2.9 Galois theory

Galois group

def: Mapping restrictions

$$\begin{align*} &\sphericalangle \\ &f: X \to Y \\ &Z \subseteq X\\ &f_{|Z}: Z \to Y, x \mapsto f(x) \\ \hline \\ &f_{|Z} - \text{restriction of } f \text{ to }Z \end{align*}$$

def: Endomorphism

$$\begin{align*} &\sphericalangle \\ &\phi: E \rightsquigarrow_F E \\ \hline \\ &\phi \in \text{End}_F(E), \phi - \text{ endomorphism on } E \end{align*}$$

def: Automorphism

$$\begin{align*} &\sphericalangle \\ &\phi: E \cong_F E \\ \hline \\ &\phi \in \text{Aut}_F(E), \phi - \text{ automorphism on } E \end{align*}$$

def: F-homomorphism

$$\begin{align*} &\sphericalangle \\ &E_1/F, E_2/F - \text{extension field} \\ &\phi: E_1 \rightsquigarrow_F E_2 \\ &\phi_{|F}: x \mapsto x \\ \hline \\ &\phi: E_1 \rightsquigarrow_{|F} E_2 - F\text{-homomorphism} \end{align*}$$

def: F-isomorphism, F-automorphism

$$\begin{align*} &\sphericalangle \\ &E_1/F, E_2/F \in \mathcal F \\ &\phi: E_1 \rightsquigarrow_{|F} E_2 \\ &\phi: E_1 \leftrightarrow E_2 \\ \hline \\ &\phi: E_1 \cong_{|F} E_2 - F\text{-isomopshism} \\ & E_1 = E_2 \implies \phi - F\text{-automorphism} \end{align*}$$

def: Galois group

$$\begin{align*} &\sphericalangle \\ &E / F \in \mathcal F \\ \hline \\ &\text{Gal}(E/F):=\{\sigma: E \cong_{|F}E\} - \text{Galois group of } E/F \end{align*}$$

note

It's pretty easy to show that Galois group is indeed a group. A composition of automorphism is still automorphism and the field $F$ is still fixed. Associativity is obvious as well the existence of the neutral element (identity automorphism) and inverse $\sigma^{-1}$

Proposition 2.9.1: F-homomorphism properties

$$\begin{align*} &\sphericalangle \\ &E_1/F, E_2/F \in \mathcal F \\ &\phi: E_1 \rightsquigarrow_{|F} E_2 \\ \hline \\ &\begin{align*} & E_1 \rightsquigarrow_{M_F} E_2 \tag{a}\\ & E=E_1=E_2, [E: F] < \infty \implies \phi: E \cong_{M_F} E \tag{b}\\ & \forall p \in F[x_1, \ldots, x_n]: \phi(p(y_1, \ldots, y_n))=p(\phi(y_1), \ldots, \phi(y_n)) \hspace{0.5cm}\tag{c} \end{align*} \end{align*}$$

Proof

a.

$\forall \alpha \in F, x \in E_1: \phi(\alpha x)=\phi(\alpha)\phi(x)=\alpha \phi(x)$

b.

By $(2.4.20)$ we know that $\ker \phi = \{0\}$. So $\dim_F(E)=\dim_F(\phi(E))$. Then $\phi(E) \subseteq E$ implies $E = \phi(E)$.

c.

$$\phi(p(x_1, \ldots, x_n))=\phi(\sum a_ix_1^{d_{i_1}}\ldots x_n^{d_{i_n}}) = \\ \sum \phi(a_i)\phi(x_1)^{d_{i_1}}\ldots \phi(x_n)^{d_{i_n}} = \\ \sum a_i\phi(x_1)^{d_{i_1}}\ldots \phi(x_n)^{d_{i_n}} = \\ p(\phi(x_1), \ldots, \phi(x_n))$$

$\square$

Proposition 2.9.2: F-automorphism is defined by its action on the generating set

$$\begin{align*} &\sphericalangle \\ &X \in \mathcal S, |X| < \infty \\ &F(X)/F \in \mathcal F \\ &\sigma, \tau \in \text{Gal}(F(X)/F) \\ &\sigma_{|X}=\tau_{|X} \\ \hline \\ &\sigma = \tau \end{align*}$$

Proof

Consider $x \in F(X)$. We know that $\exists \alpha_1, \ldots,\alpha_n \in X$ such that $x \in F(\alpha_1, \ldots, \alpha_n) \implies x = f(\alpha_1, \ldots, \alpha_n)/g(\alpha_1, \ldots, \alpha_n), f, g \in F[x_1, \ldots, x_n]$.

$$\sigma(x)=\sigma(\frac{f(\alpha_1, \ldots, \alpha_n)}{g(\alpha_1, \ldots, \alpha_n)}) \overset{(2.9.1.c)}= \frac{f(\sigma(\alpha_1), \ldots, \sigma(\alpha_n))}{g(\sigma(\alpha_1), \ldots, \sigma(\alpha_n))} \overset{\sigma_{|X}=\tau_{|X}}= \\ \frac{f(\tau(\alpha_1), \ldots, \tau(\alpha_n))}{g(\tau(\alpha_1), \ldots, \tau(\alpha_n))} \overset{(2.9.1.c)}=\tau(x)$$

$\square$

Proposition 2.9.3: F-automorphism permutes the roots of minimal polynomials

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &\sigma \in \text{Gal}(E/F) \\ &\alpha \in \overline F_E \\ &p:=\mathfrak{pol}_F(\alpha) \\ &R(\alpha) := \{x \in E: p(x) = 0\} \\ \hline \\ &\begin{align*} &f \in F[x], f(\alpha) = 0 \implies f(\sigma(\alpha))=0 \hspace{1cm}\tag{a}\\ &\sigma_{|R(\alpha)}: R(\alpha) \leftrightarrow R(\alpha) \tag{b}\\ &\forall \alpha\in E: \mathfrak{pol}_F(\alpha) = \mathfrak{pol}_F(\sigma(\alpha)) \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

a.

$$\forall \alpha \in R(f): 0 = \sigma(0) = \sigma(f(\alpha))\overset{(2.9.3.c)}=f(\sigma(\alpha))$$

b.

