$\DeclareMathOperator{\Q}{Q} \DeclareMathOperator{\Aut}{Aut}$

# COURSE YEAR 2018

Module - II : May 20, 2018 – June 01, 2018, René Schoof and Laura Geatti
Topics: Algebraically closed fields, Maps from simple extensions, Splitting fields, Multiple roots, Groups of automorphisms of fields, Separable, normal, and Galois extensions, The fundamental theorem of Galois theory
[Homeworks]

• $\mathbf{F}$ field.
• Recall the following basic fact from Module 1:
• Let $f\in \mathbf{F}[x]$ be an irreducible polynomial. Then there exists a field extension $\mathbf{F}\subset \mathbf{K}$ with the property that $f$ has a zero in $\mathbf{K}$.
• $\mathbf{K}$ $\cong$ $\mathbf{F}[x]/(f)$
• The following Proposition was stated and illustrated
• Proposition. (Milne, Prop.2.4) For every polynomial $f \in \mathbf{F}[x]$, there exists a field extension $\mathbf{F} \subset \mathbf{K}$ with the property that
1. $f$, as a polynomial in $\mathbf{K}[x]$, decomposes as $f(x)$ $=$ $c(x-\alpha_1)$ $\cdots$ $(x-\alpha_m)$, with $\alpha_i \in \mathbf{K}$, $c \in \mathbf{K};$
2. $\mathbf{F}$ $(\alpha_1$, $\cdots$, $\alpha_m)$ $=$ $\mathbf{K}$.
3. the field $\mathbf{K}$ is unique up to $\mathbf{F}$-isomorphism;
4. If $\deg(f)=n$, then $[\mathbf{K}:\mathbf{F}]\le n!$.
• By definition, $\mathbf{K}$ is a splitting field of $f$ over $\mathbf{F}$.
Note that it depends both on $f$ and on $\mathbf{F}$.
• Examples: Denote by $\mathbf{F}_f$ a spiltting field of $f$ over $\mathbf{F}$.
$f=X^2-2$, $\mathbf{Q}_f=\mathbf{Q}(\sqrt 2)$;
$f=X^2-2$, $\mathbf{R}_f=\mathbf{R}$;
$f=X^2+1$, $\mathbf{Q}_f=\mathbf{Q}(i)$;
$f=X^2+1$, $\mathbf{R}_f=\mathbf{R}(i)=\mathbf{C}$;
$f=X^3-1$, $\mathbf{Q}_f=\mathbf{Q}(\omega)$, where $\omega=e^{i2\pi/3}$;
$f=X^3-1$, $\mathbf{R}_f=\mathbf{R}(\omega)=\mathbf{C}$;
$f=X^3-2$, $\mathbf{Q}_f=\mathbf{Q}(\root{3} \of {2},\omega)$.
• Excercise. Consider the real number $\alpha=\sqrt 2+ \root{3} \of {2}$. Determine the minimal polynomial of $\alpha$ in $\mathbf{Q}[x]$.

• Proof of items (a), (b), (d) of the following proposition. Item (c) will be addressed later.
• Proposition. (Milne, Prop.2.4) For every polynomial $f\in\mathbf{F}[x]$, there exists a field extension $\mathbf{F} \subset \mathbf{K}$ with the property that
1. $f$, as a polynomial in $\mathbf{K}[x]$, decomposes as $f(x)$ $=$ $c(x-\alpha_1)$ $\cdots$ $(x-\alpha_m)$, with $\alpha_i \in \mathbf{K}$, $c \in \mathbf{K}$
2. $\mathbf{F}$ $(\alpha_1$, $\cdots$, $\alpha_m)$ $=$ $\mathbf{K}$.
3. the field $\mathbf{K}$ is unique up to $\mathbf{F}$-isomorphism;
4. If $\deg(f)=n$, then $[\mathbf{K}:\mathbf{F}] \le n!$.
• Definition An algebraically closed field is a field $\mathbf{F}$ with the property that every polynomial $f \in \mathbf{F}[x]$ splits as $f(x)$ $=$ $c(x-\alpha_1)$ $\ldots$ $(x-\alpha_m)$, with $\alpha_i$, $c \in \mathbf{F}$.
In other words, all zeroes of every polynomial $f\in \mathbf{F}[x]$ are in $\mathbf{F}$.
• Example. The field of complex numbers $\mathbf{C}$ is algebraically closed: by the Fundamental of Algebra, every non-constant polynomial in $\mathbf{C}[x]$ has a zero in $\mathbf{C}$. This implies that every $f \in \mathbf{C}[x]$ splits as $f(x)$ $=$ $c(x-\alpha_1)$ $\ldots$ $(x-\alpha_m)$, with $\alpha_i$ in $\mathbf{C}[x]$.
• Example. The fields $\mathbf{Q}$ and $\mathbf{R}$ are not algebraically closed: for example the polynomial $f(x)=x^2+1$ has no zero in $\mathbf{Q}$, nor in $\mathbf{R}$.
• Equivalent characterizations of an algebraically closed field (Milne, Prop.1.4.2 & Def. 1.4.3(a)).
• Definition. An algebraic closure of a field $\mathbf{F}$ is an extension $\overline{\mathbf{F}}$ of $\mathbf{F}$ which is algebraically closed over $\mathbf{F}$ and consists of algebraic elements over $\mathbf{F}$.
• Fact. (Milne Chapter 6) For every field $\mathbf{F}$ there exists an extension $\mathbf{F}\subset \mathbf{L}$ which is algebraically closed.
• Proposition. (Milne, Prop. 1.45) Every field $\mathbf{F}$ admits an algebraic closure $\overline{\mathbf{F}}$.
$\overline{\mathbf{F}}$ is unique, but only up to isomorphism and the isomorphism is generally not unique. Colloquially we often speak of the algebraic closure of a given field, but it is not correct.
Let $\mathbf{L}$ be an algebraically closed extension of $\mathbf{F}$. Then one defines

