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Galois Correspondence

In this section it is assumed that all fields are perfect.


Let $K/k$ be a finite extension with $K\subset \Omega$, where $\Omega$ is algebraically closed and $\Aut_k(K)$ is a group that acts on $K$. Let $x\in\Omega$ be algebraic on $k$ with minimal polynomial $P$, then the roots of $P$ in $\Omega$ will be called conjugates of $x$ or $k$ conjugates of $x$.
  Setting
$k_f$ $k_P=\dfrac{K[X]}{\lr{P(X)}}$ where $P(X)$ is irreducible polynomial
$=K$ we know from the K005 corollary that $K/k$ is galois if and only if $\deg P=$ number of roots, or the conjugates of $x$ on $k$ are in $K$. The assumption of perfect fields ensures that the polynomials defined over these fields are separable (see section Perfect). This leads to the following definition.

Definition The extension $K/k$ is Galois if it is algebraic and if the conjugates of an arbitrary element of $K$ are in $K$.

This is equivalent to the definition of galois extension being separable and normal. The separable part comes from the fact that all algebraic extensions of perfect fields are separable and normal is equivalent to having all conjugates.



Proof $\Rightarrow:$ Follows from the fact that the extension $K/k$ is Galois if it is algebraic and if the conjugates of an arbitary element of $K$ are in $K$.
$\Leftarrow:$ Let $x\in K$, then the conjugates of $x$ on $k$ are of the form $\sigma(x)$ with $\sigma\in\Hom_k(K,\Omega)$. But, by assumption $x=P(x_1,\ldots,x_n)$ with $P\in k[X_1,\ldots, X_n]$ a polynomial in $n$ variables, and thus $\sigma(x)=P(\sigma(x_1),\ldots,\sigma(x_n))$. Since, by hypothesis $\sigma(x_i)\in K$ thus, $\sigma(x)\in K$.$\qquad\blacksquare$



Proof: The minimal polynomial of $x\in K$ on $k$ has its coefficients in $k$ and thus in $E\supset k$, hence, $x$ is also algebraic on $E$. Its minimal polynomial on $k$ is divisible by the minimal polynomial of $x$ on $E$. Hence, all the $E$ conjugates of $x$ are also $k$ conjugates, thus are in $K$ by hypothesis.$\qquad\blacksquare$
 In general however, with the same hypothesis $E/k$ is not necessarily Galois. For example, if $k=\Q, E=\Q(2^{1/3}), K=\Q(2^{1/3},\omega)$ where $\omega$ is the third root of unity and we are working with minimal polynomial $X^3-2$.



 Recall that the roots or conjugates $x\in K$ are also embedding in $\Omega$ as $\sigma(x)$ with $\sigma\in \Hom_k(K,\Omega)$. Since $K\subset \Omega$ we have $$\Aut_k(K)=\Hom_k(K,K)\hookrightarrow\Hom_k(K,\Omega) $$ and with an automorphism $\varsigma\in\Aut_k(K)$ one can associate $\sigma \in\Hom_k(K,\Omega)$ given as $$\sigma:K\xrightarrow{\varsigma}K\hookrightarrow \Omega.$$ This permits identification $\sigma\leftrightarrow\varsigma.$
Proof
  1. Suppose $\sigma(K)\subset K$ and $x_1,\ldots, x_n$ the $n$ conjugates of $x_1\in K$ all in $K$ by hypothesis. $\sigma(x_i)$ is a conjugate of $x_i$, thus of $x_1$ because $x_i$ and $x_1$ have the same polynomial. Thus, $\sigma$ restricted to $S=\{x_1,\ldots, x_n\}$ is invariant by hypothesis. It is injective since maps between fields is injective, and it induces a bijection on $S$ (since $S$ is finite), thus there is an $x_i\in S$ such that $x_1=\sigma(x_i)$. As $x_i\in K$ by hypothesis, $\sigma$ is surjective.
  2. $ \Aut_k(K)=\Hom_k(K,\Omega)\Rightarrow$ All conjugates of $x\in K$ can be written as $\sigma(x)$ for $\sigma\in\Hom_k(K,\Omega)$. But, $\sigma\in\Aut_k(K)$ thus $\sigma(x)\in K$ proving that $K/k$ is galois.
    $\Leftarrow$ Let $K/k$ be galois, and $\sigma\in\Hom_k(K,\Omega)$ and $x\in K$. Then $\sigma(x)$ is a conjugate of $x$ and thus is in $K$. As $x$ is arbitrary we have $\sigma(K)\subset K$ and hence $\sigma\in\Aut_k(K)$ by the first part of Lemma 3. $\qquad\blacksquare$
Remark For a galois extension $K/k$ we have $\gal(L/k)=\Aut_k(K)=\Hom_k(K,\Omega)$. The conjugates of $x\in K$ are $\sigma(x),\sigma\in\Hom_k(K,\Omega)$, if $K/k$ is galois extension with galois group $G$, the conjugates of $x$ are $g(x), g\in G$.
For example, $|\gal(\C/\RR)|=2$ where $\C=\RR[\ci]$ and it determined by the image of $\ci$ which is $\ci$ or $\wbr{\ci}=-\ci$. Thus, the galois group has identity and conjugation as the automorphisms.



