$ be a finite group of automorphisms of a field $K$, and let $F=K^{G}$ be the fixed field. Let $\{b_1,\dots,b_r\}$ be the orbit of an element $b=b_1\in K$ under the action of $G$. Then $b$ is algebraic over $F$, its degree is $r$, and its minimal polynomial over $F$ is $g(x) = (x-b_1)\dots (x-b_r)$.
Proof.
By Vieta's Formulas, we have
Proposition. Let $K\supset F$ be a field extension, and suppose $\alpha \in K$ is algebraic over $F$. Then for any $\sigma \in G(K/F)$, $\sigma(\alpha)$ is a root of the minimal polynomial of $\alpha$ over $F$ (hence a root of all polynomials over $F$ that has $\alpha$ as a root).
Remark. The proposition implies that $G(K/F)$ acts on the set $X$ of roots of the (minimal) polynomial so any $\sigma \in G(K/F)$ permutes roots in $X$.
Proof.
If $\alpha$ satisfies $\alpha^n+c_{n-1}\alpha^{n-1}+\dots + c_1\alpha+c_0=0$ with $c_i\in F$, then since $\sigma$ is an automorphism that fixes $F$, we have $\sigma(\alpha)^n+c_{n-1}\sigma(\alpha)^{n-1}+\dots + c_1 \sigma(\alpha)+c_0=0$, hence $\sigma(\alpha)$ is also a root of the same (minimal) polynomial.
See examples on page 3/(3-31).
Definition. Let $G\le AutK$ be a subgroup, the fixed field $K^G$ of $G$ is the subfield of $K$ given by
$K^G=\{x\in K\ |\ \sigma(x)=x,\ \forall \sigma\in G\}$
See examples on page 5/(3-31).
Definition. Let $K\supset F$ be a field extension, $K$ is said to be a Galois extension over $F$ is $K\supset F$ is algebraic and $K^{G}=F$, where $G=G(K/F)$.
Remark. In this course, we only consider finite field extensions so the algebraic extension condition is automatically satisfied.
We now state the two main theorems concerning Galois extensions.
Theorem 1. Let $K\supset F$ be a finite field extension, then $|G(K/F)| \le [K:F]$. Moreover, the following are equivalent:
(1) $K\supset F$ is Galois
(2) $|G(K/F)| = [K:F]$
If $Char F = 0$, then (1) and (2) are equivalent to
(3) $K\supset F$ is normal
(4) $K$ is the splitting field of a polynomial $f$ in $F[x]$
Theorem 2. (Fundamental Theorem of Galois Theory)
Let $K\supset F$ be a Galois extension, and set $G=G(K/F)$. Then there is a bijection between subfields $E$ with $F\subset E\subset K$ and subgroups $H$ with $1\subset H \subset G$ given by the correspondences
Let $K\supset F$ be a Galois extension, and set $G=G(K/F)$. Then there is a bijection between subfields $E$ with $F\subset E\subset K$ and subgroups $H$ with $1\subset H \subset G$ given by the correspondences
$E\quad \mapsto\quad G(K/E)$
$H \quad \mapsto \quad K^H$
which are inverses of each other.
Furthermore, we have
(1) the correspondences are inclusion reversing
(2) [K:E] = |H| and [E:F]=[G:H] if $E$ corresponds to $H$.
(3) $K\supset E$ is Galois, with Galois group $H=G(K/E)$
(4) $E\supset F$ is Galois if and only if $H\lhd G$, and in this case, we have $G(E/F)\cong G/H$
See remarks on page 8/(4-02)
Remark. Let $f\in F[x]$, and $K=F(\alpha_1,\dots, \alpha_2)$ be the splitting field of $f$ over $F$. We have seen that $G=G(K/F)$ acts on the set of roots $X=\{\alpha_1,\dots,\alpha_n\}$. Therefore, we have a permutation representation (recall this here) which is a group homomorphism:
$\varphi: G\to S_X$
$\sigma \mapsto \sigma_{|X}$
This permutation representation is faithful, i.e., $\varphi$ is injective and $G\hookrightarrow S_X$ ($G$ embeds in $S_X$).
See examples on page 9-12/(4-02)
The following example gives a field extension of degree $n!$ with Galois group $S_n$ for any $n\ge 1$.
Let $F$ be a field, $x_1,\dots,x_n$ be indeterminants. Let $K=F(x_1,\dots,x_n)$ be the field of all rational functions in variables $x_1,\dots,x_n$. Let $S_n$ act on $K$ by permuting the variables, i.e., $\sigma(f(x_1,\dots,x_n)=f(x_{\sigma(1)},\dots,x_{\sigma(n)})$, then this gives an $F$-automorphism of $K$.
Consider the fixed field $S = K^{S_n}=F(x_1,\dots,x_n)^{S_n}$ of all rational functions $f$ which are symmetric in $x_1,\dots,x_n$, that is, $f(x_{\sigma(1)},\dots,x_{\sigma(n)})=f(x_1,\dots,x_n)$ for all $\sigma \in S_n$, we want to study the extension $K\supset S$.
Note that the elementary symmetric polynomials $e_1,\dots,e_n$ are in $S$, which are defined by:
The following example gives a field extension of degree $n!$ with Galois group $S_n$ for any $n\ge 1$.
Let $F$ be a field, $x_1,\dots,x_n$ be indeterminants. Let $K=F(x_1,\dots,x_n)$ be the field of all rational functions in variables $x_1,\dots,x_n$. Let $S_n$ act on $K$ by permuting the variables, i.e., $\sigma(f(x_1,\dots,x_n)=f(x_{\sigma(1)},\dots,x_{\sigma(n)})$, then this gives an $F$-automorphism of $K$.
Consider the fixed field $S = K^{S_n}=F(x_1,\dots,x_n)^{S_n}$ of all rational functions $f$ which are symmetric in $x_1,\dots,x_n$, that is, $f(x_{\sigma(1)},\dots,x_{\sigma(n)})=f(x_1,\dots,x_n)$ for all $\sigma \in S_n$, we want to study the extension $K\supset S$.
Note that the elementary symmetric polynomials $e_1,\dots,e_n$ are in $S$, which are defined by:
$e_1=\sum_i x_i, e_2=\sum_{i<j} x_ix_j, e_3= \sum_{i<j<k} x_ix_jx_k, \dots, e_n = x_1x_2\dots x_n$
Note also that if $t$ is a formal variable and
$f(t)=(t-x_1)\dots (t-x_n)$
is a polynomial in variable $t$ with roots $x_1,\dots,x_n$, then its expansion has form:
Proposition:
(1) $S = F(e_1,\dots,e_n)$ and $[K:S]=n!$
(2) $K\supset S$ is Galois, with Galois group $S_n$
Proof.
$S_n\le G(K/S)$, so $|G(K/S) \ge |S_n| = n!$. Consider the polynomial
$f(t)=t^n-e_1t^{n-1}+e_2t^{n-2}+\dots +(-1)^ne_n$ (Vieta's formulas)
Proposition:
(1) $S = F(e_1,\dots,e_n)$ and $[K:S]=n!$
(2) $K\supset S$ is Galois, with Galois group $S_n$
Proof.
$S_n\le G(K/S)$, so $|G(K/S) \ge |S_n| = n!$. Consider the polynomial
$f(t)=\prod_{i=1}^{n} (t-x_i) = t^n-e_1t^{n-1}+\dots+(-1)^ne_n$ in $S[t]$
$K=F(x_1,\dots,x_n)$ is clearly the splitting field of $f$ over $F(e_1,\dots,e_n)$, and $S\supset F(e_1,\dots,e_n)$, so $[K:S]\le [K:F(e_1,\dots,e_n)]\le n!$ (by an early proposition). Thus, $n!\le |G(K/S)|\le [K:S]\le n!$ by theorem 1, so $|G(K/S)|=[K:S]=n!$. By theorem 1, $K\supset S$ is Galois, with Galois group $S_n$. Moreover, we obtain the fundamental theorem on Symmetric Functions:
$F(x_1,\dots,x_n)^{S_n} = F(e_1,\dots,e_n)$
Corollary. If $G$ is a finite group, the $G$ is the Galois group of some field extension.
Proof.
By Cayley's theorem, $G$ is a subgroup of $S_n$ for some $n\ge 1$. Let $K\subset F$ be a Galois extension with Galois group $S_n$ (which exists by last proposition), then by theorem 2, $K\supset K^{G}$ is Galois with $G(K/K^{G}) = G$.
Definition. Let $f\in F[x]$ be a polynomial, the Galois group of $f$ over $F$ is $G(K/F)$, where $K$ is the splitting field of $f$ over $F$.
Theorem. Let $f\in \mathbb{Q}[x]$ be irreducible with deg $f=p$, a prime, if $f$ has precisely two non-real roots in $\mathbb{C}$, then the Galois group of $f$ over $\mathbb{Q}$ is $S_p$. In particular, the splitting field of $f$ over $\mathbb{Q}$ has degree $p!$.
Proof.
Let $K$ be the splitting field of $f$ over $\mathbb{Q}$. If $\alpha$ is a root of $f$ in $K$, then $[\mathbb{Q}(\alpha):\mathbb{Q}]=p$. Therefore,
$[K:\mathbb{Q}]=[K:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=p[K:\mathbb{Q}(\alpha)]$
Since $f$ has two non-real roots, they must be complex conjugates $\{z,\overline{z}\}$, let $\alpha_3, \dots, \alpha_p$ be the other real roots of $f$, then $K=\mathbb{Q}(z,\overline{z},\alpha_3,\dots,\alpha_p)$. The conjugation map $\sigma: z\mapsto \overline{z}$ is an element of $G(K/\mathbb{Q})$ of order 2, and must map onto a transposition $(ij)$ in $S_p$ under the embedding because it only permutes two roots.
Now, $G(K/S)$ as a subgroup of $S_p$ contains a transposition $(ij)$ and a $p$-cycle $(a_1 \dots a_p)$, so they generate the whole group $S_p$ (by homework 8).
Proposition. Let $G$ be a finite group of automorphisms of a field $K$, and let $F=K^{G}$ be the fixed field. Let $\{b_1,\dots,b_r\}$ be the orbit of an element $b=b_1\in K$ under the action of $G$. Then $b$ is algebraic over $F$, its degree is $r$, and its minimal polynomial over $F$ is $g(x) = (x-b_1)\dots (x-b_r)$.
Proof.
By Vieta's Formulas, we have
$g(x) = x^r - e_1(b_1,\dots,b_r)x^{r-1}+\dots+(-1)^re_r(b_1,\dots,b_r)$
Since any automorphism $\sigma\in G$ permutes the elements in the orbit $O_b=\{b_1,\dots,b_r\}$, the coefficients of $g$ are invariant under $G$, so they belong to the fixed field $F=K^{G}$, and $g\in F[x]$.
Let $f$ be the minimal polynomial of $b$ over $F$, then $f\ |\ g$, but $\sigma(b)$ is also a root of $f$, for all $\sigma\in G$, so $b_i$ is a root of $f$ for all $i$, so $g\ |\ f$, hence $f=g$.
