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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 $K\,(\forall \alpha\in L-K).$
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 Messages In This Thread Notes on 【Galois Theory】 - elim - 02-20-2017 11:47 AM 1.2 Classification of simple algebraic extensions - elim - 02-20-2017, 12:08 PM 1.3 Tests for irreducibility - elim - 02-27-2017, 11:18 PM 1.4 The degree of an extension - elim - 02-27-2017, 11:25 PM 1.5 Splitting fields - elim - 02-28-2017, 02:54 PM

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