Galois Theory week 1.2

Proposition.

$$\begin{array}{}\text{(1)}& {\mathrm{Aut}}_{\mathrm{Field}}(K)\subseteq {\mathrm{Aut}}_{\mathrm{Set}}(K),{\mathrm{Aut}}_F(K)\subseteq {\mathrm{Aut}}_{\mathrm{Field}}(K)\end{array}$$

Lemma. Let $\mathcal{C}$$\mathcal C$ be a concrete category over a locally small category $\mathcal{D}$$\mathcal{D}$. That is, there is a faithful functor $\mathcal{F}$$\mathcal F$​​

$$\begin{array}{}\text{(2)}& \mathcal{F}:\mathcal{C}\to \mathcal{D}\end{array}$$

Let $A\in \mathrm{Ob}(\mathcal{C}),B\in \mathrm{Ob}(\mathrm{Set})$$A\in\mathrm{Ob}(\mathcal C),B\in\mathrm{Ob}(\mathrm{Set})$,

$$\begin{array}{}\text{(4)}& \mathcal{F}:{\mathrm{Aut}}_{\mathcal{C}}(A)\to {\mathrm{Aut}}_{\mathcal{D}}(B)\end{array}$$

is an injective group homomorphism.

Proof. By definition. $◻$$\square$

Hence we know that if ${\mathrm{Aut}}_{\mathrm{Field}}(K)\subseteq {\mathrm{Aut}}_{\mathrm{Set}}(K),{\mathrm{Aut}}_F(K)\subseteq {\mathrm{Aut}}_{\mathrm{Field}}(K)$$\mathrm{Aut}_{\mathrm{Field}}(K)\subseteq\mathrm{Aut}_{\mathrm{Set}}(K),\mathrm{Aut}_F(K)\subseteq\mathrm{Aut}_{\mathrm{\mathrm{Field}}}(K)$. Here $F$$F$ means category of field extension from $F$$F$. $◻$$\square$

For the question what is the Galois group of $\mathbb{Q}[2^{\frac{1}{3}}]$$\mathbb Q[2^{\frac{1}{3}}]$, we can use the functorial points induced by Grothendieck.

Let $f_1,...,f_n\in R[X_1,...,X_n]$$f_1,...,f_n\in R[X_1,...,X_n]$ and $L$$L$ be an $R$$R$ algebra, $\psi :L\to L^{\prime }$$\psi:L\to L'$ be a $R-$$R-$alg homomorphism.

Let $X_f(L)$$X_f(L)$ be the solution of $f_1,...,f_n$$f_1,...,f_n$ in $L$$L$, then $X_f(-):R-\mathrm{Alg}\to \mathrm{Set}$$X_f(-):R-\mathrm{Alg}\to\mathrm{Set}$ is a functor.

Since if $a\in X_f(L)$$a\in X_f(L)$ then $\psi (a)$$\psi(a)$ is a solution of $f_1,...,f_n$$f_1,...,f_n$ in $L^{\prime }$$L'$.

We claim that $X_f(L)\cong {\mathrm{Hom}}_{R-\mathrm{Alg}}({\displaystyle \frac{R[X_1,...,X_n]}{(f_1,...,f_n)}},L)$$X_f(L)\cong \mathrm{Hom}_{R-\mathrm{Alg}}\left(\dfrac{R[X_1,...,X_n]}{(f_1,...,f_n)},L\right)$.

Indeed, that is a representable functor,

$$\begin{array}{}\text{(5)}& X_f(-)\cong {\mathrm{Hom}}_{R-\mathrm{Alg}}({\displaystyle \frac{R[X_1,...,X_n]}{(f_1,...,f_n)}},-)\end{array}$$

Proof.

For $a\in X_f(L)$$a\in X_f(L)$, there exists a universal evaluation map ${\mathrm{ev}}_a:R[X_1,...,X_n]\to L$$\mathrm{ev}_a:R[X_1,...,X_n]\to L$ by the universal property of the polynomial ring.

We know that $\mathrm{Ker}({\mathrm{ev}}_a)\supseteq (f_1,...,f_n)$$\mathrm{Ker}(\mathrm{ev}_a)\supseteq(f_1,...,f_n)$. By the universal property of the quotient ring, we know that ${\mathrm{ev}}_a$$\mathrm{ev}_a$ uniquely factors through ${\displaystyle \frac{R[X_1,...,X_n]}{(f_1,...,f_n)}}$$\dfrac{R[X_1,...,X_n]}{(f_1,...,f_n)}$. Conversely, let $\psi \in {\mathrm{Hom}}_{R-\mathrm{Alg}}({\displaystyle \frac{R[X_1,...,X_n]}{(f_1,...,f_n)}},L)$$\psi\in \mathrm{Hom}_{R-\mathrm{Alg}}\left(\dfrac{R[X_1,...,X_n]}{(f_1,...,f_n)},L\right)$ and consider the canonical projection:

$$\begin{array}{}\text{(6)}& \pi :R[X_1,...,X_n]\to {\displaystyle \frac{R[X_1,...,X_n]}{(f_1,...,f_n)}}.\end{array}$$

We claim that $\psi \circ \pi [X_i]$$\psi\circ\pi[X_i]$ is the solution of $f_1,...,f_n$$f_1,...,f_n$ in $L$$L$.

