Click on the colored portions to expand or collapse content.

$k$ homomorphisms and Galois Extensions


Consider the following map $\phi$ between two field extensions $L/k$ and $L'/k$ which becomes identity when restricted to $k$. Such a map is called a $k$ homomorphism. The set of all $k$ homomorphisms is denoted as $\Hom_k(L,L')$. Since, $\phi$ is a non zero map between fields it is necessarily injective and we get field extensions $L'/\phi(L)/k$.
L L' k k Identity map ϕ
If a $k$ homomorphism is from $L\ra L$ (same field) and is also an isomorphism, it is called a $k$ automorphism. The set of all $k$ automorphisms is denoted as $\Aut_k(L)$. The set $\Aut_k(L)$ is a group with operation as function composition.
L L k k Identity map ϕ an iso.
Since $L/k$ are vector spaces one should think of $\Hom_k$ and $\Aut_k$ as matrices over $k$.



Since any non zero $k$-homomorphism $\phi: L\to L'$ is an injection, $[L:K] = [\phi(L):K]$, in particular if $[L:k] = [L':k] < \infty$ we get $[L:K] = [\phi(L):K]=[L':k]$. Since, $\phi(L)$ is a subfield of $L'$ the equality of dimension of field extensions forces $L'=\phi(L)$ making $\phi$ an isomorphism. As $\phi \in\Aut_k(L)$ is an isomorphism, this gives, $ \Aut_k(L) = \Hom_k(L, L)$.




Example Let $E=\RR[X]/\lr{X^2+1}$ (or concretely $\C$), by construction $\wbr{X}$ is the root of $X^2+1\in E[X]$, (or more concretely $\ci$) and the basis is given as $\{1,\wbr{X}\}$ over $\RR$. But, $-\wbr{X}$ is also a root now since it will also satisfy $X^2+1\in E[X]$, thus giving the basis as $\{1,-\wbr{X}\}$. Consider both the basis $\{1,\pm\wbr{X}\}$ with the following two isomorphisms $E\to E$ $$ \begin{align*} \text{Identity Map }:\{1,\wbr{X}\}\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} &\ra \{1,\wbr{X}\}\\ \text{Identity Map }:\{1,-\wbr{X}\}\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} &\ra \{1,-\wbr{X}\}\\ \phi:~~\{1,\wbr{X}\}\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} &\ra \{1,-\wbr{X}\}\\ \phi:~~\{1,-\wbr{X}\}\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} &\ra \{1,\wbr{X}\} \end{align*} $$ The maps are isomorphisms because determinant of the matrices are non zero. The identity maps basis to itself, and $\phi$ switches the basis giving $\Aut_{\RR}E=\Aut_{\RR}\C=\{1,\phi\}$.



Example Let $E=\Q[X]/\lr{X^2-2}$ (or concretely $\Q(\sqrt{2})$), by construction $\wbr{X}$ is the root of $X^2-2\in E[X]$, (or more concretely $\sqrt{2}$) and the basis is given as $\{1,\wbr{X}\}$ over $\Q$. But, $-\wbr{X}$ is also a root now since it will also satisfy $X^2-2\in E[X]$, thus giving the basis as $\{1,-\wbr{X}\}$. Consider both the basis $\{1,\pm\wbr{X}\}$ with the following two isomorphisms $E\to E$ $$ \begin{align*} \text{Identity Map }:\{1,\wbr{X}\}\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} &\ra \{1,\wbr{X}\}\\ \text{Identity Map }:\{1,-\wbr{X}\}\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} &\ra \{1,-\wbr{X}\}\\ \phi:~~\{1,\wbr{X}\}\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} &\ra \{1,-\wbr{X}\}\\ \phi:~~\{1,-\wbr{X}\}\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} &\ra \{1,\wbr{X}\} \end{align*} $$ The maps are isomorphisms because determinant of the matrices are non zero. The identity maps basis to itself, and $\phi$ switches the basis giving $\Aut_{\Q}E=\Aut_{\Q}\Q(\sqrt{2})=\{1,\phi\}$.