We now begin proving theorem 1. We start by showing that if $G$ is a finite group of automorphisms of a field $K$, and $F=K^{G}$ is the fixed field, then $[K:F]=[K:K^{G}]=|G|$.
Proposition. Let $K$ be a field, and $\sigma_1,\dots,\sigma_n\in Aut\ K$ be distinct. Suppose that $a_1,\dots, a_n\in K$ are such that
Proof.
Suppose that, on the contrary, there exist relations (1) with $a_i$'s in $K$, not all zero, by reindexing the $a_i$'s and $\sigma_i$'s, we can assume that the relation of minimal length is
Now $\sum_{i=1}^{m} a_i\sigma_i(cu)=0\Rightarrow \sum_{i=1}^m a_i\sigma_i(c)\sigma_i(u)=0$. Also, $\sigma_1(c)\sum_{i=1}^m a_i\sigma_i(u)=0$. Subtracting gives
But if we let $b_i=a_i(\sigma_i(c)-\sigma_1(c))$ for $i\ge 2$, this gives us a relation (1) with shorter length, a contradiction.
Theorem. Let $G=\{\sigma_1=1,\sigma_2,\dots,\sigma_n\}$ be a finite subgroup of $Aut\ K$, and set $F=K^{G}$, then $[K:F]=|G|$.
Proof.
Suppose $[K:F]< n$, let $w_1,\dots,w_m$ be a basis of $K$ over $F$, $m<n$. Consider the system of equations (3):
Now suppose $[K:G]>n$, then there are more than $n$ $F$-linearly independent elements, say $a_1,\dots,a_{n+1}$. The system of equations (4)
has a solution $b_1,\dots,b_{n+1}$, with $b_i$'s not all zero. Moreover, at least one $b_i$ is in $F$, because otherwise, the first equation ($\sigma_1=1$) contradicts that $a_i$'s are linearly independent.
Choose a solution of (4) with the minimal possible number of non-zero $b_i$'s, say $r$. After reordering, we may assume one solution is $x_1=b_1,\dots,x_r=b_r, x_j=0$ for $j>r$. Dividing by $b_r$, we may assume $b_r=1$. The solution now reads as follows:
Now if $K\supset F$ is a field extension, and $G=G(K/F)$, then $F\subset K^{G} \subset K$. Hence, assuming $K\supset F$ is finite, we have $|G|=[K:K^{G}]\le [K:F]$ (note that $|G|$ whenever $K\supset F$ is finite by homework 7, problem 1), with equality if and only if $F=K^{G}$, that is, when $K\supset F$ is Galois. This proves (i) $\Leftrightarrow$ (ii).
Separable Extensions
Recall that in positive characteristic, there are fields $F$ and irreducible polynomials $f\in F[x]$ with multiple roots in a splitting field (recall in this notes). We have to exclude those situations in order for theorem 1 to hold when $char\ F=p$.
Definition. Let $F$ be a field, a polynomial $f\in F[x]$ is called separable if it has no repeated roots in a splitting field. Let $K\supset F$ be a field extension, an algebraic element $\alpha\in K$ is called separable over $F$ if its minimal polynomial over $F$ is separable. The extension $K\supset F$ is called a separable extension if every element in it is separable over $F$.
Definition. A field $F$ is called perfect if every irreducible polynomial in $F[x]$ is separable.
Suppose that $char\ F=p>0$, recall that the $p$-th power map $x\overset{\varphi}{\mapsto} x^p$ is an injective field homomorphism. The image $\varphi(F)=F^{p}$ is therefore a subfield of $F$.
Theorem. The field $F$ is perfect if and only if char $F=0$ or char $F=p$ and $F=F^p$.
Proof.
Suppose char $F=0$, let $f\in F[x]$ be irreducible, then $f' \neq 0$, and deg $f'<$ deg $f$, so $(f',f)=1$, so $f$ does not have multiple roots, so $f$ separable and hence $F$ perfect.
If char $F=p$, suppose char $F=F^p$ and $f\in F[x]$ irreducible. If $f'\neq 0$, then $f$ is separable as above. Otherwise, we must have
Proof.
If $F$ is finite with char $F = p$, the $p$-th power map is both 1-1 and onto, so $F=Im\ \varphi=F^p$.
Example.
Let $t$ be a variable and $F=\mathbb{F}_2(t)$, then the polynomial $f(x)=x^2-t\in F[x]$ is irreducible but not separable. Indeed, $f$ has no root in $F$, because if $\alpha = \frac{p(t)}{q(t)}$ is a root, then $p^2(t)=tq^2(t)$, which is impossible since the LHS has even degree whereas the RHS has odd degree. However, let $\sqrt{t}$ denote a root of $f$ in some extension of $F$, then
Proposition. Suppose $K\supset F$ is a Galois finite extension, then $K\supset F$ is both normal and separable.
Proof.
Choose any $\alpha\in K-F$, by a proposition before, since $F=K^{G}$ with $G=G(K/F)$, we have the minimal polynomial of $\alpha$ over $F$ is
$g(x)=\prod_{j=1}^r(x-\sigma_j(\alpha))$
, where $\sigma_1(\alpha),\dots,\sigma_r(\alpha)$ are distinct elements of the orbit $G\cdot \alpha=\{\sigma(\alpha)\ |\ \sigma\in G\}$. Thus, $\alpha$ is separable over $F$. Moreover, all conjugates $\sigma_j(\alpha)$ also lie in $K$, so $K\supset F$ is both normal and separable.
We are now ready to prove theorem 1.
Theorem 1. Let $K\supset F$ be a finite field extension, then $|G(K/F)|\le[K:F]$. Moreover, the following are equivalent:
(i) $K\supset F$ is Galois extension.
(ii) |G(K/F)|=[K:F].
(iii) $K\supset F$ is normal and separable.
(iv) $K$ is the splitting field of a separable polynomial in $f\in F[x]$.
Proof.
Let $G=G(K/F)$. We always have $F\subset K^{G}\subset K$, therefore, $|G|=[K:K^{G}]<[K:F]$
(i) $\Leftrightarrow$ (ii): already proved.
(i) $\Rightarrow$ (iii): already proved.
(iii) $\Rightarrow$ (iv): we already proved that if $K\supset F$ normal, then $K$ is the splitting field of a polynomial in $F[x]$. If $K\supset F$ is also separable, then we can take $f$ to be separable.
It remains to show (iv) $\Rightarrow$ (ii). Recall the following result:
Theorem. Let $\sigma: F\to F'$ be a field isomorphism, $f\in F[x]$, and $\sigma f\in F'[x]$. Let $K$ be the splitting field of $f$ over $F$, and $K'$ the splitting field of $\sigma f$ over $F'$. Then there exists an isomorphism $\widetilde{\sigma}: K\to K'$ which extends $\sigma$.
(iv) $\Rightarrow$ (ii): Suppose $K$ is the splitting field of the separable polynomial $f\in F[x]$. We will prove $K\supset F$ is Galois by induction on $[K:F]$. Clearly, the base case is true, so we may assume $[K:F]>1$.
Let $p(x)$ be an irreducible factor of $f(x)$ of degree $>1$ (if all irreducible factors are of degree 1, then all roots are in $F$, hence $[K:F]=1$), let $\alpha$ be a root of $p$ and set $E=F(\alpha)$, so $F\subsetneq E\subset K$. Now, $K$ is also the splitting field of $f$ over $E$, and $f$ is separable over $E$, but $[K:E]<[K:F]$, so by induction hypothesis, $K\supset E$ is Galois. Let $H=G(K/E)\le G(K/F)=G$
Let $\alpha_1,\dots,\alpha_r$ be the (distinct) roots of $p$, so deg $p=r$ and hence $[E:F]=r$. Apply previous theorem to the isomorphism:
Recall that $\sigma_i$ extends $F$ and sends $\alpha$ to $\alpha_i$. There exist $\tau_1,\dots,\tau_r\in G$ such that $\tau_i(\alpha)=\alpha_i$ for $1\le i\le r$.
Claim: The $H$-cosets $\tau_i H$ are distinct.
Proof. $\tau_i H=\tau_j H \Rightarrow \tau_i^{-1}\tau_j\in H\Rightarrow \tau_i^{-1}\tau_j(\alpha)=\alpha\Rightarrow \tau_j(\alpha)=\tau_i(\alpha)\Rightarrow \alpha_j=\alpha_i$ $\Rightarrow i=j$.
We conclude that $[K:F]=[K:E][E:F]=|H|r\le |H||G:H|=|G|\le [K:F]$. Thus, $|G|=[K:F]$, as desired.
Theorem 2. Let $K\supset F$ be a finite Galois extension and set $G=G(K/F)$. Then there is a bijection
$K$ $1$
, where the two maps are inverse to each other. Moreover, we have
(i) The above correspondence are inverse reversing.
(ii) $[K:E]=|H|$ and $[E:F]=|G:H|$ if $E\leftrightarrow H$.
(iii) $K\supset E$ is always Galois, with $H=G(K/E)$
(iv) $E\supset F$ is Galois if and only if $H\lhd G$. In this case, $G(E/F)\cong G/H$.
Proof (i)-(iii).
Consider the map $\varphi$ from subgroups of $G$ to subfields of $K$ defined by $\varphi(H)-K^H$.
$\varphi$ is 1-1: Suppose $K^{H_1}=K^{H_2}=E$, then each $H_i$ is a subgroup of $G(K/E)$, while $|G(K/E)|=|G(K/K^{H_i})|=|H_i|\Rightarrow $H_1=H_2=G(K/E)$, so $\varphi$ is 1-1.
$\varphi$ is onto: Since $K\supset F$ is Galois, $K$ is a splitting field of a separable $f\in F[x]$. Therefore, for any field $E$ with $F\subset E\subset K$, $K$ is a splitting field of $f\in E[x]$, so $K\supset E$ is Galois. It also follows that
To prove (iv), we need some preliminary results
Lemma. Let $F$ be an infinite field and $V$ a vector space over $F$. If $H_1,\dots,H_n$ are proper subspaces of $V$, then $V\neq \cup_{i=1}^n H_i$.
Proof.
We may assume without loss of generality that $H_1\not\subset \cup_{i=2}^n H_i$ (otherwise, we just omit $H_1$ in the union).
Choose $h_0\notin H_1$, $h_1\in H_1$, but $h_1\notin \cup_{i=2}^n H_i$.
Claim 1: The line $L=\{h_0+\lambda h_1\mid \lambda\in F\}=h_0+Fh_1$ is disjoint from H_1. [Indeed, if $h_0+\lambda h_1\in H_1$, then $h_0\in H_1$, a contradiction].
Claim 2: $L$ meets the subspace $H_j$ in at most one vector for $i\ge2$ . [If $h_0+\lambda h_1, h_0+\mu h_1$ both meets $H_j$ for some $j\ge 2$, then their difference $(\lambda-\mu)h_1\in H_j\Rightarrow h_1\in H_j$, a contradiction].