From $(a)$ it follows that $\sigma_{|R(\alpha)}$ is well-defined (indeed maps to $R(\alpha)$). The rest follows from $\sigma$ being bijective in $E$.

c.

$$\forall \alpha \in E: p(\sigma(\alpha)) = 0 \implies \\ \mathfrak{pol}_F(\sigma(\alpha)) \mid p \overset{p \in F[x]^-}\implies \mathfrak{pol}_F(\sigma(\alpha)) = p$$

$\square$

Example: Galois group for the complex numbers

We claim that $\text{Gal}(\mathbb C / \mathbb R)=\{\text{id}, \sigma\}$, where $\forall a, b \in R: \text{id}(a+bi)=a+bi, \sigma(a+bi) = a-bi$. Indeed $\mathbb C = \mathbb R(i)$. So by $(2.9.2)$ an element of $\text{Gal}(\mathbb C / \mathbb R)$ is fully defined by it's action on $i$. By $(2.9.3)$ The only available actions on $i$ are $\text{id}: i \mapsto i$ and $\sigma: i \mapsto -i$.

Example: Galois group for $\mathbb Q(\sqrt[3]2)/\mathbb Q$

Consider minimal polynomial $p(x):= \mathfrak {pol}(\sqrt[3]2)=x^3-2$. The roots of this polynomial are $\sqrt[3]2,\omega\sqrt[3]2, \omega^2\sqrt[3]2$, where $\omega = e^{2\pi i/3}$. As in the example above any element of Galois group is defined by its action on $\sqrt[3]2$. But the only root of $x^3-2$ in $\mathbb Q(\sqrt[3]2)$ is $\sqrt[3]2$. So $\text{Gal}(\mathbb Q(\sqrt[3]2)/\mathbb Q)=\{\text{id}\}$.

Example: Galois group for $(\mathbb F_2[x] / \lang x^2+x+1 \rang_I)/\mathbb F_2$

Consider $F:=\mathbb F_2 \equiv \Z_2$ and polynomial $p(x):=x^2+x+1$. This polynomial is irreducible since it has no roots in $\mathbb F_2$. So $E:=\mathbb F_2[x] / \lang p(x) \rang_I$ is a field. Let $\alpha$ and $\beta$ be the roots of $p(x)$ in $E$. Then we have:

$$(x-\alpha)(x-\beta)=x^2+x+1 \implies -\alpha-\beta=1 \implies \\ \beta = -1-\alpha = 1 + \alpha$$

So again we have $E = F(\alpha)$ and $\text{Gal}(E/F)=\{\text{id}, \sigma\}$, $\sigma(\alpha) = 1+\alpha$

Example: Galois group for $\mathbb F_2(x) / \mathbb F_2(x^2)$

Let $F:=\mathbb F_2(x^2), E:=\mathbb F_2(x)$. Then we have $E/F$ and $E=F(x)$. We know that $x \in \overline F_E, \mathfrak {pol}_F(x) = t^2-x^2$. Note that in $\mathbb F_2: t^2-x^2=t^2-2tx+x^2=(t-x)^2$, so $x$ is the only root and $\text{Gal}(E/F) = \{\text{id}\}$.

Corrolary 2.9.4: Galois group of a finite extension is finite

$$\begin{align*} &\sphericalangle \\ &E / F \in \mathcal F \\ &[E:F] < \infty \\ \hline \\ &|\text{Gal}(E/F)| < \infty \end{align*}$$

Proof

Since $[E:F]<\infty$, we can write $E=F(\alpha_1, \ldots, \alpha_n), \alpha_i \in \overline F_E$. From $(2.9.3)$ we know that for each alpha there's only finite possibilites for mapping into some element of $E$ (the image of $\alpha_i$ is limited to the set of roots of $\mathfrak{pol}_F(\alpha)$ which is finite). On the other hand, from $(2.9.2)$ we know that $\sigma$ is defined by the image of the set $\{\alpha_1, \ldots, \alpha_n\}$.

$\square$

So far we converted field extensions to Galois group, a subset of all automorphisms. Now we will make the reverse conversion - from automorphism to a subfield of a field:

def: Fixed subfield

$$\begin{align*} &\sphericalangle \\ & E \in \mathcal F \\ &S \subseteq \text{Aut}_F(E) \\ \hline \\ &\text{Fix}_S(E):=\{x \in E: \forall \sigma \in S: \sigma(x)=x\} \end{align*}$$

note

It's easy to check that it is indeed a subfield of $E$

Proposition 2.9.5: Fixed fields and galois groups dual properties

$$\begin{align*} &\sphericalangle \\ &E \in \mathcal F \\ \hline \\ &\begin{align*} &F_1 \subseteq_F F_2 \subseteq_F E \implies \text{Gal}(E/F_1) \supseteq \text{Gal}(E/F_2) \tag{a}\\ &S_1 \subseteq S_2 \subseteq \text{Aut}(E) \implies \text{Fix}_{S_1}(E) \supseteq \text{Fix}_{S_2}(E) \tag{b} \\ &E/F \implies F \subseteq \text{Fix}_{\text{Gal(E/F)}}(E) \tag{c}\\ &S \subseteq \text{Aut}(E) \implies S \subseteq \text{Gal}(E / \text{Fix}_S(E)) \tag{d} \\ &E/F, \exists S \subseteq \text{Aut}(E): F = \text{Fix}_S(E) \implies F = \text{Fix}_{\text{Gal}(E/F)}(E) \hspace{0.5cm} \tag{e}\\ &S \subseteq \text{Aut}(E), \exists F \subseteq_F E : S = \text{Gal}(E/F) \implies S = \text{Gal}(E/\text{Fix}_S(E)) \hspace{0.1cm} \tag{f}\\ \end{align*} \end{align*}$$

Proof

a., b.

