Book: Fields and Galois theory. Morandi

Lemma

If $K$ is a finite extension of $F$, then $K$ is algebraic and finitely generated over $F$.

Untitled

Definition

Let $K$ be a field extension of $F$. The Galois group $Gal(K/ F)$ is the set of all $F$-automorphisms of $K$

Lemma 2.3

Let $\tau: K\to L$ be an $F$-homomorphism and let $\alpha\in K$ be algebraic over $F$. If $f(x)$ is a polynomial over $F$ with $f(\alpha)=0$, then $f(\tau(\alpha))=0$. Therefore, $\tau$ permutes the roots of $\min(F,\alpha)$. Also, $\min (F,\alpha)=\min (F,\tau(\alpha))$

Untitled

Lemma 2.9 [Morandi]

Let $K$ be a field

Let $S$ be a subset of $\mathrm{Aut}(K)$, and set $\mathcal{F}(S) = \{a\in K: \tau(a)=a\text{ for all } \tau \in S\}$, then $\mathcal{F}(S)$ is a subfield of $K$, called the fixed field of $S$.

  1. If $L_1 \subseteq L_2$ are subfields of $K$, then $\operatorname{Gal}\left(K / L_2\right) \subseteq \operatorname{Gal}\left(K / L_1\right)$.
  2. If $L$ is a subfield of $K$, then $L \subseteq \mathcal{F}(\operatorname{Gal}(K / L))$.
  3. If $S_1 \subseteq S_2$ are subsets of $\operatorname{Aut}(K)$, then $\mathcal{F}\left(S_2\right) \subseteq \mathcal{F}\left(S_1\right)$.
  4. If $S$ is a subset of $\operatorname{Aut}(K)$, then $S \subseteq \operatorname{Gal}(K / \mathcal{F}(S))$.
  5. If $L=\mathcal{F}(S)$ for some $S \subseteq \operatorname{Aut}(K)$, then $L=\mathcal{F}(\operatorname{Gal}(K / L))$.
  6. If $H=\operatorname{Gal}(K / L)$ for some subfield $L$ of $K$, then $H=\operatorname{Gal}(K / \mathcal{F}(H))$.

Definition

If $G$ is a group and if $K$ is a field, then a character is a group homomorphism from $G$ to $K^*$

Dedekind’s Lemma

Let $\tau_1,…,\tau_n$ be distinct characters from $G$ to $K^*$. Then $\tau_i$ are linearly independent over $K$, that is, if $\sum c_i\tau_i(g)=0$ for all $g\in G$, where $c_i\in K$, then all $c_i=0$.

Proposition 2.14

Let $G$ be a finite group of automorphisms of $K$ with $F=\mathcal{F}(G)$. Then $|G|=[K: F]$ and so $G=Gal(K/ F)$

Untitled

Definition Galois extension

Let $K$ be an algebraic extensio of $F$, then $K$ is Galois over $F$ if $F=\mathcal{F}(Gal(K/ F))$

Let $K$ be a finite extension( $\iff$ finitely generated algebraic extension ) of $F$, then $K/ F$ is Galois $\iff$ $|Gal(K/F)|=[K:F]$

Untitled

Definition 3.2,3.4 split

If $K$ is an extension field of $F$ and if $f(x)\in F[x]$, then

  1. $f$ splits over $K$ if $f(x)=a\prod_i (x-\alpha_i)\in K[x]$ fo some $\alpha_1,…,\alpha_n\in K$ and $a\in F$. In other words, $f$ splits over $K$ if $f$ factors completely into linear factors in $K[x]$
  2. $K$ is a splitting field of $f$ over $F$ if $f$ splits over $K$ and $K=F(\alpha_1,…,\alpha_n)$ where $\alpha_1,…,\alpha_n$ are the roots of $f$
  3. If $S$ is a set of nonconstant polynomial over $F$, then $K$ is a splitting field of $S$ over $F$ if each $f\in S$ splits over $K$ and $K=F(X)$ where $X$ is the set of all roots of all $f\in S$

Theorem 3.3

Let $f(x) \in F[x]$ have degree $n$. There is an extension field $K$ of $F$ with $[K : F]\leq n$ such that $K$ contains a root of $f$. In addition, there is a field $L$ containing $F$ with $[L : F] \leq n!$ such that $f$ splits over $L$.