Proof
  1. From Proposition 2 $E/k$ is galois, the elements of $\gal(K/E)$ are $E$ linear automorphisms of $K$ while those of $\gal(K/k)$ are the automorphisms of $K$ that are $k$ linear. As $E$ contains $k$, we deduce an inclusion $\gal(K/E)\ra\gal(K/k)$, which respects composition and identity.
  2. Consider the diagram below, where the map in the first row is restriction and is surjective since there is a $k$ algebra embedding $E\hookrightarrow\Omega$ and $K\hookrightarrow\Omega$. The top row can be identified with bottom row, using part 2 of Lemma 3. (writing $\Aut$ as $\gal$) $$ \require{AMScd} \begin{CD} \Hom_k(K,\Omega) @>{}>> \Hom_k(E,\Omega);\\ @AAA @AAA \\ \gal(K/k) @>{}>> \gal(E/k) \end{CD} $$ The elements of the $\Ker:\gal(K/k)\ra \gal(E/k)$ are by definition the automorphisms of $K$ fixing $E$, thus, the elements of $\gal(K,E)$.



Proof
  1. Suppose $K/k$ is galois with $x\in K$, and we write a conjugate $y$ of $x$ as $\sigma(x)$ for $\sigma\in\Hom_k(K,\Omega)$. Since $\Aut_k(K)=\Hom_k(K,\Omega)$ from Lemma 3 part 2, the action of $\Aut_k(K)$ is transitive.
    Inversely, let $x$ generate the extension $K/k$ (using the primitive element theorem) with degree $[K:k]$, thus there are a total of $\deg_k(x)=[K:k]$ conjugates which are transitive, thus $$[K:k]\leq |\Aut_k(K)|\leq |\Hom_k(K,\Omega)|=[K:k]$$ Thus, from Lemma 3 part 2 we get Galois Extension.
  2. Using primitive element theorem let $x$ generate $K$. Let $x_i$ be the conjugates, we know that the roots of the minimimal polynomial $P$ are in $K$ by hypothesis. Hence we have $$K=k[x]\subset k[x_i] \subset K$$ and thus, $K=k[x_i]$ is the field of roots of polynomial.
    Inversely, if $K=k[x_i]$ where $x_i$ is the root of a polynomial $P$, then $\sigma\in\Hom_k(K,\Omega)$ permutes $x_i$, since $P=P^\sigma$ (the polynomial over $k$ does not change, as $k$ is fixed by $\Hom_k$). Thus, we deduce that it sends $K=k[x_i]$ to itself so that $\sigma\in\Aut_k(K)$, that is $\sigma(K)\subset K$ and by Lemma 3 part 1 $\sigma(K)=K$. Hence, $K$ contains all the conjugates and thus is a galois extension.
$\blacksquare$



This is just translation of a Theorem FOT1 from finite groups. The subextensions of $\F_{q^n}/F_q$ are of the form $\F_{q^m}/F_q$ with $m|n$ are in bijection with set of subgroups of $\gal(F_{q^n}/F_q)$, that are of the form $\{(n/m)\Z/n\Z\simeq\Z/m\Z\}$ with $m|n$.


Fixed Points


A fixed point (or invariant element) of $E$ under the action of a group $G$ is an element $x\in E$ whose stabilizer is $G$. In other words the orbit of $x$ under the action to $G$ is just $\{x\}$ or $g\cdot x=x$ for all $x\in G$.
  If $H\leq G$, the set $E^H$ denotes the elements invariant under the action of $H$ $$E^H=\{x\in E\text{ such that for all }h\in H, h\cdot x=x\}\subset E $$



Proof The first part ($|G|=[K:k]$) has been proved in the corollary K0005. For the second part, let $x$ be fixed under the action of $G$. The conjugates are of the form $\sigma(x)$ with $\sigma \in G$, and it is equal to $x$. Since $G$ is Galois the action of the group is transitive. As the minimal polynomial $P\in k[X]$ of $x$ is separable by irreducibility, it is equal to $X-x$, hence $x=-P(0)\in k$.