Follows directly from the definition

c.

$\text{Gal}(E/F)$ definitely fixes $F$ but can fix more elements, that's why $F \subseteq \text{Fix}_{\text{Gal}(E/F)}(E)$

d.

Each element of $\text{Fix}_S(E)$ is fixed by $S$ but can be fixed by other elements of $\text{Aut}(E)$ so $S \subseteq \text{Gal}(E / \text{Fix}_S(E))$

e.

$$(c) \implies F \subseteq \text{Fix}_{\text{Gal}(E/F)}(E) \\ (d) \implies S \subseteq \text{Gal}(E / \text{Fix}_S(E)) \overset{(b)}\implies \\ F=\text{Fix}_S(E) \supseteq \text{Fix}_{\text{Gal}(E / \text{Fix}_S(E))}(E)=\text{Fix}_{\text{Gal}(E / F)}(E) \\$$

f.

$$(d) \implies S \subseteq \text{Gal}(E / \text{Fix}_S(E)) \\ (c) \implies F \subseteq \text{Fix}_{\text{Gal}(E/F)}(E) \overset{(a)}\implies \\ S=\text{Gal}(E/F) \supseteq \text{Gal}(E/\text{Fix}_{\text{Gal}(E/F)}(E)) = \text{Gal}(E/\text{Fix}_S(E))$$

$\square$

Normal extensions

def: Normal extension

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &\exists P \subseteq F[x] \setminus F: E = F(\parallel P) \\ \hline \\ &E/_{\lhd}F - E \text{ is a normal extension of }F \end{align*}$$

Proposition 2.9.6: Normal extension criteria

$$\begin{align*} &\sphericalangle \\ &E/_AF \in \mathcal F \\ \hline \\ &\text{The following are equivalent:} \\ &\begin{align*} &E/_{\lhd}F \hspace{0.5cm} \tag{a}\\ &\tau: E \rightsquigarrow_{|F} \overline E \implies \tau(E)=E \hspace{0.5cm} \tag{b}\\ &N/E/L/F \in \mathcal F, \sigma: L \rightsquigarrow_{|F} N \implies \sigma(L) \subseteq E, \exists \tau \in \text{Gal}(E/F), \tau|_L=\sigma \hspace{0.5cm} \tag{c}\\ & p \in F[x]^{-}, \exists \alpha \in E: p(\alpha) = 0 \implies p \parallel E \hspace{0.5cm} \tag{d}\\ \end{align*} \end{align*}$$

Proof

$(a) \implies (b)$

Since $E/_{\lhd}F$ there exists $P \subseteq F[x]:E = F(\parallel P)$. Now in the context of $(2.8.13)$ consider $F_1=F_2=F, E_1 = E, E_2 = \phi(E), \sigma=\text{id}$, we have $\tau(E)=F(\parallel P^{\sigma})=F(\parallel P)=E$.

$(b) \implies (c)$

First note that by $(2.4.20)$ $\sigma$ is injective so

$$L \overset{\sigma}\cong \sigma(L)$$

Next, obviously $L \subseteq E, E/_AF \implies L/_AF$. Since $\sigma$ is $F$-homomorphism this implies $\sigma(L)/_AF$. So we have $E/_AF, L/_AF, \sigma(L)/_AF$, it means that

$$\overline E = \overline L= \overline {\sigma(L)}=\overline F = F(\parallel F[x] \setminus F)$$

Now consider under $(2.8.11)$ $F_1:=L, F_2:=\sigma(L), E_1=E_2 = \overline E$, there exists $\rho: \overline E \cong_F \overline E$, $\rho|_L=\sigma$. Note that

$\rho|_F = (\rho|_L)|_F = \sigma|_F=\text{id}$

Now define $\tau:=\rho|_E$. Note that $\sigma=\rho|_L=(\rho|_E)|_L=\tau|_L$. By $(b)$ we have $\tau(E)=E$ so $\sigma(L)=\tau(L)\subseteq \tau(E)=E$.

Finally we have $\tau: E \rightsquigarrow_F E$, $\tau|_F=(\rho|_E)|_F = \rho|_F=\text{id}$, so $\tau \in \text{Gal}(E/F)$

$(c) \implies (d)$

Let $p \in F[x]^{-}, \alpha \in E: p(\alpha) = 0$. Assume $L:=F(\alpha), N:=\overline E$.

Consider $\beta \in \overline E: p(\beta) = 0$ and $\sigma: F(\alpha) \rightsquigarrow_{|F} \overline E: \alpha \mapsto \beta$. By $(c)$ we have $\sigma(F(\alpha))\subseteq E \implies \beta \in E$. Since $\beta$ was arbitrary it means $p \parallel E$.