Untitled

Lemma 3.10

If $K$ is a field, then the following statements are equivalent ( algebraically closed ):

  1. There are no algebraic extension of $K$ other than $K$ itself
  2. There are no finite extensions of $K$ other than $K$ itself
  3. If $L$ is a field extension of $K$, then $K=\{a\in L: a \text{ is algebraic over } K\}$
  4. Every $f(x)\in K[x]$ splits over $K$
  5. Every $f(x)$ has a root in $K$
  6. Every irreducible polynomial over $K$ has degree 1

If $K$ is an algebraic extension of $F$ and is algebraically closed, then $K$ is said to be an algebraic closure of $F$

Untitled

Theorem 3.14 [Morandi] existence of algebraic closure, splitting field

Let $F$ be a field, then $F$ has an algebraic closure

Let $S$ be a set of nonconstant polynomials over $F$. The $S$ has a splitting field over $F$.

Let $K$ be an algebraic closure of $F$. Then each $f(x)\in S$ splits over $K$. Let $X$ be the set of roots of all $f \in S$. Then $F(X) \subseteq K$ is a splitting field for $S$ over $F$, since each $f$ splits over $F(X)$ and this field is generated by the roots of all the polynomials from $S$.

If $F$ is a field, then the splitting field of the set of all nonconstant polynomials over $F$ is an algebraic closure of $F$.

Lemma 3.17 [Morandi]

Let $\sigma : F\to F'$ be a field isomorphism. Let $f(x)\in F[x]$ be irreducible, let $\alpha$ be a root of $f$ in some extension field $K$ of $F$, and let $\alpha'$ be a root of $\sigma (f)$ in some extension $K'$ of $F'$. Then there is an isomorphism $\tau : F(\alpha)\to F'(\alpha')$ with $\tau(\alpha)=\alpha'$ and $\tau|_F=\sigma$

Untitled

Lemma 3.18

Let $\sigma : F\to F'$ be a field isomorphism, let $K$ be a field extension of $F$, and let $K'$ be a field extension of $F'$. Suppose that $K$ is a splitting field of $\{f_i\}$ over $F$ and that $\tau : K \to K'$ is a homomorphism with $\tau |_F = \sigma$. If $f_i'= \sigma(f_i)$, then $\tau(K)$ is a splitting field of $\{f_i'\}$ over $F'$.

Untitled

Theorem 3.19

Let $\sigma : F \to F'$ be a field isomorphism, let $f(x)\in F[x]$, and let $\sigma(f)\in F'[x]$ be the corresponding polynomial over $F'$. Let $K$ be the splitting field of $f$ over $F$, and let $K'$ be the splitting field of $\sigma(f)$ over $F'$. Then there is an isomorphism $\tau : K \to K'$ with $\tau| _F = \sigma$. Furthermore, if $\alpha\in K$ and if $\alpha'$ is any root of $\sigma( \min( F, \alpha))$ in $K'$, then $\tau$ can be chosen so that $\tau( \alpha) = \alpha'$.