The two claims above imply that $L\cap (\cup_{i=1}^n H_i) $ is a finite set. But since $F$ is an infinite field, $L$ has infinitely many elements. We conclude that $V\neq \cup_{i=1}^n H_i$.
Proposition. Let $K\supset F$ be a finite field extension, then there exists $\alpha\in K$ such that $K=F(\alpha)$ if and only if there are only finitely many fields $E$ such that $F\subset E\subset K$. In this case, $\alpha$ is called a primitive element.
Proof.
$(\Leftarrow)$
If $F$ is finite, then so is $K$. We have shown that $(K^*,\cdot)$ is generated by some single element $\alpha$, then $K=F(\alpha)$.
We may therefore assume $F$ is infinite, suppose there are finitely many fields $E_1,\dots, E_d$ with $F\subset E_i\subsetneq K$, the previous lemma implies that there exists an $\alpha\in K-\cup_{i=1}^d E_i$. This $\alpha$ is a primitive element, because we have $F(\alpha)\neq E_i$ for all $i$.
$(\Rightarrow)$
Assume $K=F(\alpha)$ for some $\alpha\in K$, and let $f$ be the minimal polynomial of $\alpha$ over $F$. Given any subfield $E$ with $F\subset E\subset K$, let $g_E(x)$ be the minimal polynomial of $\alpha$ over $E$. Then $g_E\mid f$ in $E[x]$.
Clearly, there are only finitely many such $g_E$. We thus get a mapping:
Let $E$ be an intermediate field, and $E_0$ denote the subfield of $E$ generated over $F$ by coefficients of $g_E$. Then $g_E$ has coefficients in $E_0$ and is irreducible in $E_0$ since it is irreducible in $E[x]$. Therefore $[E_0(\alpha):E_0]=\text{deg }g_E=[E(\alpha):E]$. But we have $F(\alpha)=K=E_0(\alpha)=E(\alpha)\supset E\supset E_0$, hence $[E(\alpha):E]=[E_0(\alpha):E]$, now we compute
Corollary (Primitive Element Theorem). Suppose $K\supset F$ is a finite Galois extension, then $K=F(\alpha)$ for some $\alpha\in K$.
Proof.
According to the previous proposition, it suffices to show there are only finitely many intermediate fields $E$ with $F\subset E\subset K$. By the Galois correspondence, such subfields $E$ are in correspondence with subgroups of $G(K/F)$, but since $G(K/F)$ is finite, there are only finitely many subgroups.
Remark. The theorem holds for any finite separable extension $K\supset F$ in general, as any such extension is contained in a Galois extension of $F$. Since $K\supset F$ is a finite extension, $K=F(a_1, \dots, a_k)$ for some $a_i\in F$. Let $f_i$ be their respective
minimal polynomials over $F$ and set $f=\prod f_i$. Since $K\supset F$ separable,each $f_i$ does not have multiple roots. We can assume $f$ is separable by removing repeated $f_i$'s. Let $L$ be the splitting field of $f$ over $F$, then $L\supset K\supset F$, and $L$ is Galois over $F$. By previous corollary, there are only finitely many intermediate fields between $F$ and $L$, hence so does $K\supset F$, and thus $K=F(\alpha)$ for some $\alpha\in F$.
We are now ready to prove part (iv). Let $K\supset F$ be a finite Galois extension, with $G=G(K/F)$, and $K\supset E\supset F$.
Lemma. $E\supset F$ is Galois if and only if $\forall\ \sigma\in G$, $\sigma(E)\subset E$ (i.e. $E$ is stable under the action of $G$).
Proof.
($\Rightarrow$) If $E\supset F$ is Galois, it is a finite separable extension. The primitive element theorem implies that $E=F(\alpha)$. Moreover, $E\supset F$ is normal, so the minimal polynomial $f$ of $\alpha$ over $F$ has all roots in $E$. But for any $\sigma\in G$, $\sigma(\alpha)$ is a root of $f$. Hence, $\sigma(E)=\sigma(F(\alpha))=F(\sigma(\alpha))\subset E$.
($\Leftarrow$) Since $K\supset F$ is separable, $E\supset F$ is also separable. By the primitive element theorem, $E=F(\alpha)$ for some $\alpha\in E$. Our assumption gives that $\sigma(\alpha)\in E$ for all $\sigma\in G$.
Let $g(x)=\prod_{\sigma\in G} (x-\sigma(\alpha))\in F[x]$. Then $\alpha=id(\alpha)$ is a root of $g$ and all roots of $g$ lie in $E$. So $E$ is a splitting field of $g$ over $F$, and hence $E\supset F$ is normal (by a corollary). Since $E\supset F$ is both normal and separable, we conclude that $E\supset F$ is Galois.
Proof (iv).
Set $H=G(K/E)$, we want to show $E\supset F$ is normal $\Leftrightarrow H\lhd G$, and in this case, $G(E/F)\cong G/H$.
As above, since $K\supset F$ separable, so is $E\supset F$, and hence $E=F(\alpha)$ for some $\alpha\in E$. Using the previous lemma, $E\supset F$ is Galois if and only if $\sigma(\alpha)\in E,\ \forall\ \sigma\in G$. Since $E=K^H=F(\alpha)$, this occurs iff
Solvable Groups
In this section, we recall some theorems regarding solvable and simple groups. We omit the proofs and examples (one can refer to notes from Math 600 or 404).
Definition. A group is solvable if there exists a chain of subgroups $G=G_0\rhd G_1\rhd\dots\rhd G_k={1}$, where each $G_i\lhd G_{i-1}$ and each quotient $G_i/G_{i+1}$ is abelian. The latter condition is equivalent to having a series with cyclic quotients.
Proposition. (i) $G'\lhd G$ and $G/G'$ is abelian, with $G'=[G,G]$.
(ii) If $H\lhd G$ with $G/H$ abelian, then $G'\subset H$
Proposition. $G$ is solvable if and only if $G^{(k)}=1$ for some $k$, with $G^{(0)}=G, G^{(1)}=G'$, $G^{(2)}=G^{(1)}$$'$, $\dots$
Corollary. Subgroups and quotients of solvable groups are solvable.
Theorem. $A_n$ is simple for $n\ge 5$.
Corollary. $S_n$ is not solvable for $n\ge 5$.
Proof.
If $S_n$ solvable, then so is $A_n$, $A_n'\lhd A_n$ with $A_n' \subsetneq A_n$, hence $A_n'=1$, so $A_n$ must be abelian, which is absurd.
Applications of Galois Theory
We have seen that there are formulas for roots of quadratic and cubic polynomials which involve radicals. We now explore a more general situation, and will show that a polynomial equation $f(x)=0$ "is solvable by radicals", which we will make precise soon, if and only if the Galois group of $f$ is a solvable group.
We restrict our discussion to fields of characteristic 0. First we will consider simple radical extensions, which are the extensions obtained by adjoining to a field $F$ the $n$-th root $\sqrt[n]{\alpha}$ for some $\alpha\in F$. As all roots of $x^n-a,\ a\in F$ differ by factors which are n-th root of unity, such an extension will be Galois if and only if it contains n-th roots of unity.
Definition. An extension $K\supset F$ is cyclic if it is Galois with cyclic Galois group.
Proposition A. Let $F$ be a field which contains the n-th roots of unity, and $a\in F$, then the extension $F(\sqrt[n]{a})$ is cyclic, with degree dividing $n$.
Proof.
$K=F(\sqrt[n]{a})\supset F$ is Galois since $K$ is the splitting field of $x^n-a\in F[x]$. For any $\sigma\in G(K/F)=G$, $\sigma(\sqrt[n]{a})$ is a root of $x^n-a$, hence
Also, $\ker\varphi=\{\sigma\in G\mid \sigma(\sqrt[n]{a})=\sqrt[n]{a}\}=1$, hence $\varphi$ is 1-1, and $G$ embeds in $\mu_n$, a cyclic group of order $n$.
Proposition B. Let $K\supset F$ be a cyclic extension of degree $n$ over a field $F$ which contains the n-th roots of unity. Then $K=F(\sqrt[n]{a})$ for some $a\in F$.
Proof.
For $\alpha\in F$ and $\zeta$ a root of unity in $F$, define the Lagrange resolvent $(\alpha,\zeta)\in K$ by
Now, let $\zeta$ be a primitive n-th root of unity, i.e., $\mu_n=(\zeta)$. By a previous proposition, $1,\sigma,\sigma^2,\dots,\sigma^{n-1}$ are "linearly independent", so $\exists\ \alpha\in K$ with $(\alpha,\zeta)\neq 0$, because otherwise we will have $\zeta=0$.
Iterating equation (1) we get $\sigma^i(\alpha,\zeta)=\zeta^{-i}(\alpha,\zeta)$, so $\sigma^i$ does not fix $(\alpha,\zeta)$ for any $i=1,2,\dots, n-1$. We conclude that $(\alpha,\zeta)$ does not lie in any proper subfield of $K$, because otherwise, the subfield it lies in must be the fixed field of some subgroup of $G$, so $(\alpha,\zeta)$ must be fixed by some elements of $G$. Therefore, $K=F(\alpha,\zeta)$ (recall this in the proof of the primitive element theorem), and since $a=(\alpha,\zeta)^n\in F$, $K=F(\sqrt[n]{a})$.
Definition. An element $\alpha$ which is algebraic over $F$ can be expressed in radicals if $\alpha$ lies in a field $K$ that can be obtained via a series of simple radical extensions:
From now on, we will omit most proofs. Check professor's note for detailed proofs.
Lemma 1. If $K\supset F$ is a root extension, then $K$ is contained in some root extension which is Galois over $F$ and such that each intermediate extension is cyclic.
Theorem (Galois). The polynomial $f\in F[x]$ can be solved by radicals if and only if the Galois group of $f$ is a solvable group.
The proof of the converse requires the two results below:
Lemma 2. Let $K\supset F$ be a Galois extension, and $F'\supset F$ be any finite extension, then $KF'\supset F$ is Galois, and
Corollary. Suppose $K/F$ is Galois and $F'/F$ is any finite extension, then we have
Corollary. The general equation of degree $n\ge 5$ cannot be solved by radicals.
Proof. $S_n$ is not solvable for $n\ge 5$.
Galois Group as a Permutation Group of the Roots.
Set up: Consider a monic polynomial $f\in F[x]$ of positive degree $n$, with $\alpha_1,\dots,\alpha_n$ the roots of $f$, and $K$ the splitting field of $f$ over $F$. Then the Galois group of $f$ over $F$ $G(K/F)$ is a subgroup of $S_n=S(\{\alpha_1,\dots,\alpha_n\})$.
Theorem. Suppose the $f\in F[x]$ does not have multiple roots. Then $f$ is irreducible over $F$ if and only if $G$ is a transitive group of the roots $\alpha_i$, $i=1,2,\dots,n$, that is, for any two roots $\alpha_i,\alpha_j$, there exists $\sigma\in G$ with $\sigma(\alpha_i)=\sigma(\alpha_j)$.
Examples.