$(d) \implies (a)$ We know that forall $\alpha \in E: \mathfrak {pol}_F(\alpha) \parallel E$. So for $P:=\{\mathfrak {pol}_F(\alpha), \alpha \in E\}$ we have $E = F(\parallel P)$.

$\square$

Example: Normal extension

$\mathbb C / \mathbb R$ is normal since $\mathbb C=\mathbb R(\parallel x^2+1)$.

Example: Not normal extension

$\mathbb Q(\sqrt[3]2)/\mathbb Q$ is not normal since $\mathfrak {pol}_{\mathbb Q}(\sqrt[3]2) = x^3-2$ and $\mathbb Q(\parallel x^3-2)=\mathbb Q(\sqrt[3]2, \sqrt{-3})$.

Separable extensions

def: Multiplicity of root

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &p \in F[x] \\ &\alpha \in F(\parallel p) \\ &(x-\alpha)^m \mid p, (x-\alpha)^{m+1} \nmid p \\ \hline \\ &\text{ord}_{\alpha}p:=m, \text{ root } \alpha \text{ has multiplicity } m \end{align*}$$

Example: Root multiplicity of 1

$x^3 - 2$ has $\text{ord}_{\sqrt[3]2}(x^3-2)=1$ in $\mathbb Q$. Because $x-\sqrt[3]2 \mid (x^3-2)$ and $(x-\sqrt[3]2)^2 \nmid (x^3-2)$

Example: Root multiplicity of 2

In $\mathbb F_2(x)/\mathbb F_2(x^2)$ we have $\text{ord}_x t^2-x^2 = 2$ because $t^2-x^2=(t-x)^2$

def: Separable polynomial

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &p \in F[x] \\ &\forall \alpha \in F(\parallel p), p(\alpha) = 0 \implies \text{ord}_{\alpha}p=1 \\ \hline \\ &p \in F_\boxminus[x] \end{align*}$$

def: Separable element

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ & \alpha \in E \\ &\mathfrak {pol}(\alpha) \in F_{\boxminus}[x] \\ \hline \\ &\alpha \in F_{\boxminus} \end{align*}$$

def: Separable extension

$$\begin{align*} &\sphericalangle \\ &E/_AF \in \mathcal F \\ &\forall \alpha \in E: \alpha \in F_{\boxminus} \\ \hline \\ &E/_{\boxminus}F, E - \text{ separable extension of } F \end{align*}$$

Example: Separable extension

$\mathbb Q(\sqrt 2)/\mathbb Q$ is separable since $\mathfrak{pol}_{\mathbb Q}(\sqrt 2)= x^2-2 = (x-\sqrt 2)(x+\sqrt 2)$

Example: Non-separable extension

$\mathbb F_2(x)/\mathbb F_2(x^2)$ is not separable as $x$ has a minimal polynomial $t^2-x^2$, and $x$ is a root of multiplicity 2 in this polynomial.

def: Polynomial derivative

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &p \in F[x], p = a_0+a_1x+ \ldots + a_nx^n \\ \hline \\ &p'(x):=a_1 + 2a_2x+\ldots na_nx^{n-1} \end{align*}$$

note

We can assume that $\deg p' = \deg p - 1$. However that's only true if $\text{char } F=0$. If, for example, $\text{char } F = n$ and $p(x)=x^n$ then $p'(x)=nx^{n-1}= 0$.

Proposition 2.9.7: Derivative properties

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &p, q \in F[x] \\ \hline \\ &\begin{align*} & \forall a, b \in F: (ap(x)+bq(x))' = ap'(x) + bq'(x) \hspace{0.5cm} \tag{a}\\ & (p(x)q(x))' = p'(x)q(x) + p(x)q'(x) \hspace{0.5cm} \tag{b}\\ & (p(q(x)))' = p'(q(x))q'(x) \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

Verified directly and left as an exercise.

$\square$

Proposition 2.9.8: Polynomial separability criterion

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &p \in F[x] \setminus F \\ \hline \\ &p \in F_{\boxminus}[x] \iff \gcd(p, p')=1 \end{align*}$$

Proof

Take some $E/F$. Let's prove that

$$\gcd(p, p')=1 \text{ (in F)} \iff \gcd(p, p')=1 \text{ (in E)}$$

If $\gcd(p, p')=1$ in $F$ then by $(2.1.6)$ (which is true for any GCD domain and $F[x]$ is a GCD domain), we have $\exists g, h \in F[x]: pg+p'h = 1$. This is also an equation in $E$ so $pg+p'h = 1 \implies \gcd(p, p')=1$ in $E$.

Conversely assume $\gcd(p, p')=1$ in $E$. Denote $d:=\gcd(p, p')$ in $F$. Then $d \mid p, d \mid p'$ in $F$. But this is also true in $E$. This implies that $d \mid \gcd(p, p')$ in $E$. So $d \mid 1 \implies d = 1$.

Now we fix for the rest of the proof $E:=F(\parallel \{p, p'\})$. Let's prove

$$p \in F_{\boxminus}[x] \iff \nexists \alpha \in E: p(\alpha)=p'(\alpha)=0$$

If $\exists \alpha \in E: p(\alpha)=p'(\alpha)=0$ then $p(x)=(x-\alpha)^mq(x), q(\alpha)\ne 0, p'(x)=m(x-\alpha)^{m-1}q(x)+(x-\alpha)^{m}q'(x)=(x-\alpha)^{m-1}(mq(x)+(x-\alpha)q'(x))$. Now if $m=1$ then $p'(\alpha)=q(\alpha)\ne0$ which is a contraction. So $m>1$ and $p \notin F_{\boxminus}[x]$. Conversely if $p \notin F_{\boxminus}[x]$ then $p = (x-\alpha)^mg(x), m>1$ and so $p'(\alpha) = 0$.