Since

$$\begin{array}{}\text{(7)}& f_i(\psi \circ \pi (X_1),...,\psi \circ \pi (X_n))=\psi \circ \pi (f_i(X_1,...,X_n))=\psi (0)=0◻\end{array}$$

Let $f=x^3-2$$f=x^3-2$. The extension $\mathbb{Q}\subseteq \mathbb{Q}[2^{\frac{1}{3}}]$$\mathbb{Q} \subseteq \mathbb{Q}[2^{\frac{1}{3}}]$ makes $\mathbb{Q}[2^{\frac{1}{3}}]$$\mathbb{Q}[2^{\frac{1}{3}}]$ become a $\mathbb{Q}-$$\mathbb{Q}-$algebra.

Since $X_f(\mathbb{Q}[2^{\frac{1}{3}}])=\{2^{\frac{1}{3}}\}\cong {\mathrm{Hom}}_{\mathbb{Q}-\mathrm{Alg}}({\displaystyle \frac{\mathbb{Q}[X]}{(X^3-2)}},\mathbb{Q}[2^{\frac{1}{3}}])\cong {\mathrm{Hom}}_{\mathbb{Q}-\mathrm{Alg}}(\mathbb{Q}[2^{\frac{1}{3}}],\mathbb{Q}[2^{\frac{1}{3}}])\supseteq \mathrm{Gal}(\mathbb{Q}[2^{\frac{1}{3}}]/\mathbb{Q})\supseteq \{e\}.$$X_f(\mathbb{Q}[2^{\frac{1}{3}}])=\{2^{\frac{1}{3}}\}\cong\mathrm{Hom}_{\mathbb{Q}-\mathrm{Alg}}\left(\dfrac{\mathbb{Q}[X]}{(X^3-2)},\mathbb{Q}[2^{\frac{1}{3}}]\right)\cong\mathrm{Hom}_{\mathbb{Q}-\mathrm{Alg}}\left(\mathbb{Q}[2^{\frac{1}{3}}],\mathbb{Q}[2^{\frac{1}{3}}]\right)\supseteq\mathrm{Gal}(\mathbb{Q}[2^{\frac{1}{3}}]/\mathbb{Q})\supseteq\{e\}.$

Hence the Galois group of $\mathbb{Q}[2^{\frac{1}{3}}]/\mathbb{Q}$$\mathbb{Q}[2^{\frac{1}{3}}]/\mathbb{Q}$​ is trivial.

There is a question ask you to prove $K^S=K^{⟨S⟩}$$K^S=K^{\langle S\rangle}$, it is a part Galois connection.

Definition. Let $S\subseteq \mathrm{Gal}(K/L)$$S\subseteq\mathrm{Gal}(K/L)$, and define

$$\begin{array}{}\text{(8)}& F(S)=K^S=\{a\in K:\sigma (a)=a\mathrm{\forall }\sigma \in S\}=\bigcap _{\sigma \in S}\{a\in K:\sigma (a)=a\}=\bigcap _{\sigma \in S}\mathrm{Ker}(\sigma -I)\end{array}$$

Easy to see $F(S)$$F(S)$ is a subfield since $L\subseteq F(S)$$L\subseteq F(S)$ hence $F(S)\ne \mathrm{\varnothing }$$F(S)\ne\empty$.Each $\mathrm{Ker}(\sigma -I)$$\mathrm{Ker}(\sigma-I)$ is a subvector spcae and closed under multiplication since $\sigma (ab)=\sigma (a)\sigma (b)=ab.$$\sigma(ab)=\sigma(a)\sigma(b)=ab.$ It is closed under inverse since $\sigma (a^{-1})=\sigma (a{)}^{-1}=a^{-1}$$\sigma(a^{-1})=\sigma(a)^{-1}=a^{-1}$​.

Easy to see that $S\subseteq S^{\prime }⟹F(S)\supseteq F(S^{\prime })$$S\subseteq S'\implies F(S)\supseteq F(S')$.

Proposition. Consider an intermediate field $M$$M$ such that $L\subseteq M\subseteq K$$L\subseteq M\subseteq K$. We could consider $\mathrm{Gal}(K/M)$$\mathrm{Gal}(K/M)$.

Easy to see that $M\subseteq M^{\prime }⟹\mathrm{Gal}(K/M)\supseteq \mathrm{Gal}(K/M^{\prime })$$M\subseteq M'\implies\mathrm{Gal}(K/M)\supseteq\mathrm{Gal}(K/M')$.

Proposition. $F(-)⊣\mathrm{Gal}(K/-)$$F(-)⊣ \mathrm{Gal}(K/-)$ . That is, $F$$F$ is the left adjoint of $\mathrm{Gal}(K/-)$$\mathrm{Gal}(K/-)$ or

$$\begin{array}{}\text{(9)}& F(S)\supseteq M⟺S\subseteq \mathrm{Gal}(K/M)\end{array}$$

For convinient, we denote $\mathrm{Gal}(K/-)$$\mathrm{Gal}(K/-)$ as $G$$G$.

Proof. By the equivalent definition of Galois connection, we only need to prove that

$$\begin{array}{}\text{(10)}& S\subseteq GF(S),M\subseteq FG(M)\end{array}$$

By definition, $GF(S)$$GF(S)$ contains all the automorphism of $K$$K$ whcih fix $F(S)$$F(S)$, hence $S\subseteq GF(S)$$S\subseteq GF(S)$.

Similarly for $FG(M)$$FG(M)$.

Hence $G(M\vee M^{\prime })=G(M)\cap G(M^{\prime }),F(S\cup S^{\prime })=F(S)\cap F(S^{\prime })$$G(M\vee M')=G(M)\cap G(M'),F(S\cup S')=F(S)\cap F(S')$ by the property of Galois connection.