1) Suppose $f\in F[x]$ is a cubic irreducible polynomial, then its Galois group is a transitive subgroup of $S_3$, hence either $A_3$ or $S_3$.
2) It's an exercise (see homework) to show that the only transitive subgroups of $S_4$ are
(i) $S_4$ (ii) $A_4$ (iii) $V$, the Klein 4-group
(iv) $C:=\{1,(1234),(13)(24),(1432)\}\cong C_4$ and its conjugates
(v) $D:= V\cup \{(12),(34),(1423),(1324)\}\cong D_8$, which is a 2-sylow subgroup of $S_4$, and its conjugates.
Theorem. Let $F$ be a field with Char $F\neq 2$, and $f\in F[x]$ with distinct roots $\{\alpha_1,\dots,\alpha_n\}$ in a splitting field $K/F$. Let $G=G(K/F)$ and
Corollary. The Galois group of $f$ over $F$ is a subgroup of $A_n$ if and only if the discriminant $d=\prod_{1\le i<j\le n}(\alpha_i-\alpha_j)^2$ is the square of an element in $F$.
In addition, we have the formula of $d$ as below
See examples here (one another application of the discriminant is to detect multiple roots of $f$).
Cyclotomic Polynomials and Extensions
Definition. Let $\mu_n$ denote the group of n-th roots of unity in $\mathbb{C}$, then $\mu_n\cong C_n$. Let $\mathbb{Q}(\zeta_n)$ be the field by adjoining $\mathbb{Q}$ the elements of $\mu_n$, so the splitting field of $x^n-1=0\in \mathbb{Q}[x]$. Here $\zeta_n=e^{2\pi i/n}$. We call $\mathbb{Q}(\zeta_n)$ the n-th cyclotomic field.
Remark. $\mu_d\mid \mu_n \Leftrightarrow d\mid n$.
Definition. The n-th cyclotomic polynomial $\Phi_n(x)$ is the polynomial whose roots are the primitive n-th roots of unity (the generators of $\mu_n$), each with multiplicity one, so
Clearly, deg $\Phi_n=\varphi(n)$. Recall that $\varphi$ is multiplicative and $\varphi(p^k)=p^{k-1}(p-1)$.
Proposition. $x^n-1 = \prod_{d\mid n} \Phi_d(x)$. Comparing degrees, we have $n=\sum_{d\mid n} \varphi(d)$.
Remark. Using this formula, we can compute $\Phi_n(x)$ recursively, see examples on page 3 here.
One example is that we have $\Phi_p(x)=\frac{x^p-1}{x-1}$, recall this is irreducible by Eisenstein's criterion.
Proposition. The cyclotomic polynomials $\Phi_n(x)$ is a monic polynomial of degree $\varphi(n)$ with integer coefficients.
Remark. It is known that there exist cyclotomic polynomials with arbitrary large coefficients (so not just $\pm 1$).
Theorem. $\Phi_n(x)$ is irreducible in $\mathbb{Z}[x]$ hence in $\mathbb{Q}[x]$, so it is the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$.
Corollary. $[\mathbb{Q}(\zeta_n)):\mathbb{Q}]=\varphi(n)$, for any $n\ge 1$.
Theorem. The Galois group $G(\mathbb{Q}(\zeta_n))/\mathbb{Q})$ is isomorphic to the group of units $(\mathbb{Z}/n\mathbb{Z})^{\times}$, given by
Remark. The theorem implies that $\mathbb{Q}(\zeta_n)\supset \mathbb{Q}$ is abelian extension. In fact, cyclotomic extensions are rich enough to contain all finite abelian extensions of $\mathbb{Q}$.
Theorem. Let $G$ be any finite abelian group, then there exists a subfield $K$ of a cyclotomic field with $G(K/\mathbb{Q})\cong G$.
The proof uses two facts (see p7 here):
1) structure theorem for finite abelian groups.
2) there are infinitely many primes with $p\equiv 1$ (mod m) for any $m\ge 2$.
Theorem (Kronecher-Weber). Let $K/\mathbb{Q}$ be a finite abelian Galois extension, then $K$ is contained in a cyclotomic extension of $\mathbb{Q}$
Constructibility of a regular n-gon Cont'd
Theorem (Gauss). A regular n-gon can be constructed if and only if $\varphi(n)=2^k$ for some $k\ge 1$.
Corollary. The regular n-gon is constructible if and only if $n=2^rp_1p_2\dots p_m$ with $p_i$'s distinct Fermat primes.
Algebraic Independence and Transcendence Degree
Definition. Let $K\supset F$ be a field extension and suppose $\alpha_i$ for $1\le i\le n$ are $n$ elements of $K$. We have an evaluation map (in fact a homomorphism):
Example. A singleton set $\{\alpha\}$ is algebraically independent iff $\alpha$ is transcendental over $F$.
We say a subset $S$ of $K$ is algebraically independent over $F$ if any finite subset of $S$ is algebraically independent over $F$.
Proposition. If $K\supset F$ is a field extension and $A=\{\alpha_1,\dots,\alpha_n\}\subset K$ is algebraically independent over $F$, then the evaluation map can be extended to a unique field isomorphism:
Lemma 1. Let $K\supset F$ is a field extension and $\alpha_1,\dots,\alpha_n\in K$. Let $F_0=F$, and $F_i=F(\alpha_1,\dots,\alpha_i)$ for $1\le i\le n$, then $A=\{\alpha_1,\dots,\alpha_n\}$ is algebraically independent over $F$ if and only if $\alpha_i$ is transcendental over $F_{i-1}$ for all $1\le i\le n$.
Definition. Let $K$ be a field, and $F\subset K$ a subfield. A transcendental basis of $K$ over $F$ is an algebraically independent set $\mathcal{B}$ of $K/F$ which is maximal under inclusion.
Proposition. Suppose $K\supset F$ is a field extension and $S\subset K$ is any subset. Then $S$ is a transcendental basis of $K/F$ iff $S$ is algebraically independent over $F$ and $K\supset F(S)$ is an algebraic extension.
Theorem. Let $K\supset F$ be a field extension, $A\subset K$ a subset such that the extension $K\supset F(A)$ is algebraic, and $C\subset A$ a subset that is algebraically independent over $F$. Then there exists a transcendental basis $\mathcal{B}$ of $K/F$ with $C\subset \mathcal{B}\subset A$.
Definition. Let $K\supset F$ be any field extension, we say $K$ is finitely generated over $F$ if there exist finitely many $\alpha_i\in K$, $1\le i\le n$, such that $K=F(\alpha_1,\dots,\alpha_n)$. $K$ has finite transcendental degree over $F$ if there exists a subfield $E$ with $K\supset E\supset F$ such that $K\supset E$ is algebraic and $E\supset F$ is finitely generated.
Theorem. Suppose $K\supset F$ is a field extension, and $C=\{c_1,\dots,c_r\}$ is a subset of $K$ that is algebraically independent over $F$. Assume that $A=\{\alpha_1,\dots,\alpha_s\}\subset K$ is such that $K\supset F(A)$ is algebraic. Then $r\le s$ and there exists a set $D$ with $C\subset D\subset A\cup C$ such that $|D|=s$ and $K\supset F(D)$ is algebraic.
Corollary. If $K\supset F$ is a field extension, and $\mathcal{B}_1$ and $\mathcal{B}_2$ are two transcendental bases of $K/F$, then either $\mathcal{B}_1$ and $\mathcal{B}_2$ are both infinite, or $|\mathcal{B}_1|=|\mathcal{B}_2|$.
Theorem (Luroth). Suppose $F(x)\supset F$ is a simple extension and $x$ is transcendental over $F$. Let $E$ be a subfield with $F\subset E\subset F(x)$. Then $E\supset F$ is a simple extension.
See more examples here.
Traces and Norms
Let $K\supset F$ be a field extension, with $[K:F]=n$. Then any element $\alpha\in K$ defines an $F$-linear map $T_{\alpha}: K\to K$ by $T_{\alpha}(x)=\alpha x,\forall x\in K$.
Definition. The trace, norm and characteristic polynomial of $\alpha$ (in the setting of $K/F$) are defined as below
See examples on page7-8 here.
Theorem (Hilbert's theorem 90). Let $K$ be a cyclic extension of $F$ and let $G(K/F)=(\sigma)$. If $\alpha\in K$, then
(i) $N(\alpha)=1\Leftrightarrow\ \exists \beta\in K$ with $\alpha = \frac{\beta}{\sigma(\beta)}$.
(ii) $Tr(\alpha)=0\Leftrightarrow\ \exists \gamma\in K$ with $\alpha =\gamma - \sigma(\gamma)$.
Proposition. Let $K\supset F$ be a finite extension of fields, $\alpha\in K$ and $f$ be the minimal polynomial of $f$ over $F$. If $[K:F(\alpha)]=m$, then we have $p_{\alpha}(x)=f(x)^m$.
Corollary. Suppose that the roots of the minimal polynomial of $\alpha$ are $\alpha_1,\dots,\alpha_n$ in some splitting field containing $K$, and $[K:F(\alpha)]=m$, then we have
See an example on page 10 here.
Suppose $K\supset F$ is a finite, separable extension, and let $\overline{F}$ be an algebraic closure of $F$. Recall that an $F$-isomorphism $\sigma: K\to \overline{F}$ is a map such that $\sigma |_F=id$ (also recall that any homomorphism of fields is injective). We claim that the set $\Sigma$ of all $F$-homomorphisms is finite. Indeed, since $K\supset F$ is finite separable, by the theorem of primitive element, $K=F(\alpha)$ for some $\alpha\in K$. Let $f\in F[x]$ be the minimal polynomial of $\alpha$ over $F$, with $n$ distinct roots $\alpha_1=\alpha, \dots, \alpha_n$, then $[K:F]=n$. Then any $\sigma \in \Sigma$ is determined by $\sigma(\alpha)$, and we must have $\sigma(\alpha)=\alpha_i$ for some $1\le i\le n$. Moreover, any such choice is an element of $\Sigma$, so in fact $|\Sigma|=n$.
Proposition. Let $K\supset F$ and $\Sigma$ as above, then for any element $\alpha\in K$, we have
Fact. Suppose $K\supset F$ finite separable, and $E$ is a field with $K\supset E\supset F$, then
We now use the norm to get a nice formula for the discriminant of a polynomial.
Proposition. Let $f(x)$ be a monic irreducible polynomial in $F[x]$, and $\alpha$ a root of $f$ in some splitting field of $f$, then
We now begin proving theorem 1. We start by showing that if $G$ is a finite group of automorphisms of a field $K$, and $F=K^{G}$ is the fixed field, then $[K:F]=[K:K^{G}]=|G|$.
Proposition. Let $K$ be a field, and $\sigma_1,\dots,\sigma_n\in Aut\ K$ be distinct. Suppose that $a_1,\dots, a_n\in K$ are such that
$\sum_{i=1}^{n}a_i\sigma_{i}(u)=0,\quad \forall\ u\in K$ (1)
then $a_1=a_2=\dots=a_n=0$ ($\sigma_i(u)$'s are $K$-linearly independent for any $u\in K$)Proof.