$\implies$

Denote $d:=\gcd(p, p'), d \in E[x]$. Obviously $d \parallel E$ and $\forall \alpha \in E: d(\alpha) = 0 \implies p(\alpha)=p'(\alpha)=0$. But $p \in F_{\boxminus}[x] \implies \nexists \alpha \in E: p(\alpha)=p'(\alpha)=0$, so $\nexists \alpha \in E: d(\alpha)=0$. So $d \in E^* = E[x]^*$ and thus $\gcd(p, p')=1$ in $E$ (remember by $(2.4.11)$ gcd is specified up to a unit). Consequently $\gcd(p, p')=1$ in $F$.

$\impliedby$

If $\exists \alpha \in E: p(\alpha)=p'(\alpha)=0$ then $\gcd(p, p') \ne 1$. So $\nexists \alpha \in E: p(\alpha)=p'(\alpha)=0 \implies p \in F_{\boxminus}[x]$

$\square$

note

This theorem is trivial if we consider $\gcd(p, p')$ in a splitting field of $p$. The value added by this theorem is $\gcd(p, p')$ is considerd in the base field.

Proposition 2.9.9: Irreducible polynomial separability

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &f \in F[x]^- \\ &p \in \mathfrak P \\ \hline \\ &\begin{align*} &\text{char }F = 0 \implies f \in F_{\boxminus}[x] \hspace{0.5cm} \tag{a}\\ &\text{char }F = p > 0 \implies f \in F_{\boxminus}[x] \iff f' \ne 0 \iff f \notin F[x^p] \hspace{0.5cm} \tag{b}\\ &\text{char }F = p > 0 \implies \exists g \in F[x]^- \cap F_{\boxminus}[x], m \ge 0: f(x)=g(x^{p^m}) \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

Note that $f \in F[x]^- \implies (\gcd(f, f')=1) \vee (\gcd(f, f')=f)$ (since $f$ cannot have any non-constant factors).

a.

If $\text{char}F=0$ then $\deg f' = \deg f - 1 \implies f \nmid f' \implies \gcd(f, f')=1 \overset{(2.9.8)}\implies f \in F_{\boxminus}[x]$.

b.

If $\text{char}F=p>0$ then $f \notin F_{\boxminus}[x] \iff \gcd(f, f')=f \iff f \mid f' \overset{\deg f' < \deg f}\iff f'=0 \iff f \in F[x^p]$ (otherwise if there's a least one term $x^n, n \ne pq$ then the derivative cannot be $0$ as it will be the only term with this power and it's not zero).

c.

Assume $m$ is maximal such that $f \in F[x^{p^m}]$ (note that $f \in F[x^{p^0}]$ and the degree of $f$ is bounded so we can always pick such $m$). So assume $f = g(x^{p^{m}}), g \in F[x]$. First, $g \in F[x]^-$, otherwise if $g = h\cdot k \implies f(x)=h(x^{p^{m}})\cdot k(x^{p^{m}})$ would be reducible which is a contradiction. Next, if $g \in F[x^p]$ then it will contradict the maximality of $m$ so $g \notin F[x^p] \overset{(b)}\implies g \in F_{\boxminus}[x]$.

$\square$

Purely inseparable extensions

def: Purely inseparable element

$$\begin{align*} &\sphericalangle \\ & E/F \in \mathcal F \\ &\alpha \in E \\ &Z(p):=\{a \in E: p(a)=0\} \\ &|Z(\mathfrak {pol}_F(\alpha))|=1 \\ \hline \\ &\alpha \in F_\Box \end{align*}$$

def: Purely inseparable extension

$$\begin{align*} &\sphericalangle \\ & E/_AF \in \mathcal F \\ &\forall \alpha \in E: \alpha \in F_\Box \\ \hline \\ &E/_\Box F \end{align*}$$

Example: Purely inseparable extension

If $F \in \mathcal F, \text{char }F = p \in \mathfrak P$ then $F(\sqrt[p]x)/F(x)$ is purely inseparabe extension. This is because $\mathfrak{pol}_{F(x)}(\sqrt[p]x)=t^p-x = (t-x)^p$. (the latter equality follows from $(2.10.3)$ but can also be derived as an exercise).

Proposition 2.9.10: Purely inseparabe element criteria

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &\text{char }F = p \in \mathfrak P \\ & \alpha \in \overline F \\ \hline \\ &\begin{align*} & \alpha \in F_\Box \iff \exists n \ge 0: \alpha^{p^n} \in F\hspace{0.5cm} \tag{a}\\ & \exists n \ge 0: \alpha^{p^n} \in F \implies \mathfrak{pol}_F(\alpha)=(x-\alpha)^{p^n}\hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a., b

$\implies$

Define $f(x):=\mathfrak{pol}_F(\alpha)$. $(2.9.9.c) \implies \exists g \in F[x]^- \cap F_{\boxminus}[x], m \ge 0: f(x)=g(x^{p^n})$. Then $f(x)=(x^{p^n}-b_1)\cdot \ldots \cdot (x^{p^n}-b_n), b_i \in F(\parallel g)$. Additionally $g \in F_\boxminus[x] \implies b_i \ne b_j$. Since $\alpha \in F_\Box$ we know that $\alpha$ is the only root of $f(x)$ so $f(x)=x^{p^n}-b_1=(x-\alpha)^{p^n}$. Note that $f(x)$ is defined over $F$ so $b_1 \in F$ and $\alpha^{p^n}=b_1 \in F$. We also proved $(b)$ because $f(x) = (x-\alpha)^{p^n}$.

