Notes on 【Galois Theory】 - Printable Version +- OMath! (http://math.elinkage.net) +-- Forum: Math Forums (/forumdisplay.php?fid=4) +--- Forum: Algebra (/forumdisplay.php?fid=5) +--- Thread: Notes on 【Galois Theory】 (/showthread.php?tid=721) Notes on 【Galois Theory】 - elim - 02-20-2017 11:47 AM 1 Revision from Groups, Rings and Fields 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 $\small\varphi(a/b)=\varphi(a)/\varphi(b).$ 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 $K\,(\forall \alpha\in L-K).$ 1.2 Classification of simple algebraic extensions - elim - 02-20-2017 12:08 PM 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 $\qquad\qquad\qquad\qquad\begin{matrix}K & \rightarrow & K[X]\\ & \searrow & \downarrow\\ & & K(\alpha)\end{matrix}$ 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$\small\;K[x]$. 1.3 Tests for irreducibility - elim - 02-27-2017 11:18 PM 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 $K[X]$. 1.4 The degree of an extension - elim - 02-27-2017 11:25 PM 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. 1.5 Splitting fields - elim - 02-28-2017 02:54 PM 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 $\qquad\qquad\qquad\qquad\qquad\begin{matrix}L & \overset{\tilde{\theta}}{\longrightarrow} & L'\\ \uparrow & & \uparrow\\ K & \overset{\theta}{\longrightarrow}& K'\end{matrix}$