Suppose that, on the contrary, there exist relations (1) with $a_i$'s in $K$, not all zero, by reindexing the $a_i$'s and $\sigma_i$'s, we can assume that the relation of minimal length is
$a_1\sigma_1(u)+\dots +a_m\sigma_m(u)=0$
Clearly, $m\ge 2$, because otherwise, we have $a_1\sigma_1(u)=0\ \forall u\in K\Rightarrow a_1=0$. There exists a $c\in K$ such that $\sigma_1(c)\neq \sigma_m(c)$ since the automorphisms are distinct.Now $\sum_{i=1}^{m} a_i\sigma_i(cu)=0\Rightarrow \sum_{i=1}^m a_i\sigma_i(c)\sigma_i(u)=0$. Also, $\sigma_1(c)\sum_{i=1}^m a_i\sigma_i(u)=0$. Subtracting gives
$\sum_{i=2}^{m} a_i(\sigma_i(c)-\sigma_1(c))\sigma_i(u)=0,\quad \forall\ u\in K $ (2)
But if we let $b_i=a_i(\sigma_i(c)-\sigma_1(c))$ for $i\ge 2$, this gives us a relation (1) with shorter length, a contradiction.
Theorem. Let $G=\{\sigma_1=1,\sigma_2,\dots,\sigma_n\}$ be a finite subgroup of $Aut\ K$, and set $F=K^{G}$, then $[K:F]=|G|$.
Proof.
Suppose $[K:F]< n$, let $w_1,\dots,w_m$ be a basis of $K$ over $F$, $m<n$. Consider the system of equations (3):
$\sigma_1(w_1)x_1+\dots+\sigma_n(w_1)x_n=0$
$\vdots$
$\sigma_1(w_m)x_1+\dots+\sigma_n(w_m)x_n=0$
There are more unknowns than equations, so there is a non-zero solution, say $b_1,\dots,b_n,\ b_i\in K$. Given any $u\in K$, write $u=\sum_1^m a_iw_i$ for some $a_i\in F$. Multiplying the $i$-th equation in (3) by $a_i$ and adding them up gives
$\sigma_1(u)b_1+\dots+\sigma_n(u)b_m=0,$ $\forall\ u\in K$
with $b_i$'s not all zero, a contradiction to the previous proposition. So $[K:F]\ge n$.Now suppose $[K:G]>n$, then there are more than $n$ $F$-linearly independent elements, say $a_1,\dots,a_{n+1}$. The system of equations (4)
$\sigma_1(a_1)x_1+\dots+\sigma_1(a_{n+1})x_{n+1}=0$
$\vdots$
$\sigma_1(a_1)x_1+\dots+\sigma_n(a_{n+1})x_{n+1}=0$
has a solution $b_1,\dots,b_{n+1}$, with $b_i$'s not all zero. Moreover, at least one $b_i$ is in $F$, because otherwise, the first equation ($\sigma_1=1$) contradicts that $a_i$'s are linearly independent.
Choose a solution of (4) with the minimal possible number of non-zero $b_i$'s, say $r$. After reordering, we may assume one solution is $x_1=b_1,\dots,x_r=b_r, x_j=0$ for $j>r$. Dividing by $b_r$, we may assume $b_r=1$. The solution now reads as follows:
$\sigma_i(a_1)b_1+\dots+\sigma_i(a_{r})=0$ for $i=1,\dots,n$ (5)
Say $b_1\notin F$, then there exists $k$ such that $\sigma_k(b_1)\neq b_1$ (otherwise, $b_1$ is fixed by $G$, hence in $F=K^{G}$). Applying $\sigma_k$ to equation (5) gives
$\sigma_k\sigma_i(a_1)\sigma_k(b_1)+\dots+\sigma_k\sigma_i(a_{r})=0$ for $i=1,\dots,n$ (6)
But $\{\sigma_k\sigma_1,\dots,\sigma_k\sigma_n\}$ is equal to $\{\sigma_1,\dots,\sigma_n\}$ since $G$ is a group. Therefore, by reindexing $j\leftrightarrow ki$, we have (6) written as
$\sigma_j(a_1)\sigma_k(b_1)+\dots+\sigma_j(a_{r})=0$ for $j=1,\dots,n$ (7)
Subtracting (7) from (5) gives
$\sigma_i(a_1)(b_1-\sigma_k(b_1))+\dots+\sigma_i(a_{r-1})(b_{r-1}-\sigma_k(b_{r-1})=0$ for $i=1,\dots,n$ (8)
Now equation (8) exhibits a solution to the system (4) of length shorter than $r$, not identically zero, because $b_1-\sigma_k(b_1)\neq 0$, a contradiction. Therefore, $[K:F]=|G|=n$.Now if $K\supset F$ is a field extension, and $G=G(K/F)$, then $F\subset K^{G} \subset K$. Hence, assuming $K\supset F$ is finite, we have $|G|=[K:K^{G}]\le [K:F]$ (note that $|G|$ whenever $K\supset F$ is finite by homework 7, problem 1), with equality if and only if $F=K^{G}$, that is, when $K\supset F$ is Galois. This proves (i) $\Leftrightarrow$ (ii).
Separable Extensions
Recall that in positive characteristic, there are fields $F$ and irreducible polynomials $f\in F[x]$ with multiple roots in a splitting field (recall in this notes). We have to exclude those situations in order for theorem 1 to hold when $char\ F=p$.
Definition. Let $F$ be a field, a polynomial $f\in F[x]$ is called separable if it has no repeated roots in a splitting field. Let $K\supset F$ be a field extension, an algebraic element $\alpha\in K$ is called separable over $F$ if its minimal polynomial over $F$ is separable. The extension $K\supset F$ is called a separable extension if every element in it is separable over $F$.
Definition. A field $F$ is called perfect if every irreducible polynomial in $F[x]$ is separable.
Suppose that $char\ F=p>0$, recall that the $p$-th power map $x\overset{\varphi}{\mapsto} x^p$ is an injective field homomorphism. The image $\varphi(F)=F^{p}$ is therefore a subfield of $F$.
$F^p = \{x^p\ |\ x\in F\}$
An element of $F$ is in $F^p$ if and only if it's a $p$-th root in $F$.Theorem. The field $F$ is perfect if and only if char $F=0$ or char $F=p$ and $F=F^p$.
Proof.
Suppose char $F=0$, let $f\in F[x]$ be irreducible, then $f' \neq 0$, and deg $f'<$ deg $f$, so $(f',f)=1$, so $f$ does not have multiple roots, so $f$ separable and hence $F$ perfect.
If char $F=p$, suppose char $F=F^p$ and $f\in F[x]$ irreducible. If $f'\neq 0$, then $f$ is separable as above. Otherwise, we must have
$f(x)= \sum_{r=0}^n a_rx^{pr}, a_r\in F$
Because $F^p=F$, there exists $b_r^p=a_r$ for each $r$, then
$f(x)= \sum_{r=0}^n b_r^px^{pr}=\sum_{r=0}^n (b_rx^{r})^p=(\sum_{r=0}^n b_rx^{r})^p$
, which contradicts the fact that $f$ is irreducible, so we must have $f'=0$, hence $F$ perfect.
Conversely, suppose $F$ is perfect. If char $F=p$, let $a\in F$ and consider the polynomial $x^p-a\in F[x]$, if it has a root $b$, then $a=b^p\in F^p$. Otherwise, let $f$ be a monic irreducible factor of $x^p-a$ in $F[x]$, and let $K\supset F$ be an extension in which $f$ has a root $b$. In $K[x]$ we have
Conversely, suppose $F$ is perfect. If char $F=p$, let $a\in F$ and consider the polynomial $x^p-a\in F[x]$, if it has a root $b$, then $a=b^p\in F^p$. Otherwise, let $f$ be a monic irreducible factor of $x^p-a$ in $F[x]$, and let $K\supset F$ be an extension in which $f$ has a root $b$. In $K[x]$ we have
$x^p-a=x^p-b^p=(x-b)^p$
Since $f$ divides this polynomial, we have $f=(x-b)^m$ for some power $m\le p$. $m\neq 1$ because $b\notin F$. So $m>1$, but then $b$ is not a simple root of $f$, so $f$ is not separable. This contradicts the fact that $F$ is perfect. Thus, we must have $F\subset F^p$ hence $F=F^p$.
Corollary. Every finite field is perfect.Proof.
If $F$ is finite with char $F = p$, the $p$-th power map is both 1-1 and onto, so $F=Im\ \varphi=F^p$.
Example.
Let $t$ be a variable and $F=\mathbb{F}_2(t)$, then the polynomial $f(x)=x^2-t\in F[x]$ is irreducible but not separable. Indeed, $f$ has no root in $F$, because if $\alpha = \frac{p(t)}{q(t)}$ is a root, then $p^2(t)=tq^2(t)$, which is impossible since the LHS has even degree whereas the RHS has odd degree. However, let $\sqrt{t}$ denote a root of $f$ in some extension of $F$, then
$f(x)=x^2-t=x^2+t=x^2-2\sqrt{t}x+t = (x-\sqrt{t})^2$
So $f$ has a multiple root, and hence not separable.Proposition. Suppose $K\supset F$ is a Galois finite extension, then $K\supset F$ is both normal and separable.
Proof.
Choose any $\alpha\in K-F$, by a proposition before, since $F=K^{G}$ with $G=G(K/F)$, we have the minimal polynomial of $\alpha$ over $F$ is
$g(x)=\prod_{j=1}^r(x-\sigma_j(\alpha))$
, where $\sigma_1(\alpha),\dots,\sigma_r(\alpha)$ are distinct elements of the orbit $G\cdot \alpha=\{\sigma(\alpha)\ |\ \sigma\in G\}$. Thus, $\alpha$ is separable over $F$. Moreover, all conjugates $\sigma_j(\alpha)$ also lie in $K$, so $K\supset F$ is both normal and separable.
We are now ready to prove theorem 1.
Theorem 1. Let $K\supset F$ be a finite field extension, then $|G(K/F)|\le[K:F]$. Moreover, the following are equivalent:
(i) $K\supset F$ is Galois extension.
(ii) |G(K/F)|=[K:F].
(iii) $K\supset F$ is normal and separable.
(iv) $K$ is the splitting field of a separable polynomial in $f\in F[x]$.
Proof.
Let $G=G(K/F)$. We always have $F\subset K^{G}\subset K$, therefore, $|G|=[K:K^{G}]<[K:F]$
(i) $\Leftrightarrow$ (ii): already proved.
(i) $\Rightarrow$ (iii): already proved.
(iii) $\Rightarrow$ (iv): we already proved that if $K\supset F$ normal, then $K$ is the splitting field of a polynomial in $F[x]$. If $K\supset F$ is also separable, then we can take $f$ to be separable.
It remains to show (iv) $\Rightarrow$ (ii). Recall the following result:
Theorem. Let $\sigma: F\to F'$ be a field isomorphism, $f\in F[x]$, and $\sigma f\in F'[x]$. Let $K$ be the splitting field of $f$ over $F$, and $K'$ the splitting field of $\sigma f$ over $F'$. Then there exists an isomorphism $\widetilde{\sigma}: K\to K'$ which extends $\sigma$.