Artin's Lemma


Proof Verify that $K^G$ is perfect. There is no problem if the $\Char K=p>0$ (by defn $\Char 0$ is perfect). Let $x\in K^G$, as $K$ is perfect, it has a $p$th root $\xi\in K$ and $x=\xi^p$. Since $x$ is invariant under the action of $G$, we have $\xi^p=g(\xi^p)$ for all $g\in G$. Since, the Frobenius morphism is injective, we deduce that $\xi$ is fixed by $G$ and thus $\xi\in K^G$ is, so that $K^G$ is perfect. In the rest of the proof set $K^G=k$
  Observe that all elements $x\in K$ is algebraic with $\deg_k(x)\leq |Gx|$ and thus $\deg\leq|G|$. In fact, the polynomial $$P_x=\prod_{\xi\in Gx}(X-\xi) $$ is invariant under the aciont of $G$ (that is, all the coefficients of the Polynomial above are invariant under $G$). Thus, $P_x\in k[X]$. Let $x\in K$ be an element of maximum degree on $k$, then $K=k[x]$. If not, let $y=K\bs k[x]$, the extension $k[x,y]$ is generated by a single element $z$ (by the primitive element theorem) and $\deg_k z>\deg_k(x)$ leading to a contradiction.
Since, $K=k[x]$ we get $$[K:k]=\deg_k(x)\leq|G| $$ Let $\Omega$ be algebraically closed containing $K$, then we have the following $$\begin{equation}\label{art3} |G|\leq |\Aut_k(K)| \leq |\Hom_k(K,\Omega)|=[K:k]\leq |G| \end{equation} $$ The first inequality is assumed in the proposition $G\leq \Aut_k(K)$, the second one follows from the fact that $\Aut_k(K)\leq\Hom_k(K,\Omega)$ and the last has been proved above. The equality $|\Aut_k(K)|=|\Hom_k(K,\Omega)|$, follows from $\eqref{art3}$, which makes the extension $K/k$ galois.



Proof $\varphi$ and $\phi$ are inverse of each other $$\varphi\phi(L)=\varphi(\gal(K/L))=K^{\gal(K/L)}=L. $$ Since $K$ is galois over $L$ with galois group $H=\gal(L/K)$, we have $K^H=L$ from Proposition 7.. Also notice $$\phi\varphi(H)=\phi(K^H)=\gal(K/K^H)=H $$ $\gal(K/K^H)=H$ follows from Artin's Lemma. Thus, we have shown inverse and bijectiveness, the decreasing is clear and strictness follows from bijection.


Follows from Artin's Lemma.


Let $H\leq G$ and we prove surjectivity of $r_H$. All $k$ morphism $\sigma_H\in\Hom_k(K^H,\Omega)$ are embedded in $\sigma\in \Hom_k(K,\Omega)$ from theorem CL1T. As $K/k$ is galois we have $\sigma(K)=K$ (from Lemma 3 part 1), that is $\sigma \in G$ so that $r_H(\sigma)=\sigma_H$, giving surjectivity.
  Of course, $g$ and $gh$ have the same image, if $h\in H$, so that we have a surjection $$\rho_H:G/H\ra\Hom_k(K^H,\Omega)$$ We then have $$|\Hom_k(K^H,\Omega)|=[K^H:k]=[K:k]/[K:K^H]=|G|/|H|=|G/H| $$ so that $\rho_H$ is bijective.


Let $H$ be a subgroup of $G$ and $g\in G$, then $g(K^H)\subset K^{gHg^{-1}}$. This follows from the fact that $$g(x)= gHg^{-1}\circ g(x)\quad\text{for} \quad x\in K^H$$ in words, $g$ sends $K^H$ to $K^{gHg^{-1}}$. In the other direction $g^{-1}(K^{gHg^{-1}})\subset K^H$. This follows from the fact that $$g^{-1}(x)= g^{-1}\circ gHg^{-1}(x) \quad\text{for} \quad x\in K^{gHg^{-1}}.$$ Notice that $g^{-1}(x)=Hg^{-1}(x)$, showing that $g^{-1}$ sends $K^{gHg^{-1}}$ to $K^H$. Hence, the equality $$g(K^H)=K^{g H g^{-1}} $$ Suppose $K^H/k$ is galois,we then have $$K^H=g(K^H)=K^{g Hg ^{-1}} $$ and thus $H=g Hg^{-1}$, by injectivity of galois correspondence and thus $H\triangleright G$.
  Inversely, if $H\triangleright G$ we have $g(K^H)=K^{g Hg^{-1}}=K^H$ and $K/K^H$ is galois. The isomorphism and the last part come from Proposition 4.


See Proposition 4.


The reference for this page is INTRODUCTION À LA THÉORIE DE GALOIS by Yves Lazlo.