$\impliedby$

If $\alpha^{p^n}=a \in F$, then if we define $f(x):=x^{p^n}-a\overset{(2.10.3)}=(x-\alpha)^{p^n}$ then $f(\alpha)=0$. So $\mathfrak {pol}_F(\alpha) \mid f(x) \implies \alpha \in F_\Box$.

$\square$

Proposition 2.9.11: Purely inseparabe and separable extensions

$$\begin{align*} &\sphericalangle \\ &E/_AF \in \mathcal F \\ \hline \\ &\begin{align*} &\alpha \in E \cap F_\boxminus \cap F_\Box \implies \alpha \in F \hspace{0.5cm} \tag{a}\\ &E /_\boxminus F, E /_\Box F \implies E= F \hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

If $\alpha$ is the only root of $\mathfrak {pol}_F(\alpha)$ and its order is $1$ then $\mathfrak {pol}_F(\alpha)=x-\alpha$ and so $\alpha \in F$

b.

Follows from $(a)$

$\square$

Perfect fields

def: Perfect field

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &E/_AF \implies E/_{\boxminus}F \\ \hline \\ &F \in \mathcal F^{\mathcal P} \end{align*}$$

Proposition 2.9.12: Perfect field types

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ \hline \\ &\begin{align*} &\overline F \in \mathcal F^{\mathcal P} \hspace{0.5cm} \tag{a}\\ &\text{char }F = 0 \implies F \in \mathcal F^{\mathcal P} \hspace{0.5cm} \tag{b}\\ &\text{char }F = p \in \mathfrak P \implies (F \in \mathcal F^{\mathcal P} \iff F^p=F) \hspace{0.5cm} \tag{c}\\ \end{align*} & \end{align*}$$

Proof

a.

There are no proper extensions of algebraically closed field so $\overline F \in \mathcal F^{\mathcal P}$

b.

Follows from $(2.9.9.a)$

c.

$\implies$

Consider $a \in F$ and $p(x)=x^p-a$. Let $\alpha \in F(\parallel p), p(\alpha) = 0$. Note that since $\text{char }F=p$ by $(2.10.3)$ we have $(x-\alpha)^p=x^p-\alpha^p=x^p-a$. We know that $\mathfrak {pol}_F(\alpha) \mid x^p-a=(x-\alpha)^p$. So $\mathfrak {pol}_F(\alpha) = (x-\alpha)^n, n \le p$. So we have:

$$F \in \mathcal F^{\mathcal P} \implies F(\alpha)/_\boxminus F \implies \\ \alpha \in F_\boxminus \implies \text{ord}_{\alpha}(\mathfrak {pol}_F(\alpha))=1 \implies (x-\alpha)^2 \nmid \mathfrak {pol}_F(\alpha) \implies \\ \mathfrak {pol}_F(\alpha) = x-\alpha \implies \alpha \in F \implies a \in F^p$$

So we proved $F \subseteq F^p$. Since fields are multiplicatively closed $F^p \subseteq F$.

$\impliedby$

Consider some $E/F$ and $\alpha \in E$, $f(x):=\mathfrak {pol}_F(\alpha)$. $(2.9.9.c) \implies \exists g \in F[x]^- \cap F_{\boxminus}[x], m \ge 0: f(x)=g(x^{p^m})$. If $g(x)=a_0+a_1x + \ldots +a_nx^n$ then $\exists b_i \in F: b_i^p=a_i$. So $f(x)=\sum_i b_i^px^{ip^m}=(\sum_i b_ix^{ip^{m-1}})^p$ which contradicts irreducibility of $f$ for any $m \ge 1$. So $m=0$ and $f(x)=g(x) \in F_\boxminus[x] \implies \alpha \in F_\boxminus$.

$\square$

Example: Perfect fields

$\mathbb Q, \mathbb R, \mathbb C$ are fields of $\text{char } F = 0$ so all of them are perfect.

Any finite field is perfect since it has $\text{char } F=p, p \in \mathfrak P$ and $F=F^p$ (this fact will be proved it the next section on finite fields)

Example: Non-perfect fields

$\mathbb F_p(x)$ is not perfect since it has a non-separable extension $\mathbb F_p(\sqrt[p]x)$

Galois extensions

def: Galois extension

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &F = \text{Fix}_{\text{Gal}(E/F)}(E) \\ \hline \\ &E/_{\text{Gal}}F, E - \text{galois extension over } F \end{align*}$$

Example: Galois extensions

$\mathbb C / \mathbb R$ is Galois, since $\text{Gal}(\mathbb C / \mathbb R) = \{\text{id}, \sigma\}, \sigma: a+bi \mapsto a-bi$. It can be seen that it fixes exactly $\mathbb R$.

$\mathbb Q(\sqrt 2)/\mathbb Q$ is also Galois (can be proved similarly).

Example: Non-Galois extensions

$\mathbb Q(\sqrt[3] 2)/\mathbb Q$ is not Galois because $\text{Gal}(\mathbb Q(\sqrt[3] 2)/\mathbb Q) = \{\text{id}\}$ so $\text{Fix}_{\text{Gal}(\mathbb Q(\sqrt[3] 2)/\mathbb Q)}(\mathbb Q(\sqrt[3] 2))=\mathbb Q(\sqrt[3] 2)$

$\mathbb F_2(x)/\mathbb F_2(x^2)$ is not Galois. From the examples above we know that $\text{Gal}(\mathbb F_2(x)/\mathbb F_2(x^2))=\text{id}$, so $\text{Fix}_{\text{Gal}(\mathbb F_2(x)/\mathbb F_2(x^2))}(\mathbb F_2(x))=\mathbb F_2(x)$.