(iv) $\Rightarrow$ (ii): Suppose $K$ is the splitting field of the separable polynomial $f\in F[x]$. We will prove $K\supset F$ is Galois by induction on $[K:F]$. Clearly, the base case is true, so we may assume $[K:F]>1$.
Let $p(x)$ be an irreducible factor of $f(x)$ of degree $>1$ (if all irreducible factors are of degree 1, then all roots are in $F$, hence $[K:F]=1$), let $\alpha$ be a root of $p$ and set $E=F(\alpha)$, so $F\subsetneq E\subset K$. Now, $K$ is also the splitting field of $f$ over $E$, and $f$ is separable over $E$, but $[K:E]<[K:F]$, so by induction hypothesis, $K\supset E$ is Galois. Let $H=G(K/E)\le G(K/F)=G$
Let $\alpha_1,\dots,\alpha_r$ be the (distinct) roots of $p$, so deg $p=r$ and hence $[E:F]=r$. Apply previous theorem to the isomorphism:
$\sigma_i:F(\alpha)\to F(\alpha_i)$
Recall that $\sigma_i$ extends $F$ and sends $\alpha$ to $\alpha_i$. There exist $\tau_1,\dots,\tau_r\in G$ such that $\tau_i(\alpha)=\alpha_i$ for $1\le i\le r$.
Claim: The $H$-cosets $\tau_i H$ are distinct.
Proof. $\tau_i H=\tau_j H \Rightarrow \tau_i^{-1}\tau_j\in H\Rightarrow \tau_i^{-1}\tau_j(\alpha)=\alpha\Rightarrow \tau_j(\alpha)=\tau_i(\alpha)\Rightarrow \alpha_j=\alpha_i$ $\Rightarrow i=j$.
We conclude that $[K:F]=[K:E][E:F]=|H|r\le |H||G:H|=|G|\le [K:F]$. Thus, $|G|=[K:F]$, as desired.
Theorem 2. Let $K\supset F$ be a finite Galois extension and set $G=G(K/F)$. Then there is a bijection
$K$ $1$
$\cup$ $\cap$
subfields $E$: $E$ $\leftrightarrow$ subgroups $H$: $H$
$\cup$ $\cap$
$F$ $G$
$E$ $\longmapsto$ $ G(K/E)$
$K^{H}$ $\newcommand{\testleftlong}{\longleftarrow\!\shortmid} \testleftlong$ $H$
(i) The above correspondence are inverse reversing.
(ii) $[K:E]=|H|$ and $[E:F]=|G:H|$ if $E\leftrightarrow H$.
(iii) $K\supset E$ is always Galois, with $H=G(K/E)$
(iv) $E\supset F$ is Galois if and only if $H\lhd G$. In this case, $G(E/F)\cong G/H$.
Proof (i)-(iii).
Consider the map $\varphi$ from subgroups of $G$ to subfields of $K$ defined by $\varphi(H)-K^H$.
$\varphi$ is 1-1: Suppose $K^{H_1}=K^{H_2}=E$, then each $H_i$ is a subgroup of $G(K/E)$, while $|G(K/E)|=|G(K/K^{H_i})|=|H_i|\Rightarrow $H_1=H_2=G(K/E)$, so $\varphi$ is 1-1.
$\varphi$ is onto: Since $K\supset F$ is Galois, $K$ is a splitting field of a separable $f\in F[x]$. Therefore, for any field $E$ with $F\subset E\subset K$, $K$ is a splitting field of $f\in E[x]$, so $K\supset E$ is Galois. It also follows that
$E=K^{G(K/E)}=\varphi(G(K/E))$
One can check that the two maps are inverses of each other and that the diagram is inclusion reversing. It remains to show (ii). If $\varphi(E)=H$, then $|H|=|G(K/E)|=[K:E]. Moreover, |G|=[K:F], so dividing these two equations gives |G:H|=[E:F].To prove (iv), we need some preliminary results
Lemma. Let $F$ be an infinite field and $V$ a vector space over $F$. If $H_1,\dots,H_n$ are proper subspaces of $V$, then $V\neq \cup_{i=1}^n H_i$.
Proof.
We may assume without loss of generality that $H_1\not\subset \cup_{i=2}^n H_i$ (otherwise, we just omit $H_1$ in the union).
Choose $h_0\notin H_1$, $h_1\in H_1$, but $h_1\notin \cup_{i=2}^n H_i$.
Claim 1: The line $L=\{h_0+\lambda h_1\mid \lambda\in F\}=h_0+Fh_1$ is disjoint from H_1. [Indeed, if $h_0+\lambda h_1\in H_1$, then $h_0\in H_1$, a contradiction].
Claim 2: $L$ meets the subspace $H_j$ in at most one vector for $i\ge2$ . [If $h_0+\lambda h_1, h_0+\mu h_1$ both meets $H_j$ for some $j\ge 2$, then their difference $(\lambda-\mu)h_1\in H_j\Rightarrow h_1\in H_j$, a contradiction].
The two claims above imply that $L\cap (\cup_{i=1}^n H_i) $ is a finite set. But since $F$ is an infinite field, $L$ has infinitely many elements. We conclude that $V\neq \cup_{i=1}^n H_i$.
Proposition. Let $K\supset F$ be a finite field extension, then there exists $\alpha\in K$ such that $K=F(\alpha)$ if and only if there are only finitely many fields $E$ such that $F\subset E\subset K$. In this case, $\alpha$ is called a primitive element.
Proof.
$(\Leftarrow)$
If $F$ is finite, then so is $K$. We have shown that $(K^*,\cdot)$ is generated by some single element $\alpha$, then $K=F(\alpha)$.
We may therefore assume $F$ is infinite, suppose there are finitely many fields $E_1,\dots, E_d$ with $F\subset E_i\subsetneq K$, the previous lemma implies that there exists an $\alpha\in K-\cup_{i=1}^d E_i$. This $\alpha$ is a primitive element, because we have $F(\alpha)\neq E_i$ for all $i$.
$(\Rightarrow)$
Assume $K=F(\alpha)$ for some $\alpha\in K$, and let $f$ be the minimal polynomial of $\alpha$ over $F$. Given any subfield $E$ with $F\subset E\subset K$, let $g_E(x)$ be the minimal polynomial of $\alpha$ over $E$. Then $g_E\mid f$ in $E[x]$.
Clearly, there are only finitely many such $g_E$. We thus get a mapping:
$E\overset{\varphi}{\longmapsto} g_E$
It suffices to show this map is 1-1, which implies that there are only finitely many intermediate fields $E$.Let $E$ be an intermediate field, and $E_0$ denote the subfield of $E$ generated over $F$ by coefficients of $g_E$. Then $g_E$ has coefficients in $E_0$ and is irreducible in $E_0$ since it is irreducible in $E[x]$. Therefore $[E_0(\alpha):E_0]=\text{deg }g_E=[E(\alpha):E]$. But we have $F(\alpha)=K=E_0(\alpha)=E(\alpha)\supset E\supset E_0$, hence $[E(\alpha):E]=[E_0(\alpha):E]$, now we compute
$[E_0(\alpha):E_0]=[E(\alpha):E]=[E_0(\alpha):E]=[E_0(\alpha):E][E:E_0]\Rightarrow E=E_0$
So $E$ is uniquely determined by the coefficients of $g_E$, hence $\varphi$ is 1-1.Corollary (Primitive Element Theorem). Suppose $K\supset F$ is a finite Galois extension, then $K=F(\alpha)$ for some $\alpha\in K$.
Proof.
According to the previous proposition, it suffices to show there are only finitely many intermediate fields $E$ with $F\subset E\subset K$. By the Galois correspondence, such subfields $E$ are in correspondence with subgroups of $G(K/F)$, but since $G(K/F)$ is finite, there are only finitely many subgroups.
Remark. The theorem holds for any finite separable extension $K\supset F$ in general, as any such extension is contained in a Galois extension of $F$. Since $K\supset F$ is a finite extension, $K=F(a_1, \dots, a_k)$ for some $a_i\in F$. Let $f_i$ be their respective
minimal polynomials over $F$ and set $f=\prod f_i$. Since $K\supset F$ separable,each $f_i$ does not have multiple roots. We can assume $f$ is separable by removing repeated $f_i$'s. Let $L$ be the splitting field of $f$ over $F$, then $L\supset K\supset F$, and $L$ is Galois over $F$. By previous corollary, there are only finitely many intermediate fields between $F$ and $L$, hence so does $K\supset F$, and thus $K=F(\alpha)$ for some $\alpha\in F$.
We are now ready to prove part (iv). Let $K\supset F$ be a finite Galois extension, with $G=G(K/F)$, and $K\supset E\supset F$.
Lemma. $E\supset F$ is Galois if and only if $\forall\ \sigma\in G$, $\sigma(E)\subset E$ (i.e. $E$ is stable under the action of $G$).
Proof.
($\Rightarrow$) If $E\supset F$ is Galois, it is a finite separable extension. The primitive element theorem implies that $E=F(\alpha)$. Moreover, $E\supset F$ is normal, so the minimal polynomial $f$ of $\alpha$ over $F$ has all roots in $E$. But for any $\sigma\in G$, $\sigma(\alpha)$ is a root of $f$. Hence, $\sigma(E)=\sigma(F(\alpha))=F(\sigma(\alpha))\subset E$.
($\Leftarrow$) Since $K\supset F$ is separable, $E\supset F$ is also separable. By the primitive element theorem, $E=F(\alpha)$ for some $\alpha\in E$. Our assumption gives that $\sigma(\alpha)\in E$ for all $\sigma\in G$.
Let $g(x)=\prod_{\sigma\in G} (x-\sigma(\alpha))\in F[x]$. Then $\alpha=id(\alpha)$ is a root of $g$ and all roots of $g$ lie in $E$. So $E$ is a splitting field of $g$ over $F$, and hence $E\supset F$ is normal (by a corollary). Since $E\supset F$ is both normal and separable, we conclude that $E\supset F$ is Galois.
Proof (iv).
Set $H=G(K/E)$, we want to show $E\supset F$ is normal $\Leftrightarrow H\lhd G$, and in this case, $G(E/F)\cong G/H$.
As above, since $K\supset F$ separable, so is $E\supset F$, and hence $E=F(\alpha)$ for some $\alpha\in E$. Using the previous lemma, $E\supset F$ is Galois if and only if $\sigma(\alpha)\in E,\ \forall\ \sigma\in G$. Since $E=K^H=F(\alpha)$, this occurs iff
$\forall\ \sigma\in G, \forall\ \tau\in H, \tau\sigma(\alpha)=\sigma(\alpha)\Leftrightarrow \sigma^{-1}\tau\sigma(\alpha)=\alpha \Leftrightarrow \sigma^{-1}\tau\sigma\in H \Leftrightarrow H\lhd G$
Finally, in the above situation, where $E\supset F$ is Galois, and $H=G(K/E)\lhd G$, we have a group homomorphism
$\psi:G(K/F)\to G(E/F)$
$\sigma\mapsto \sigma |_{E}$
, with $\ker\psi = G(K/E)=H$. Furthermore, $\psi$ is surjective, since by FIT, we have $|Im\psi|=|G|/|H|=|G:H|=[E:F]=|G(E/F)|$. Therefore, by FIT, $G/H\cong G(E/F)$.Solvable Groups
In this section, we recall some theorems regarding solvable and simple groups. We omit the proofs and examples (one can refer to notes from Math 600 or 404).