$\overline {\mathbf{F}}$ $=$ $\{ x\in \Omega ~|~$ $x$ is algebraic over $\mathbf{F} \}$.

The proof that $\overline{\mathbf{F}}$ satisfies the required properties, i.e. is a field and it is algebraically closed, is based on the following two facts:
1. $\mathbf{F} \subset \mathbf{K}$ fields, $\alpha\in \mathbf{K}$. Then $\alpha$ is algebraic over $\mathbf{F}$ if and only if there exists a finite extension $\mathbf{E}$ of $\mathbf{F}$ such that $\mathbf{F}$ $\subset$ $\mathbf{E}$ $\subset$ $\mathbf{K}$.
2. $\mathbf{F}$ $\subset$ $\mathbf{K}$ $\subset$ $\mathbf{L}$ fields. If $\mathbf{K}$ is algebraic over $\mathbf{F}$ and $\mathbf{L}$ is algebraic over $\mathbf{K}$, then $\mathbf{L}$ is algebraic over $\mathbf{F}$.
• Example. $\overline {\mathbf{R}} \cong \mathbf{C}$;
In $\mathbf{C}$ there is an algebraic closure $\overline {\mathbf{Q}}$ of $\mathbf{Q}$: it consists of the complex numbers which are algebraic over $\mathbf{Q}$. It is a proper subset of $\mathbf{C}$ because it is countable: for example it does not contain $\pi$.

• $\mathbf{F}$ field, $\mathbf{F} \subset \mathbf{E}$ finite extension, $\mathbf{F} \subset \mathbf{L}$ arbitrary extension.
(Hom)$_\mathbf{F}(E,L)$ $=$ $\{\phi\colon\mathbf{E}$ $\to$ $\mathbf{L},$ field homomorphisms such that $\phi_{|\mathbf{F}}:$ $=$ $id_{|\mathbf{F}}\}$
• Goal:estimate the cardinality of (Hom)$_\mathbf{F}(E,L)$.
Started from special case $\mathbf{E}=\mathbf{F}(\alpha)$, where $\alpha\in\mathbf{E}$, with minimal polynomial $f\in \mathbf{F}[x]$. In this case, an $\mathbf{F}$-homomorphism $\phi$ is completely determined by $\phi(\alpha)$.
• Lemma. (Milne, Prop.2.1) (Hom)$_\mathbf{F}(\mathbf{F}(\alpha),\mathbf{L})$ is in 1-1 correspondence with the set $\mathbf{Z}(f)$ $\cap$ $\mathbf{L}$ $=$ $\{z\in$ $\mathbf{L} ~|~ f(z)$ $= 0\}$. The correspondence is given by $\phi \mapsto \phi(\alpha)$.
• Corollary. #(Hom) $_\mathbf{F}(\mathbf{F}(\alpha)$, $\mathbf{L}) \le \deg(f)$.
• In general (Milne, Prop.2.7)
• Proposition. #(Hom)$_\mathbf{F}(\mathbf{E}$, $\mathbf{L}) \le$ $[\mathbf{E}:\mathbf{F}]$.
• Example. $\mathbf{F}=\mathbf{Q}$, $\alpha=\sqrt 2$, $\mathbf{L}=\mathbf{R}$.
Then #(Hom)$_\mathbf{Q}(\mathbf{Q}(\sqrt 2),\mathbf{R})=2$.
There are 2 possibilities: $\phi(\sqrt 2)=\pm \sqrt 2$.
• Example. $\mathbf{F}$ $=$ $\mathbf{Q}$,$\alpha$ $=$ $\root{3}\of{2}$, $\mathbf{L}=\mathbf{R}$.
Then #(Hom)$_\mathbf{Q}(\mathbf{Q}$$(\root{3}\of{2})$,$\mathbf{R}) = 1$.
There is only 1 possibility: $\phi(\root{3}\of{2})$ $=$ $\root{3}\of{2}$.
• Example. $\mathbf{F}=\mathbf{Q}$, $\alpha=\root{3}\of{2}$, $\mathbf{L}=\mathbf{C}$.
Then #(Hom)$_\mathbf{Q}$ $(\mathbf{Q}(\root{3}\of{2})$, $\mathbf{C})$ $=3$.
There are 3 possibilities: $\phi(\root{3}\of{2})$ $=$ $\root{3}\of{2}$, $\root{3}\of{2} \omega$, $\root{3}\of{2} \omega^2$, where $\omega=e^{2\pi i/3}$.