Example Let $E=\Q[X]/\lr{X^2+X+1}$ (or concretely $\Q(\omega)$, where $\omega$ is third root of unity), by construction $\wbr{X}$ is the root of $X^2+X+1\in E[X]$, (or more concretely $\omega$) and the basis is given as $\{1,\wbr{X}\}$ over $\Q$. But, $\wbr{X}^2$ is also a root now since it will also satisfy $X^2+X+1\in E[X]$, thus giving the basis as $\{1,\wbr{X}^2\}$ and also $\wbr{X}^2+\wbr{X}+1=0\in E[X]$. Consider both the basis with the following two isomorphisms $E\to E$. $$ \begin{align*} \text{Identity Map }:\{1,\wbr{X}\}\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} &\ra \{1,\wbr{X}\}\\ \text{Identity Map }:\{1,\wbr{X}^2\}\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} &\ra \{1,\wbr{X}^2\}\\ \phi:~~\{1,\wbr{X}\}\begin{pmatrix} -1&1\\ -1&0 \end{pmatrix} &\ra \{\wbr{X}^2,1\}\\ \phi:~~\{1,\wbr{X}^2\}\begin{pmatrix} -1&1\\ -1&0 \end{pmatrix} &\ra \{\wbr{X},1\} \end{align*} $$ Thus, we get $\Aut_\Q E=2$ given by the two matrices above. This is just permuting the roots of the polynomial or the basis defined by the roots.



Example Let $E=\Q[X]/\lr{X^3-2}$, by construction $\wbr{X}$ is the root of $X^3-2\in E[X]$, (or more concretely $\sqrt[3]{2}$) and the basis is given as $b_1=\{1,\wbr{X},\wbr{X}^2\}$ over $\Q$. Let $\omega$ be the third root of unity ($\omega^3=1$) then, $\omega \wbr{X}$ is also a root since it will also satisfy $X^3-2\in E[X]$, thus giving the basis as $b_2=\{1,\omega\wbr{X},\omega^2\wbr{X}^2\}$, similarly $\omega^2\wbr{X}$ is a root too giving a basis $b_3=\{1,\omega^2\wbr{X},\omega\wbr{X}^2\}$. $$\begin{align*} \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}\text{ preserves all the basis }\begin{pmatrix} b_1&b_2&b_3\\ b_1&b_2&b_3\\ \end{pmatrix}\\ \begin{pmatrix} 1&0&0\\ 0&\omega&0\\ 0&0&\omega^2 \end{pmatrix}\text{ permutes the basis }\begin{pmatrix} b_1&b_2&b_3\\ b_2&b_3&b_1\\ \end{pmatrix}\\ \begin{pmatrix} 1&0&0\\ 0&\omega^2&0\\ 0&0&\omega \end{pmatrix}\text{ permutes the basis }\begin{pmatrix} b_1&b_2&b_3\\ b_3&b_1&b_2\\ \end{pmatrix}\\ \end{align*} $$ Notice that only the identity matrix is defined over $\Q$, in order to carry the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ to another basis with $\omega$ we have to define a matrix over $\Q(\omega)$. Hence, $\Aut_{\Q}E=1$ for $E$ as any of the three realizations $\Q(\sqrt[3]{2}),\Q(\omega\sqrt[3]{2}),\Q(\omega^2\sqrt[3]{2}) $. But, the story is different over $\Q(\omega)$, we could use all three matrices giving $\Hom_{\Q(\omega)}(E,\C)=3$ where $E\neq \Q(\sqrt[3]{2})$.



Continuing the previous examples $\Hom_E(L,\C)$ is given by the matrices $$\begin{align*} \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}\qquad \begin{pmatrix} 1&0&0\\ 0&\omega&0\\ 0&0&\omega^2 \end{pmatrix}\qquad \begin{pmatrix} 1&0&0\\ 0&\omega^2&0\\ 0&0&\omega \end{pmatrix} \end{align*} $$ where the permutation is $\sqrt[3]{2}\ra\omega\sqrt[3]{2}$ and $\sqrt[3]{2}\ra\omega^2\sqrt[3]{2}$ with $\omega\ra \omega$, that is working with the basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$.
$\Hom_\Q(E,\C)$ is given by the matrices $$\begin{equation} \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}\qquad \begin{pmatrix} -1&1\\ -1&0 \end{pmatrix} \end{equation}$$ which maps $\omega\ra\omega^2$ (and $\sqrt[3]{2}\ra\sqrt[3]{2}$, since we are only considering the basis $\{1,\omega\}$ or $\{1,\omega^2\}$). Hence, there are a total of $2\times 3$ maps in $L/\Q$ ( 2 maps $E/\Q$ and 3 maps $L/E$) giving $|\Aut_{\Q}L|=6$, where $L=\Q(\omega,\sqrt[3]{2})$ and $E=\Q(\omega)$.
   In other words, the above corresponds to permutation of roots $(\sqrt[3]{2},\omega\sqrt[3]{2}, \omega^2\sqrt[3]{2})$, that is $S_3$ which is of size $6$.