Definition. A group is solvable if there exists a chain of subgroups $G=G_0\rhd G_1\rhd\dots\rhd G_k={1}$, where each $G_i\lhd G_{i-1}$ and each quotient $G_i/G_{i+1}$ is abelian. The latter condition is equivalent to having a series with cyclic quotients.
Proposition. (i) $G'\lhd G$ and $G/G'$ is abelian, with $G'=[G,G]$.
(ii) If $H\lhd G$ with $G/H$ abelian, then $G'\subset H$
Proposition. $G$ is solvable if and only if $G^{(k)}=1$ for some $k$, with $G^{(0)}=G, G^{(1)}=G'$, $G^{(2)}=G^{(1)}$$'$, $\dots$
Corollary. Subgroups and quotients of solvable groups are solvable.
Theorem. $A_n$ is simple for $n\ge 5$.
Corollary. $S_n$ is not solvable for $n\ge 5$.
Proof.
If $S_n$ solvable, then so is $A_n$, $A_n'\lhd A_n$ with $A_n' \subsetneq A_n$, hence $A_n'=1$, so $A_n$ must be abelian, which is absurd.
Applications of Galois Theory
We have seen that there are formulas for roots of quadratic and cubic polynomials which involve radicals. We now explore a more general situation, and will show that a polynomial equation $f(x)=0$ "is solvable by radicals", which we will make precise soon, if and only if the Galois group of $f$ is a solvable group.
We restrict our discussion to fields of characteristic 0. First we will consider simple radical extensions, which are the extensions obtained by adjoining to a field $F$ the $n$-th root $\sqrt[n]{\alpha}$ for some $\alpha\in F$. As all roots of $x^n-a,\ a\in F$ differ by factors which are n-th root of unity, such an extension will be Galois if and only if it contains n-th roots of unity.
Definition. An extension $K\supset F$ is cyclic if it is Galois with cyclic Galois group.
Proposition A. Let $F$ be a field which contains the n-th roots of unity, and $a\in F$, then the extension $F(\sqrt[n]{a})$ is cyclic, with degree dividing $n$.
Proof.
$K=F(\sqrt[n]{a})\supset F$ is Galois since $K$ is the splitting field of $x^n-a\in F[x]$. For any $\sigma\in G(K/F)=G$, $\sigma(\sqrt[n]{a})$ is a root of $x^n-a$, hence
$\sigma(\sqrt[n]{a})=\zeta_{\sigma}\sqrt[n]{a}$
for some n-th root of unity $\zeta_{\sigma}$. This gives a map
$\varphi:G\to \mu_n=\{x\in F\mid x^n=1\}$
$\sigma\mapsto \zeta_{\sigma}$
Since $\mu_n\subset F$ and $F$ is fixed by all $\sigma\in G$, $\sigma\tau(\sqrt[n]{a})=\sigma(\zeta_{\tau}\sqrt[n]{a})=\zeta_{\tau}\sigma(\sqrt[n]{a})=\zeta_{\tau}\zeta_{\sigma}\sqrt[n]{a}\Rightarrow \zeta_{\tau}\zeta_{\sigma}=\zeta_{\sigma\tau}\Rightarrow \varphi(\sigma\tau)$ $=\varphi(\sigma)\varphi(\tau)$. So $\varphi$ is a group homomorphism.Also, $\ker\varphi=\{\sigma\in G\mid \sigma(\sqrt[n]{a})=\sqrt[n]{a}\}=1$, hence $\varphi$ is 1-1, and $G$ embeds in $\mu_n$, a cyclic group of order $n$.
Proposition B. Let $K\supset F$ be a cyclic extension of degree $n$ over a field $F$ which contains the n-th roots of unity. Then $K=F(\sqrt[n]{a})$ for some $a\in F$.
Proof.
For $\alpha\in F$ and $\zeta$ a root of unity in $F$, define the Lagrange resolvent $(\alpha,\zeta)\in K$ by
$(\alpha,\zeta)=\alpha+\zeta\sigma(\alpha)+\zeta^2\sigma^2(\alpha)+\dots+\zeta^{n-1}\sigma^{n-1}(\alpha)=\sum_{k=0}^{n-1}\zeta^k\sigma^k(\alpha)$
, where $G(K/F)=(\sigma)$.
Observe that
$\sigma(\alpha,\zeta)=\sigma(\alpha)+\zeta\sigma^2(\alpha)+\dots+\zeta^{-1}\alpha = \zeta^{-1}(\alpha,\zeta)$ (1)
Hence $\sigma(\alpha,\zeta)^n=(\alpha,\zeta)^n$, so $(\alpha,\zeta)^n$ is fixed by $G=G(K/F)$, so $(\alpha,\zeta)\in F=K^G$ for all $\alpha\in K$.Now, let $\zeta$ be a primitive n-th root of unity, i.e., $\mu_n=(\zeta)$. By a previous proposition, $1,\sigma,\sigma^2,\dots,\sigma^{n-1}$ are "linearly independent", so $\exists\ \alpha\in K$ with $(\alpha,\zeta)\neq 0$, because otherwise we will have $\zeta=0$.
Iterating equation (1) we get $\sigma^i(\alpha,\zeta)=\zeta^{-i}(\alpha,\zeta)$, so $\sigma^i$ does not fix $(\alpha,\zeta)$ for any $i=1,2,\dots, n-1$. We conclude that $(\alpha,\zeta)$ does not lie in any proper subfield of $K$, because otherwise, the subfield it lies in must be the fixed field of some subgroup of $G$, so $(\alpha,\zeta)$ must be fixed by some elements of $G$. Therefore, $K=F(\alpha,\zeta)$ (recall this in the proof of the primitive element theorem), and since $a=(\alpha,\zeta)^n\in F$, $K=F(\sqrt[n]{a})$.
Definition. An element $\alpha$ which is algebraic over $F$ can be expressed in radicals if $\alpha$ lies in a field $K$ that can be obtained via a series of simple radical extensions:
$F=K_0\subset F_1\subset\dots \subset K_s=K$
where $K_{i+1}=K_i(\sqrt[n_i]{a_i})$ for some $a_i\in K_i$. Here $\sqrt[n_i]{a_i}$ denotes some root of $x^{n_i}-a_i$. Such a field $K$ is called a root extension of $F$. A polynomial $f\in F[x]$ can be solved by radicals if the splitting field of $f$ over $F$ lies in a root extension over $F$.From now on, we will omit most proofs. Check professor's note for detailed proofs.
Lemma 1. If $K\supset F$ is a root extension, then $K$ is contained in some root extension which is Galois over $F$ and such that each intermediate extension is cyclic.
Theorem (Galois). The polynomial $f\in F[x]$ can be solved by radicals if and only if the Galois group of $f$ is a solvable group.
The proof of the converse requires the two results below:
Lemma 2. Let $K\supset F$ be a Galois extension, and $F'\supset F$ be any finite extension, then $KF'\supset F$ is Galois, and
$G(KF'/F')\cong G(K/K\cap F')$,
so $G(KF'/F')$ is naturally a subgroup of $G(K/F)$.Corollary. Suppose $K/F$ is Galois and $F'/F$ is any finite extension, then we have
$[KF':F]=\frac{[K:F][F':F]}{[K\cap F':F]}$.
Corollary. The general equation of degree $n\ge 5$ cannot be solved by radicals.
Proof. $S_n$ is not solvable for $n\ge 5$.
Galois Group as a Permutation Group of the Roots.
Set up: Consider a monic polynomial $f\in F[x]$ of positive degree $n$, with $\alpha_1,\dots,\alpha_n$ the roots of $f$, and $K$ the splitting field of $f$ over $F$. Then the Galois group of $f$ over $F$ $G(K/F)$ is a subgroup of $S_n=S(\{\alpha_1,\dots,\alpha_n\})$.
Theorem. Suppose the $f\in F[x]$ does not have multiple roots. Then $f$ is irreducible over $F$ if and only if $G$ is a transitive group of the roots $\alpha_i$, $i=1,2,\dots,n$, that is, for any two roots $\alpha_i,\alpha_j$, there exists $\sigma\in G$ with $\sigma(\alpha_i)=\sigma(\alpha_j)$.
Examples.
1) Suppose $f\in F[x]$ is a cubic irreducible polynomial, then its Galois group is a transitive subgroup of $S_3$, hence either $A_3$ or $S_3$.
2) It's an exercise (see homework) to show that the only transitive subgroups of $S_4$ are
(i) $S_4$ (ii) $A_4$ (iii) $V$, the Klein 4-group
(iv) $C:=\{1,(1234),(13)(24),(1432)\}\cong C_4$ and its conjugates
(v) $D:= V\cup \{(12),(34),(1423),(1324)\}\cong D_8$, which is a 2-sylow subgroup of $S_4$, and its conjugates.
Theorem. Let $F$ be a field with Char $F\neq 2$, and $f\in F[x]$ with distinct roots $\{\alpha_1,\dots,\alpha_n\}$ in a splitting field $K/F$. Let $G=G(K/F)$ and
$D:=\prod_{1\le i<j\le n} (\alpha_i-\alpha_j)$
Then the subfield of $K/F$ corresponding to $G\cap A_n$ in the Galois correspondence is $F(D)$.Corollary. The Galois group of $f$ over $F$ is a subgroup of $A_n$ if and only if the discriminant $d=\prod_{1\le i<j\le n}(\alpha_i-\alpha_j)^2$ is the square of an element in $F$.
In addition, we have the formula of $d$ as below
$d=D^2= \begin{vmatrix} p_{2n-2}& p_{2n-3}&\dots& p_{n-1}\\ p_{2n-3}& p_{2n-4}&\dots& p_{n-2}\\ \vdots &\vdots&\dots& \vdots \\ p_{n-1}& p_{n-2}&\dots& p_{0} \end{vmatrix}$
where $p_k=\sum_{j=1}^n \alpha_j^k$, with $p_0=n$. Then using Newton's identities (see homework), we can express $d^2$ using $e_i$'s, hence the coefficients of $f$.See examples here (one another application of the discriminant is to detect multiple roots of $f$).
Cyclotomic Polynomials and Extensions
Definition. Let $\mu_n$ denote the group of n-th roots of unity in $\mathbb{C}$, then $\mu_n\cong C_n$. Let $\mathbb{Q}(\zeta_n)$ be the field by adjoining $\mathbb{Q}$ the elements of $\mu_n$, so the splitting field of $x^n-1=0\in \mathbb{Q}[x]$. Here $\zeta_n=e^{2\pi i/n}$. We call $\mathbb{Q}(\zeta_n)$ the n-th cyclotomic field.