• Concluded the proof of Proposition 2.7, in Milne:
Proposition 2.7(b) Let $\mathbf{F}$ be a field, let $f\in\mathbf{F}[x]$ be a polynomial. If $\mathbf{K}$ and $\mathbf{K}'$ are splitting fields of $f$ over $\mathbf{F}$, then there exists a field $\mathbf{F}$-isomorphism $\phi\colon \mathbf{K}\to\mathbf{K}'$.
• Definition. Let $\mathbf{F}$ be a field. A polynomial $f\in \mathbf{F}[x]$ is called separable if in a splitting field of $f$ the zeroes of $f$ are distinct.
• Examples:
1. $f(x)$ $=$ $x(x-1)$ $\in$ $\Q[x]$ is separable;
2. $f(x)=$ $(x-1)^2$ $\in\Q[x]$ is not separable;
3. $f(x)=$ $x^2+1$ $\in \mathbf{R}[x]$ is separable: the splitting field of $f$ is $\mathbf{R}_f=\mathbf{C}$ and there $f$ has distinct zeros $\pm i$.
4. $f(x)=$ $x^2+1=$ $(x+1)^2$ $\in \mathbf{Z}/$ $2\mathbf{Z}[x]$ is not separable.
• Definition. Let $\mathbf{F}$ be a field. The derivative of a polynomial $f(x)=$ $\sum_{n=0}^d$ $a_nx^n$ in $\mathbf{F}[x]$ is defined as follows

$f'(x)=$ $\sum_{n=1}^d$ $na_nx^{n-1}$

(here $n=$ $1+\ldots$ $+1$, where $1\in\mathbf{F}$ and $nx=x+$ $\ldots +x$).

• Excercise. The derivative defined above is linear and satisfies the Leibnitz rule.
• Proposition. (Prop.2.13, Milne) Let $\mathbf{F}$ be a field. A polynomial $f\in \mathbf{F}[x]$ is separable if and only if $\gcd(f,f')=1$.

• Defined the characteristic of a field $\mathbf{F}$. The characteristic of a field is either equal to 0 or to a prime $p$.
• Examples of fields of characteristic $0$: $\mathbf{Q}$, $\mathbf{R}$, $\mathbf{C}$, $\mathbf{Q}(\alpha)$, $\ldots$ Every field of characteristic 0 contains a copy of $\mathbf{Q}$.
• Examples of of fields of characteristic $p$: the field of $p$ elements $\mathbf{F}_p =$ $\mathbf{\mathbf{Z}} /p \mathbf{Z}$, the field of $p^n$ elements $\mathbf{F}_{p^n}\cong$ $\mathbf{F}_p[x]/(f)$, where $f$ is an irreducible polynomial of degree $n$ in $\mathbf{F}_p[x]$. Every field of characteristic $p$ contains a copy of $\mathbf{\mathbf{Z}}/p\mathbf{\mathbf{Z}}$.
Definition. A field $\mathbf{F}$ is perfect if either $char(\mathbf{F})=0$ or $char(\mathbf{F})=p$ and $\mathbf{F}^p=\mathbf{F}$.
• Examples of perfect fields: $\mathbf{Q}$, $\mathbf{R}$, $\mathbf{C}$, $\mathbf{Q}(\alpha)$, $\mathbf{F}_p$, $\mathbf{F}_{p^n}$.
• Examples of non-perfect fields: $\mathbf{F}=\mathbf{F}_p(T).$
• Recall that a polynomial is separable if it has distinct zeros in any splitting field.
Proposition (Milne, Prop 2.12) Let $\mathbf{F}$ be a field and let $f\in\mathbf{F}[x]$ be an irreducible polynomial. Then
1. If $char(\mathbf{F})$ $=0$, then $f$ is separable;
2. If $char(\mathbf{F})$ $=p$, then there exists an irreducible separable polynomial $g\in\mathbf{F}[x]$ such that $f(x)$ $=$ $g(x^{p^e})$, for some $e\in \mathbf{Z}_{\ge 1}$;
3. If $\mathbf{F}$ is perfect, then $f$ is separable.
• Let $\mathbf{F}\subset \mathbf{E}$ be a field extension.
$\Aut_\mathbf{F}(\mathbf{E})$ $=$ $\{$ $\phi\colon$ $\mathbf{E}\to$ $\mathbf{E}$, $\mathbf{F}$ -field isomorphisms $\}$.
If the extension has finite degree $[\mathbf{E}:\mathbf{F}]$ $<\infty$, then $\Aut_\mathbf{F}(\mathbf{E})$ $=$ $\hom_\mathbf{F}$ $(E,E)$.