Untitled

Theorem 3.20 Isomorphism Extension Theorem

Let $\sigma : F \to F'$ be a field isomorphism. Let $S = \{f_i (x)\}$ be a set of polynomials over $F$, and let $S' = \{\sigma(f_i)\}$ be the corresponding set over $F'$. Let $K$ be a splitting field for $S$ over $F$, and let $K'$ be a splitting field for $S'$ over $F'$. Then there is an isomorphism $\tau : K \to K'$ with $\tau |_F = \sigma$. Furthermore, if $\alpha\in K$ and $\alpha'$ is any root of $\sigma(\min( F, \alpha))$ in $K'$, then $\tau$ can be chosen so that $\tau( \alpha) = \alpha'$.

proof:

We prove this with a Zorn's lemma argument. Let $\mathcal{S}$ be the set of all pairs $(L, \varphi)$ such that $L$ is a subfield of $K$ and $\varphi: L \rightarrow K^{\prime}$ is a homomorphism extending $\sigma$. This set is nonempty since $(F, \sigma) \in \mathcal{S}$. Furthermore, $\mathcal{S}$ is partially ordered by defining $(L, \varphi) \leq\left(L^{\prime}, \varphi^{\prime}\right)$ if $L \subseteq L^{\prime}$ and $\left.\varphi^{\prime}\right|_L=\varphi$. Let $\left\{\left(L_i, \varphi_i\right)\right\}$ be a chain in $\mathcal{S}$. If $L=\bigcup_i L_i$ and $\varphi: L \rightarrow K^{\prime}$ is defined by $\varphi(a)=\varphi_i(a)$ if $a \in L_i$, then it is not hard to see that $L$ is a field extension of all the $L_i$ and $\varphi$ is a homomorphism extending $\sigma$. Thus, $(L, \varphi)$ is an upper bound in $\mathcal{S}$ for this chain. Therefore, by Zorn's lemma there is a maximal element $(M, \tau)$ in $\mathcal{S}$. We claim that $M=K$ and $\tau(M)=K^{\prime}$. If $M \neq K$, then there is an $f \in S$ that does not split over $M$. Let $\alpha \in K$ be a root of $f$ that is not in $M$, and let $p(x)=\min (F, a)$. Set $p^{\prime}=\sigma(p) \in F^{\prime}[x]$ and let $\alpha^{\prime} \in K^{\prime}$ be a root of $p^{\prime}$. Such an $\alpha^{\prime}$ exists since $p^{\prime}$ divides $f^{\prime}$ and $f^{\prime}$ splits over $K^{\prime}$. By Lemma 3.17, there is a $\rho: M(\alpha) \rightarrow \tau(M)\left(\alpha^{\prime}\right)$ that extends $\tau$. Then $(M(\alpha), \rho) \in S$ is larger than $(M, \tau)$, a contradiction to the maximality of $(M, \tau)$. This proves that $M=K$. The equality $\tau(K)=K^{\prime}$ follows immediately from Lemma 3.18 , since $\tau(K) \subseteq K^{\prime}$ is a splitting field for $S^{\prime}$ over $F^{\prime}$.

Corollary 3.21

Let $F$ be a field, and let $S$ be a subset of $F[x]$. Any two splitting fields of $S$ over $F$ are $F$-isomorphic. In particular, any two algebraic closures of $F$ are $F$-isomorphic.

(Any algebraic closure of $F$ is a splitting field of the set of all nonconstant polynomials in $F[x]$)

Corollary 3.22

Let $F$ be a field, and let $N$ be an algebraic closure of $F$. If $K$ is an algebraic extension of $F$, then $K$ is isomorphic to a subfield of $N$.

Definition 3.23 Normal extension

If $K$ is a field extension of $F$, then $K$ is normal over $F$ if $K$ is a splitting field of a set of polynomials over $F$

If $[K: F]=2$, then $K$ is normal over $F$.

For, if $a\in K - F$, then $K = F(a)$, since $[K: F] = 2$. If $p(x) = \min(F, a)$, then $p$ has one root in $K$; hence, since $deg(p) = 2$, this polynomial factors over $K$. Because $K$ is generated over $F$ by the roots of $p(x)$, we see that $K$ is a splitting field for $p(x)$ over $F$.

abstract algebra - Field extension of degree two is normal - Mathematics Stack Exchange