Next we'll state several proposition without the proofs.

Proposition 2.9.13: Galois extension criterion

$$\begin{align*} &\sphericalangle \\ &E / F \\ &[E:F] < \infty \\ \hline \\ &E/_{\text{Gal}}F \iff |\text{Gal}(E/F)|=[E:F] \end{align*}$$

Proposition 2.9.14: Galois extension criterion for simple extensions

$$\begin{align*} &\sphericalangle \\ &E / F \\ &\alpha \in \overline F_E \\ &V := \{y \in F(\alpha): \mathfrak {pol}(\alpha)(y)=0\} \\ \hline \\ &F(\alpha)/_{\text{Gal}}F \iff |V| = \deg \alpha \end{align*}$$

Proposition 2.9.15: Galois extension criteria

$$\begin{align*} &\sphericalangle \\ &E/_AF \in \mathcal F \\ \hline \\ &\text{The following are equivalent:} \\ &\begin{align*} &E/_{\text{Gal}}F \hspace{0.5cm} \tag{a}\\ &E/_{\lhd}F, E/_{\boxminus}F \hspace{0.5cm} \tag{b}\\ &\exists P \subseteq F[x]\setminus F, p_i \in F_{\boxminus}[x], E = F(\parallel P) \hspace{0.5cm} \tag{c} \end{align*} \end{align*}$$

Proof

$(a) \implies (b)$

Take some $\alpha \in E$. Let

$$X:=\{\sigma(\alpha), \sigma \in \text{Gal}(E/F)\}$$

Note that by $(2.9.3.b)$ any $\sigma$ just permutes the roots of $\mathfrak {pol}(\alpha)$, so $X$ is finite and we can assume

$$X=\{\alpha_1, \ldots, \alpha_n\}$$

Denote

$$p(x):=\prod_i(x-\alpha_i) \in E[x]$$

Note that $\forall \sigma \in \text{Gal}(E/F): p^{\sigma}=p$ because by $(2.9.3.b)$ $\sigma$ just permutes the roots of polynomial. So $p\in \text{Fix}_{\text{Gal}(E/F)}(E)[x]=F[x]$. Next $\mathfrak {pol}_F(\alpha) \mid p \implies \mathfrak {pol}_F(\alpha) \parallel E, \mathfrak {pol}_F(\alpha) \in F_{\boxminus}[x] \implies \alpha \in F_{\boxminus}$. So we proved that:

$$\forall \alpha \in E: \mathfrak {pol}_F(\alpha) \parallel E, \alpha \in F_{\boxminus}$$

It immediately follows that $E/_{\boxminus}F$. Finally, let's use $(2.9.6)$. If $f \in F[x]^-, \alpha \in E: f(\alpha)=0$, then $f(x)=u\cdot\mathfrak {pol}_F(\alpha), u \in F^*$. Since $\mathfrak {pol}_F(\alpha) \parallel E$ so is $f \parallel E$ and so $E/_{\lhd}F$.

$(b) \implies (c)$

Define $P:=\{\mathfrak{pol}_F(\alpha):\alpha \in E\}$, since by $(2.9.6)$ for every $p \in P$ we have $p \parallel E$ and so all the roots of $P$ will be exactly $E$, so by definition of a splitting field: $E = F(\parallel P)$. Since $E/_{\boxminus}F$ we have $\forall p \in P: p \in F_{\boxminus}[x]$.

$(c) \implies (a)$

We will skip the proof, it can be found in Patrick Morandi "Field and Galois theory".

$\square$

Fundamental theorem of Galois theory

Theorem 2.9.16: Fundamental theorem of Galois theory

$$\begin{align*} &\sphericalangle \\ &E/_{\text{Gal}}F \in \mathcal F \\ &[E:F] < \infty \\ &\mathcal F_{E:F} := \{L: F \subseteq_F L \subseteq_F E\} \\ &\mathcal G_{E:F}:= \{H: H \subseteq_G \text{Gal}(E/F)\}\\ &\phi: \mathcal F_{E:F} \to \mathcal G_{E:F}, L \mapsto \text{Gal}(E/L)\\ &\phi^{-1}: \mathcal G_{E:F} \to \mathcal F_{E:F}, H \mapsto \text{Fix}_H(E) \\ &L \in \mathcal F_{E:F}, H \in \mathcal G_{E:F} \\ \hline \\ &\begin{align*} & \phi: \mathcal F_{E:F} \leftrightarrow \mathcal G_{E:F}, \phi \phi^{-1}=\text{id}, \phi^{-1} \phi =\text{id} \hspace{0.5cm} \tag{a}\\ &[E:L]=|H|, [L:F]=[\text{Gal}(E/F):H] \hspace{0.5cm} \tag{b} \\ &H \lhd_G \text{Gal}(E/F) \iff L/_{\text{Gal}}F \hspace{0.5cm} \tag{c} \\ \end{align*} \end{align*}$$ $$\begin{CD} E @>\text{Gal}(E/E)>> \{\text{id}\} @>\text{Fix}_{\{\text{id}\}}(E)>> E \\ @AAA @VVV @AAA \\ L @>\text{Gal}(E/L)>> H @>\text{Fix}_{H}(E)>> L \\ @AAA @VVV @AAA \\ F @>\text{Gal}(E/F)>> \text{Gal}(E/F) @>\text{Fix}_{\text{Gal}(E/F)}(E)>> F \end{CD}$$

Example: Intermediate fields in a splitting field of $x^3-2$ over $\mathbb Q$.