• $\mathbf{F} \subset \mathbf{E}$ finite field extension; $\Aut_\mathbf{F}(\mathbf{E})$ the group of $\mathbf{F}$-field homomorphisms.
• Let $G$ be a subgroup of $\Aut_\mathbf{F}(\mathbf{E})$. Denote by $\mathbf{E}^G=$ $\{$ $x\in \mathbf{E}$ $\vert$ $phi(x)=x$, $\forall \phi\in G$ $\}$ the subset of invariant elements. Then $\mathbf{E}^G$ is a subfield of $E$ (Exercise).
• The next two theorems estimate the cardinality of $\Aut_\mathbf{F}(\mathbf{E})$ both from above and from below.
• Theorem 1. (Milne, Thm.3.2)
1. $\mathbf{F}\subset \mathbf{E}$ finite field extension. Then $\#\Aut_\mathbf{F}(\mathbf{E})$ $\le$ $[\mathbf{E}:\mathbf{F}]$;
2. If $\mathbf{E}=\mathbf{F}_f$ is the splitting field of a separable polynomial $f\in \mathbf{F}[x]$, then $\#\Aut_\mathbf{F}(\mathbf{E})$ $=$ $[\mathbf{E}:\mathbf{F}]$
• Theorem 2. (Artin's Lemma) (Milne, Thm.3.4) Let $\mathbf{E}$ be a field, and let $G$ be a finite subgroup of $\Aut(\mathbf{E})$. Then $\#G\ge$ $[\mathbf{E}:\mathbf{E}^G]$.

Note: There was a lecture before this lecture, in which, we mainly solve the students' problems. Hence, concerning lecture counter, lecture-7 means lecture-8.

• Corollary. (Milne, Cor.3.5) $\mathbf{E}$ field. For any finite group $G\subset$ $\Aut(\mathbf{E})$, one has $G=$ $\Aut_{\mathbf{E}^G}(\mathbf{E})$.
In particular, if $\mathbf{F} \subset \mathbf{E}$ is a finite degree extension, then $[\mathbf{E}:\mathbf{E}^G]$ $=\# G.$
Let $\mathbf{F}\subset \mathbf{E}$ be an algebraic extension.
• Definition. $\mathbf{F}\subset \mathbf{E}$ is called separable if the minimum polynomial over $\mathbf{F}$ of every element in $\mathbf{E}$ is separable.
• Definition. $\mathbf{F}\subset \mathbf{E}$ is called normal if the minimum polynomial over $\mathbf{F}$ of every element in $\mathbf{E}$ splits in $\mathbf{E}[x]$.
In particular, an algebraic extension $\mathbf{F}\subset \mathbf{E}$ is both separable and normal if the minimum polynomial $f$ over $\mathbf{F}$ of every element in $\mathbf{E}$ splits in $\mathbf{E}[x]$ as $f(x)=$ $c(x-\alpha_1)$ $\ldots$ $(x-\alpha_n)$, with $c,\alpha_i\in$ $\mathbf{E}$ and $\alpha_i\not=\alpha_j$, for $i\not=j$.
• Theorem. (Milne, thm.3.10) Let $\mathbf{F} \subset \mathbf{E}$ be an extension. Then the following four statements are equivalent:
1. $\mathbf{F}$ is the invariant subfield of $\Aut_\mathbf{F}(\mathbf{E})$;
2. $\mathbf{F}=\mathbf{E}^G$, for some finite subgroup $G\subset$ $\Aut(\mathbf{E})$;
3. $\mathbf{E}$ is a finite degree separable and normal extension of $\mathbf{F}$;
4. $\mathbf{E}$ is the splitting field of a separable polynomial in $\mathbf{F}[x]$.
• Definition. If an extension $\mathbf{F}\subset \mathbf{E}$ satisfies any of the above statements, then ot is called a Galois extension and $G:=$ $\Aut_\mathbf{F}(\mathbf{E})$ is its Galois group.
• Example. (see Milne, Example 3.21) $\mathbf{Q} \subset$ $\mathbf{Q}(\zeta_7)$, where $\zeta_7$ is a primitive $7^{th}$ root of 1.