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 Notes on 【Galois Theory】
02-20-2017, 11:47 AM (This post was last modified: 02-25-2020 01:33 PM by elim.)
Post: #1
 elim Moderator     Posts: 578 Joined: Feb 2010 Reputation: 0
Notes on 【Galois Theory】
1 Revision from Groups, Rings and Fields Galois Theory Dr P.M.H. Wilson [Michaelmsa Term 2000].pdf (Size: 295.07 KB / Downloads: 0)

1.1 Field Extensions
Let $K,\,L$ be fields, ring homomorphism $\varphi: K\to L$ is
non-zero, then $\small\ker\varphi = \{0\},\;\varphi\,$is a homomorphism of fields
$\small\varphi(a/b)=\varphi(a)/\varphi(b).\;\;\color{grey}{0\ne\alpha\in K\implies\varphi(\alpha)\ne 0\implies \alpha\not\in\ker(\varphi)}$

1.1.1 Definition If there is a homomorphism of fields
$\varphi:K\hookrightarrow L$, then $L$ is a field extension of $K$, $\varphi$ is an
embedding of $K$ into $L$.

1.1.2 Remark If $\varphi: K\hookrightarrow L$, then $\varphi:K\to \varphi(K)$ is an
isomorphism. $L$ might be an field extension via different
homomorphism thus we simply use $L/K$ to denote that $L$
is a field extension of $K$ and assume that $K\subset_{\small F}L$.

1.1.3 Lemma $(\forall j\in J\;(K_j\subset_{\small F} L))\implies \displaystyle{\bigcap_{j\in J}K_j \subset_{\small F} L}.$
$\qquad\;(A\subset_{\small F} B\,$means $A$ is a subfield of field $B)$

If $K\subset_{\small F} L,\; S\subset L,\,$then $\small\displaystyle{K(S):=\bigcap\{M\mid K\cup S\subset M\subset_{\small F} L\} }$
is the smallest subfield of $L$ containing $K\cup S$.
Write $K(a_1,\ldots, a_n)\,$For $K(\{a_1,\ldots,a_n\})$

1.1.4 Definition $\small L/K\,$is finitely generated if $\small K(\alpha_1,\ldots,\alpha_n) = L$
(where $n\in\mathbb{N}$). If $L=K(\alpha),\,$the extension is simple.

1.1.5 Definition Given $\small L/K$, we say $\small\alpha\in L$ is algebraic over
$K$ if $\small\exists f\in K[X]\;(f(\alpha) = 0).$ Otherwise $\alpha\in L\,$is transcendental
over $K.$ If $\alpha$ is algebraic, $f: \small X^n+a_{n-1}X^{n-1}+\cdots +a_0$ is the
monic polynomial of smallest degree such that $f(\alpha) = 0$ is
called the minimal polynomial of $\alpha$. Clearly such an $f$ is
unique and irreducible.

1.1.6 Definition $L/K$ is algebraic if $\alpha$ is algebraic over $K$
$(\forall \alpha\in L)$; It is pure transcendental if $\alpha$ is transcendental
over $K\,(\forall \alpha\in L-K).$
02-20-2017, 12:08 PM (This post was last modified: 02-25-2020 04:42 PM by elim.)
Post: #2
 elim Moderator     Posts: 578 Joined: Feb 2010 Reputation: 0
1.2 Classification of simple algebraic extensions
Let $K$ be a field, $f\in K[X],\; (f) = \{\lambda f\mid \lambda\in K[x]\}$ Then
$\quad(g'_1\,=g_1+\lambda_1 f)\wedge(g'_2\,=g_2+\lambda_2 f)\implies$
$\qquad\quad(g'_1+g'_2=g_1+g_2+(\lambda_1+\lambda_2)f)\wedge$
$\qquad\quad(g'_1g'_2=g_1g_2+(\lambda_1g2+\lambda_2g_1+\lambda_1\lambda_2)f)$​
$\therefore K[X]/(f)=\{\bar{h}=\{h+(f)\}\mid h\in K[x]\}$ is a ring
$\quad($with operations $\bar{u}+\bar{v}=\overline{u+v},\;\bar{u}\bar{v}=\overline{uv}).$

Let $f\underset{\,}{\in} K[X]$ is irreducible, then $K[X]/(f)$ is a field since
$\quad\;\bar{h}\ne 0\iff h\not\in(f)\iff (\exists u,g\in K[X]\,(uf +gh =1))$
$\qquad\qquad\iff (\exists g\in K[x]\,(\bar{g}\bar{h}=1))$.