Let $L/k$ be a finite extension, it is called a Galois Extension if $|\Aut_kL|=[L:k]$.
Let $L/k$ be a Galois Extension then its Galois Group $\gal(L/k):=\Aut_kL$. In particular size of the Galois group is $[L:k]$.
 For any extension $L/k$ we can define its Galois group to be $\Aut_kL$, but Galois groups of Galois extensions have special properties.



Example The field extension $\Q(\sqrt{2})/\Q$ has degree $2$ coming from irreducible polynomial $X^2-2\in\Q[X]$ and the $|\Aut_\Q\Q(\sqrt{2})|=2$. Thus, $\Q(\sqrt{2})/\Q$ is Galois.
Example In the examples we computed that $$|\Aut_{\Q}\Q(\omega,\sqrt[3]{2})|=6=[\Q(\omega,\sqrt[3]{2}):\Q]=[\Q(\omega,\sqrt[3]{2}):\Q(\omega)][\Q(\omega):\Q],$$hence the extension $\Q(\omega,\sqrt[3]{2})/\Q$ is Galois. The degree six polynomial is $(X^3-2)(X^2+X+1)$, it is not irreducible but it is separable.
Non -Example The field extension $\Q(\sqrt[3]{2})/\Q$ has degree $3$ coming from irreducible polynomial $X^3-2\in\Q[X]$ and the $|\Aut_\Q\Q(\sqrt[3]{2})|=1$. Thus, $\Q(\sqrt[3]{2})/\Q$ is not Galois.



Consider a field extension $L/k$ and a polynomial $f\in k[X]$, then $\Root_fL$ will denote the roots of a polynomial $f$ in $L$. If $f$ is irreducible a field $k_f=k[X]/\lr{f}$ can be constructed. The field $k_f$ can be thought of as $k(r)$ where $r$ is the root of polynomial $f$.


K004 Lemma


Proof Let $\wbr{X}$ be the image of $X$ in $k_f$, then by construction it is the root of $f$ and $\Hom_k(k_f,L)$ is simply different realizations of $\wbr{X}$ in $L$. What are these realizations? Ans: Precisely the set $\Root_fL$. That is, if $\rho\in\Root_fL$ the correspondence is $\rho\leftrightarrow \wbr{X}$, in other words $\wbr{X}$ embeds in $L$ as roots of the polynomial.
Let $\sigma\in \Hom_k(k_f,L)=\Hom(k[X]/\lr{f},L)$ and by the result below it makes $f$ zero in $L$. In other words, if $\alpha=\sigma(X)$ then $\sigma(f(X))=f(\alpha)$ is zero ($\sigma$ does not change coefficients of $f$). If $B$ is a $k$ algebra and $\varphi\in\Hom_k(k[X],B)$, then $\varphi(\sum a_iX^i)=\sum a_ib^i$ where $b$ is the image of $X$ in $B$.


Given a homomorphism $f: A\ra B$ with $f(I)=0$ in $B$, the following diagram commutes
A A/I B π f
The map $\pi^{\star}$ is a bijection $$ \begin{align*} \pi^{\star}:\Hom(A/I,B)&\to\{f\in\Hom(A,B)\text{ such that }f(I)=0\}\\ \pi^{\star}(g)&=g\circ \pi \end{align*} $$ Injectivity: Let $\phi,\phi'$ be such that $\phi\circ\pi=\phi'\circ\pi$, that is to say $(\phi-\phi')\circ\pi=0$, as $\pi$ is a surjective map, $\pi-\phi'$ is zero on $\pi(A)=A/I$.
Surjectivity: Let $f\in \Hom(A,B)$ zero on $I$, and construct $\phi\in\Hom(A/I,B)$. Let $t\in A/I$: which is the class of an element $a$ well determined upto addition by an element in $I$. Since, $f(I)=0$, the images by $f$ of all elements representing $t$ are one and the same element that we call $\phi(t).$ By construction $\phi\circ pi=f$ and $\phi$ is obviously a homomorphism ( for example, if $t=\pi(a),t'=\pi(a')$ with $a,a'\in A$, this gives $$ \phi(tt')=\phi(\pi(a)\pi(a'))=\phi(\pi(aa'))=f(aa')=f(a)f(a')=\phi(\pi(a))\phi(\pi(a'))=\phi(t)\phi(t') $$ which proves the multiplicity since $\phi$ is surjective. The additivity is proved similarly.)

K005 Corollary


Proof Set $k_f=L$ in the Lemma above and note that $\Hom_k(k_f,k_f)=\Aut_kk_f$.