Nepal Algebra Project (NAP)

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

Monday, May 21, 2018 (also download from here)

$\mathbf{F}$ 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. $\mathbf{F}$ $(\alpha_1$, $\cdots$, $\alpha_m)$ $=$ $\mathbf{K}$.
  In addition,
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 $\mathbf{Q}[x]$.
You can download a few more examples from here.

Tuesady, May 22, 2018 (also download from here)

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. $\mathbf{F}$ $(\alpha_1$, $\cdots$, $\alpha_m)$ $=$ $\mathbf{K}$.
  In addition,
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 $\pi$.

Wednesday, May 23, 2018 (also download from here)

$\mathbf{F}$ 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 $\omega=e^{2\pi i/3}$.
You can download a few more examples from here.

Wednesday, May 23, 2018 (also download from here)

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 $\gcd(f,f')=1$.
You can download a few more examples from here.

Monday, 28 May, 2018 (also download from here)

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 $\Aut_\mathbf{F}(\mathbf{E})$ $=$ $\hom_\mathbf{F}$ $(E,E)$.
You can download a few more examples from here.

Tuesday, 29 May, 2018 (also download from here)

$\mathbf{F} \subset \mathbf{E}$ 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 $ $[\mathbf{E}:\mathbf{E}^G]$.
You can download a few more examples from here.

Thursday, 31 May, 2018 (also download from here)

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.
You can download a few more examples from here.

Homeworks: problem set 1 , problem set 2

Solutions: problem set 1 , problem set 2

Nepal Algebra Project (NAP) / नेपाल बीज-गणित परियोजना

COURSE YEAR 2018