Now let $\varphi: K[x]\to K[X]/(f),\; g\mapsto \bar{g} = g+(f)$ be an
epimorphism of rings, $\alpha = \varphi(X),\,$then$\,K(\alpha)\subset_{\small F}K[X]/(f)$
since field$\,K[X]/(f)\supset K\cup\{\alpha\}.\;$ Conversetly,
$\small\forall g\in K[x]\,\exists r\in K[X]\;(r = 0)\vee (f\mid g-r,\,\deg(r )<\deg(f))$ thus
$\small K[X]/(f) = \{\bar{r}(\alpha) \mid r\in K[x],\,(r = 0)\vee (\deg( r)<\deg(f))\}\subset K(\alpha)$
Again, we identify $K$ with $\varphi(K)$ here. This can also be stated
as: If $\alpha$ is algebriac, $f$ is the minimal polynomial of $\alpha$ over $K$
then we have a commutative diagram
$\qquad\qquad\qquad\qquad\small\begin{matrix}K & \rightarrow & K[X]\\ & \searrow & \downarrow\\ & & K(\alpha)\end{matrix}$
inducing an isomorphism of fields $\small K[X]/(f)\cong K(\alpha)$. Thus any
simple algebraic extension of $K$ is of the form $\small K\hookrightarrow K[X]/(f)$
for some$\small\;f\in K[X]$ irreducible up to an isomorphism.

So, classifying simple algebraic extensions of $K$ is equivalent
to classifying irreducible monic polynomials in$\small\;K[x]$.
02-27-2017, 11:18 PM
Post: #3
 elim Moderator     Posts: 578 Joined: Feb 2010 Reputation: 0
1.3 Tests for irreducibility
Let $R$ be a UFD, $K$ its field of fractions, e.g. $\small R=\mathbb{Z},\,K=\mathbb{Q}.$

1.3.1 Lemma (Gauss)
$\quad f\in R[X]$ is irreducible in $R[X]$ iff it's irreducible in $K[X].$

1.3.2 Theorem(Eisenstein's Criterion)
$\quad$If $\; f=\small a_nX^n+\cdots+a_1 X+a_0\in R[X],\;\;p\in R$ is irreducible
$\quad$such that $p\mid a_i\,(i=\overline{0,n-1}),\;\;p^2\nmid a_0,\;\; p\nmid a_n,$
$\quad$then $f$ is irreducible in $R[X]$ hence irreducible in $K[X]$.
02-27-2017, 11:25 PM
Post: #4
 elim Moderator     Posts: 578 Joined: Feb 2010 Reputation: 0
1.4 The degree of an extension
1.4.1 Definition If $L/K$ is a field extension, then $L$ has the
structure of a vector space over $\small K.\;$The dimension $\small [L:K]$ of
the vector space is called the degree of the extension. $L$ is
finite over $K$ if $\small[L:K]$ is finite.

1.4.2 Theorem $\alpha$ is algebraic over a field $K$ iff $K(a)/K$ is
finite. In such case, $[K(\alpha): K]$ is the degree of the
minimal polynomial of $\alpha$.

1.4.3 Proposition (Tower Law)
$\qquad (K\hookrightarrow L\hookrightarrow M)\implies [M:K]=[M:L][L:K].$

1.4.4 Corollary If $L =K(\alpha_1,\ldots,\alpha_n)$ and each $\alpha_j$ is
$\qquad$algebraic over $K$, then $L/K$ is a finite field extension.
02-28-2017, 02:54 PM
Post: #5
 elim Moderator     Posts: 578 Joined: Feb 2010 Reputation: 0
1.5 Splitting fields
1.5.1 Definition If $L/K$ is a field extension, we say that $f\in K[x]$ splits
(completely) over $L$ if $f = k\small (X-\alpha_1)\cdots(X-\alpha_n)$ where $k\in K,\;\alpha_j\in L$
$\,(j=\overline{1,n}).\quad L\,$ is called a splitting field for $f$ if $f$ fails to split over any
proper subfield of $L$, i.e. if $\small L = K(\alpha_1,\ldots,\alpha_n)$.

1.5.2 Remark Splitting fields always exist.
For if $g$ is an irreducible factor of $f$, then $\small K[X]/(g) = K(\alpha)$ is an extension
of $K$ for which $g(\alpha)=0\,(\alpha$ is the image of $X)$. The remainder theorem
implies that $g$(and hence $f$) splits off a linear factor. Induction implies that
there exists a splitting field $L$ for $f$, with $[L:K]\le n! \;(n = \deg(f))$ by
proposition 1.4.3.

Splitting fields are unique up to isomorphisms over $K$.

1.5.3 Proposition If $\theta:K\to K'$ is a field isomorphism with $f\in K[X]$]
corresponding to $g=\theta(f)\in K'[X]$. Then any splitting field $L$ of $f$ over
$K$ is isomorphic over $\theta$ to any splitting field $L'$ of $g$ over $K'$, and we have
the commutative diagram
$\qquad\qquad\qquad\qquad\qquad\begin{matrix}L & \overset{\tilde{\theta}}{\longrightarrow} & L'\\ \uparrow & & \uparrow\\ K & \overset{\theta}{\longrightarrow}& K'\end{matrix}$
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