Polynomial $x^3-2$ has 3 roots:

$$\sqrt[3]2, \sqrt[3]2 w, \sqrt[3]2w^2, w = \frac{i\sqrt 3-1}{2}=e^{\frac{2\pi i}{3}}$$

We know from examples in the previous section that $E:=\mathbb Q(\parallel x^3-2)=\mathbb Q(\sqrt[3]2, w)$ and $[\mathbb Q(\sqrt[3]2, w):\mathbb Q]=6$. Denote $F:=\mathbb Q$.

Note that $E/_\lhd F$ by definition and $E/_{\boxminus} F$ by $(2.9.12.b)$. So by $(2.9.15.b)$ we have $E/_\text{Gal}F$ and thus we can apply Fundamental theorem of Galois theory $(2.9.16)$. If we assume $L=F$ then $H=\text{Gal}(E/F)$ so by $(2.9.16.b)$ we have

$$|\text{Gal}(E/F)|=|[E:F]|=6$$

It can be shown that the only groups of order 6 are $\Z_6$ and $S_3$ - the group of all permutations of a set $\{1, 2, 3 \}$. Which one is the Galois group?

We know from the examples in the previous section that $\mathbb Q(\sqrt[3]2)/\mathbb Q$ is not normal hence is not Galois and so by $(2.9.16.c)$ the corresponding Galois subgroup is not normal. But that's not possible for $\Z_6$ because it's Abelian. So

$$\text{Gal}(E/F)=S_3=\{(1), (1\,2),(1\,3),(2\,3), (1\,2\,3), ((1\,3\,2))\}$$

The proper non-trivial subgroups of this group are:

$$H_1=\{(1), (1\,2)\} \\ H_2=\{(1), (1\,3)\} \\ H_3=\{(1), (2\,3)\} \\ H_4=\{(1), (1\,2\,3), (1\,3\,2)\} \\$$

We can simplify the structure if we define $\tau:=(1\,2), \sigma:=(1\,2\,3)$. Then $(1\,3)=\sigma \tau, (2\,3) = \sigma^2\tau$. So we can write:

$$H_1=\lang \tau \rang_G \\ H_2=\lang \sigma \tau \rang_G \\ H_3=\lang \sigma^2 \tau \rang_G \\ H_4=\lang \sigma \rang_G \\$$

Note that $\text{ord }\sigma = 3, \text{ord }\tau = 2$.

Now let's take a look at the $F$-automorphisms of $E$. They are defined by the action on two elements: $\sqrt[3]2$ and $w$. $\mathfrak {pol}_F(\sqrt[3]2)=x^3-2, \mathfrak {pol}_F(w)=x^3-1$. So the possibilites for mapping $\sqrt[3]2$ are $\{\sqrt[3]2, w\sqrt[3]2, w^2\sqrt[3]2\}$ and the possibilites for $w$ are $\{w, w^2\}$. Note that we cannot map $w$ to $1$, because $1 \in F$ and we're considering automorphisms fixing $F$. So it's clear that:

$$\sigma: \sqrt[3]2 \mapsto w\sqrt[3]2, w \mapsto w \\ \tau: \sqrt[3]2 \mapsto \sqrt[3]2, w \mapsto w^2 \\$$

Note that $\sigma$ makes a cyclic shift of the $\sqrt[3]2, w\sqrt[3]2, w^2\sqrt[3]2$ roots akin to $(1\,2\,3)$. On the other hand $\tau$ swaps $w$ and $w^2$ roots and $w \sqrt[3]2$ and $w^2\sqrt[3]2$ roots just like $(1\,2)$ would do.

Finally let's see what fields are fixed by Galois subgroups:

$$H_1=\lang \tau \rang_G \implies L_1=\text{Fix}_{H_1}(\mathbb Q(\sqrt[3]2, w))=\mathbb Q(\sqrt[3]2) \\ H_2=\lang \sigma\tau \rang_G \implies L_2=\text{Fix}_{H_2}(\mathbb Q(\sqrt[3]2, w))=\mathbb Q( w^2\sqrt[3]2) \\ H_3=\lang \sigma^2 \tau \rang_G \implies L_3=\text{Fix}_{H_3}(\mathbb Q(\sqrt[3]2, w))=\mathbb Q( w\sqrt[3]2) \\ H_4=\lang \sigma \rang_G \implies L_4=\text{Fix}_{H_4}(\mathbb Q(\sqrt[3]2, w))=\mathbb Q(w)$$

Let's see for example why $L_2 = \mathbb Q( w^2\sqrt[3]2)$. Consider a root $w^2\sqrt[3]2$: $\sigma \tau(w^2\sqrt[3]2)=\sigma(\tau(w^2\sqrt[3]2))=\sigma(w\sqrt[3]2)=w^2\sqrt[3]2$. We can easily check that other roots are not fixed. Also note that if $\sigma \tau$ fixes a root then all other members of $\lang \sigma \tau \rang_G$ also fix it.

On the final note: it's easy to see that $H_1 \cong H_2, \cong H_3$ and so are $L_1 \cong L_2 \cong L_3$ (as they are generates from the root of the same polynomial)

note

It might seem strange that $[\mathbb Q(w):\mathbb Q]=2$ while $w$ is a root of $x^3-1$. The thing is $w^2 = -1-w$ hence such extension degree.

note

The only normal subgroup is $\lang \sigma \rang_G$ so the only Galois extension of $\mathbb Q$ is $\mathbb Q(w)$.