Remark. $\mu_d\mid \mu_n \Leftrightarrow d\mid n$.
Definition. The n-th cyclotomic polynomial $\Phi_n(x)$ is the polynomial whose roots are the primitive n-th roots of unity (the generators of $\mu_n$), each with multiplicity one, so
$\Phi_n(x)=\prod_{\zeta\in\mu_n\\ \mu_n=(\zeta)} (x-\zeta)=\prod_{1\le k<n\\ (k,n)=1} (x-\zeta_n^k)$
Clearly, deg $\Phi_n=\varphi(n)$. Recall that $\varphi$ is multiplicative and $\varphi(p^k)=p^{k-1}(p-1)$.
Proposition. $x^n-1 = \prod_{d\mid n} \Phi_d(x)$. Comparing degrees, we have $n=\sum_{d\mid n} \varphi(d)$.
Remark. Using this formula, we can compute $\Phi_n(x)$ recursively, see examples on page 3 here.
One example is that we have $\Phi_p(x)=\frac{x^p-1}{x-1}$, recall this is irreducible by Eisenstein's criterion.
Proposition. The cyclotomic polynomials $\Phi_n(x)$ is a monic polynomial of degree $\varphi(n)$ with integer coefficients.
Remark. It is known that there exist cyclotomic polynomials with arbitrary large coefficients (so not just $\pm 1$).
Theorem. $\Phi_n(x)$ is irreducible in $\mathbb{Z}[x]$ hence in $\mathbb{Q}[x]$, so it is the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$.
Corollary. $[\mathbb{Q}(\zeta_n)):\mathbb{Q}]=\varphi(n)$, for any $n\ge 1$.
Theorem. The Galois group $G(\mathbb{Q}(\zeta_n))/\mathbb{Q})$ is isomorphic to the group of units $(\mathbb{Z}/n\mathbb{Z})^{\times}$, given by
$(\mathbb{Z}/n\mathbb{Z})^{\times}\to G(\mathbb{Q}(\zeta_n))/\mathbb{Q})$
$r\mod\ n \mapsto \sigma_r$, where $\sigma_r(\zeta_n)=\zeta_n^r$.
Remark. The theorem implies that $\mathbb{Q}(\zeta_n)\supset \mathbb{Q}$ is abelian extension. In fact, cyclotomic extensions are rich enough to contain all finite abelian extensions of $\mathbb{Q}$.
Theorem. Let $G$ be any finite abelian group, then there exists a subfield $K$ of a cyclotomic field with $G(K/\mathbb{Q})\cong G$.
The proof uses two facts (see p7 here):
1) structure theorem for finite abelian groups.
2) there are infinitely many primes with $p\equiv 1$ (mod m) for any $m\ge 2$.
Theorem (Kronecher-Weber). Let $K/\mathbb{Q}$ be a finite abelian Galois extension, then $K$ is contained in a cyclotomic extension of $\mathbb{Q}$
Constructibility of a regular n-gon Cont'd
Theorem (Gauss). A regular n-gon can be constructed if and only if $\varphi(n)=2^k$ for some $k\ge 1$.
Corollary. The regular n-gon is constructible if and only if $n=2^rp_1p_2\dots p_m$ with $p_i$'s distinct Fermat primes.
Algebraic Independence and Transcendence Degree
Definition. Let $K\supset F$ be a field extension and suppose $\alpha_i$ for $1\le i\le n$ are $n$ elements of $K$. We have an evaluation map (in fact a homomorphism):
$ev: F[x_1,\dots,x_n]\to K$
$f(x_1,\dots,x_n)\mapsto f(\alpha_1,\dots,\alpha_n)$
We say that $\{\alpha_1,\dots,\alpha_n\}$ is an algebraically independent set if the homomorphism ev is 1-1, that is, there is no polynomial relation (other than 0) such among the elements $\{\alpha_1,\dots,\alpha_n\}$. Equivalently, the set of all monomials $\{\alpha_1^{i_1}\dots\alpha_n^{i_n}\mid i_k\ge 0\}$ is linearly independent over $F$.Example. A singleton set $\{\alpha\}$ is algebraically independent iff $\alpha$ is transcendental over $F$.
We say a subset $S$ of $K$ is algebraically independent over $F$ if any finite subset of $S$ is algebraically independent over $F$.
Proposition. If $K\supset F$ is a field extension and $A=\{\alpha_1,\dots,\alpha_n\}\subset K$ is algebraically independent over $F$, then the evaluation map can be extended to a unique field isomorphism:
$F(x_1,\dots,x_n)\cong F(\alpha_1,\dots,\alpha_n)$
Lemma 1. Let $K\supset F$ is a field extension and $\alpha_1,\dots,\alpha_n\in K$. Let $F_0=F$, and $F_i=F(\alpha_1,\dots,\alpha_i)$ for $1\le i\le n$, then $A=\{\alpha_1,\dots,\alpha_n\}$ is algebraically independent over $F$ if and only if $\alpha_i$ is transcendental over $F_{i-1}$ for all $1\le i\le n$.
Definition. Let $K$ be a field, and $F\subset K$ a subfield. A transcendental basis of $K$ over $F$ is an algebraically independent set $\mathcal{B}$ of $K/F$ which is maximal under inclusion.
Proposition. Suppose $K\supset F$ is a field extension and $S\subset K$ is any subset. Then $S$ is a transcendental basis of $K/F$ iff $S$ is algebraically independent over $F$ and $K\supset F(S)$ is an algebraic extension.
Theorem. Let $K\supset F$ be a field extension, $A\subset K$ a subset such that the extension $K\supset F(A)$ is algebraic, and $C\subset A$ a subset that is algebraically independent over $F$. Then there exists a transcendental basis $\mathcal{B}$ of $K/F$ with $C\subset \mathcal{B}\subset A$.
Definition. Let $K\supset F$ be any field extension, we say $K$ is finitely generated over $F$ if there exist finitely many $\alpha_i\in K$, $1\le i\le n$, such that $K=F(\alpha_1,\dots,\alpha_n)$. $K$ has finite transcendental degree over $F$ if there exists a subfield $E$ with $K\supset E\supset F$ such that $K\supset E$ is algebraic and $E\supset F$ is finitely generated.
Theorem. Suppose $K\supset F$ is a field extension, and $C=\{c_1,\dots,c_r\}$ is a subset of $K$ that is algebraically independent over $F$. Assume that $A=\{\alpha_1,\dots,\alpha_s\}\subset K$ is such that $K\supset F(A)$ is algebraic. Then $r\le s$ and there exists a set $D$ with $C\subset D\subset A\cup C$ such that $|D|=s$ and $K\supset F(D)$ is algebraic.
Corollary. If $K\supset F$ is a field extension, and $\mathcal{B}_1$ and $\mathcal{B}_2$ are two transcendental bases of $K/F$, then either $\mathcal{B}_1$ and $\mathcal{B}_2$ are both infinite, or $|\mathcal{B}_1|=|\mathcal{B}_2|$.
Theorem (Luroth). Suppose $F(x)\supset F$ is a simple extension and $x$ is transcendental over $F$. Let $E$ be a subfield with $F\subset E\subset F(x)$. Then $E\supset F$ is a simple extension.
See more examples here.
Traces and Norms
Let $K\supset F$ be a field extension, with $[K:F]=n$. Then any element $\alpha\in K$ defines an $F$-linear map $T_{\alpha}: K\to K$ by $T_{\alpha}(x)=\alpha x,\forall x\in K$.
Definition. The trace, norm and characteristic polynomial of $\alpha$ (in the setting of $K/F$) are defined as below
$Tr(\alpha)=Tr(T_{\alpha})$
$N(\alpha)=\det(T_{\alpha})$
$p_{\alpha}(x)=p_{T_{\alpha}}(x)$
See examples on page7-8 here.
Theorem (Hilbert's theorem 90). Let $K$ be a cyclic extension of $F$ and let $G(K/F)=(\sigma)$. If $\alpha\in K$, then
(i) $N(\alpha)=1\Leftrightarrow\ \exists \beta\in K$ with $\alpha = \frac{\beta}{\sigma(\beta)}$.
(ii) $Tr(\alpha)=0\Leftrightarrow\ \exists \gamma\in K$ with $\alpha =\gamma - \sigma(\gamma)$.
Proposition. Let $K\supset F$ be a finite extension of fields, $\alpha\in K$ and $f$ be the minimal polynomial of $f$ over $F$. If $[K:F(\alpha)]=m$, then we have $p_{\alpha}(x)=f(x)^m$.
Corollary. Suppose that the roots of the minimal polynomial of $\alpha$ are $\alpha_1,\dots,\alpha_n$ in some splitting field containing $K$, and $[K:F(\alpha)]=m$, then we have
$Tr(\alpha)=m\sum_{i=1}^n \alpha_i$ and $N(\alpha) = (\prod_{i=1}^n \alpha_i)^m$
See an example on page 10 here.
Suppose $K\supset F$ is a finite, separable extension, and let $\overline{F}$ be an algebraic closure of $F$. Recall that an $F$-isomorphism $\sigma: K\to \overline{F}$ is a map such that $\sigma |_F=id$ (also recall that any homomorphism of fields is injective). We claim that the set $\Sigma$ of all $F$-homomorphisms is finite. Indeed, since $K\supset F$ is finite separable, by the theorem of primitive element, $K=F(\alpha)$ for some $\alpha\in K$. Let $f\in F[x]$ be the minimal polynomial of $\alpha$ over $F$, with $n$ distinct roots $\alpha_1=\alpha, \dots, \alpha_n$, then $[K:F]=n$. Then any $\sigma \in \Sigma$ is determined by $\sigma(\alpha)$, and we must have $\sigma(\alpha)=\alpha_i$ for some $1\le i\le n$. Moreover, any such choice is an element of $\Sigma$, so in fact $|\Sigma|=n$.
Proposition. Let $K\supset F$ and $\Sigma$ as above, then for any element $\alpha\in K$, we have
$Tr(\alpha)=\sum_{\sigma\in \Sigma}\sigma(\alpha)$, $N(\alpha)=\prod_{\sigma\in \Sigma}\sigma(\alpha)$
Remark. If $K\supset F$ is finite Galois, then $\Sigma= G(K/F)$.Fact. Suppose $K\supset F$ finite separable, and $E$ is a field with $K\supset E\supset F$, then
$Tr_{E/F}\circ Tr_{K/E}=Tr_{K/F}$ and $N_{E/F}\circ N_{K/E}=N_{K/F}$
We now use the norm to get a nice formula for the discriminant of a polynomial.
Proposition. Let $f(x)$ be a monic irreducible polynomial in $F[x]$, and $\alpha$ a root of $f$ in some splitting field of $f$, then
$dis\ f(x)= (-1)^{\frac{n(n-1)}{2}}N_{F(\alpha)/F)}(f'(\alpha))$
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