Extensions of $\mathbb{Q}$

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 02 February 2012

Extensions of $\mathbb{Q}$

• An algebraic number is an element of $\stackrel{\_}{\mathbb{Q}}$ the algebraic closure of $\mathbb{Q}$.
• An algebraic number field is a finite extension of $\mathbb{Q}$.
• Let $f\left(x\right)\in \mathbb{Q}\left[x\right].$ The discriminant of $f\left(x\right)$ is $$D^{2}whereD=\prod _{1\le i<j\le k}({\alpha }_i-{\alpha }_j),$$ where ${\alpha }_1,...,{\alpha }_k,\in ,\stackrel{\_}{\mathbb{Q}}$ are such that $f\left(x\right)=(x-{\alpha }_1)\cdots (x-{\alpha }_k)\in \stackrel{\_}{\mathbb{Q}}\left[x\right].$

Let $𝔼$ be the splitting field of $f\left(x\right)\in \mathbb{Q}\left[x\right].$ Note that if $\sigma \in \mathrm{Gal}(𝔼/𝔽)$ then $\sigma$ is a permutation of the roots and $\sigma \cdot D={\left(-1\right)}^{l\left(\sigma \right)}D.$ Thus $D^2$ is always fixed by $\sigma$ and so $D^2\in 𝔽.$ If $D\notin 𝔽$ then $\mathbb{Q}\left(D\right)$ is a degree two extension of $\mathbb{Q}$ since the minimal polynomial of $D$ is $x^2-D^2.$ If $D\in \mathbb{Q}$ then $\mathrm{Gal}(𝔼/\mathbb{Q})\subseteq A_n,$ the alternating group.

Example 1. If $f\left(x\right)=x^2+bx+c\in \mathbb{Q}\left[x\right]$ and $$f\left(x\right)=x^2+bx+c=(x-{\alpha }_1)(x-{\alpha }_2)\in \stackrel{\_}{\mathbb{Q}}\left[x\right]$$ then $$\mathbb{Q}\left({\alpha }_1\right)=\mathbb{Q}\left({\alpha }_2\right)since{\alpha }_1=b-{\alpha }_1.$$ If $\sigma \in \mathrm{Gal}(\mathbb{Q}\left({\alpha }_1\right)/\mathbb{Q})$ is the element given by $$\sigma \left({\alpha }_1\right)={\alpha }_{2}then\sigma \left({\alpha }_2\right)={\alpha }_1,$$ since ${\alpha }_2+\sigma \left({\alpha }_2\right)=\sigma ({\alpha }_1+{\alpha }_2)=\sigma \left(b\right)=b={\alpha }_1+{\alpha }_2.$ So $$\mathrm{Gal}(\mathbb{Q}\left({\alpha }_1\right)/\mathbb{Q})= {1,\sigma } \cong \mathbb{Z}/2\mathbb{Z}=S_2.$$ Now the discriminant $D^2=b^2-4c\in \mathbb{Q}$ and $D=\sqrt{b^2-4c}\in \mathbb{Q}\left({\alpha }_1\right).$ So $$\mathbb{Q}\left({\alpha }_1\right)=\mathbb{Q}\left(D\right)and\sigma \left(D\right)=-D.$$

Example 2. Let $f\left(x\right)=x^3+a_2{x}^2+a_{1}x+a_0=0.$ Change variable $x=y-\frac{a_2}{3}.$ Then $$f\left(x\right)=y^3-a_2{y}^2+\left(\frac{3y{a}_2^2}{9}\right)-\frac{a_2^3}{27}+a_2{y}^2-\left(\frac{2a_2{x}^2}{3}\right)y+\frac{a_2^3}{9}+a_{1}y-\frac{a_1{a}_2}{3}+a_0=0.$$ So assume $f\left(x\right)=x^3+bx+c$ and let $𝔼$ be the splitting field of$f\left(x\right)$. If $f\left(x\right)$ is separable and irreducible then $$D^2=-4b^3-27c^{2}and\mathrm{Gal}(𝔼/𝔽)=\{\begin{array}{ll}{S}_3,& if\; D\notin 𝔽,\\ \mathbb{Z}/3\mathbb{Z},& if\; D\in 𝔽.\end{array}\phantom{\}}$$ If $f\left(x\right)=x^3+bx+c=(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)$ then $$\begin{array}{rcl}{e}_3& =& {\alpha }_1{\alpha }_2{\alpha }_3=-c,\\ e_2& =& {\alpha }_1{\alpha }_2+{\alpha }_2{\alpha }_3+{\alpha }_1{\alpha }_3=b,\\ e_1& =& {\alpha }_1+{\alpha }_2+{\alpha }_3.\end{array}$$ $$\begin{array}{cccccccc}& & \mathbb{Q}\left({\alpha }_1,{\alpha }_2,{\alpha }_3\right)& & & & \left\{1\right\}& \\ \mathbb{Q}\left({\alpha }_1\right)& \mathbb{Q}\left({\alpha }_2\right)& \mathbb{Q}\left({\alpha }_3\right)& & {1,\left(\mathrm{1,3}\right)} & {1,\left(23\right)} & {1,\left(23\right)} & \\ & & & \mathbb{Q}\left(D\right)& & & & A_3\\ & & \mathbb{Q}& & & & S_3& \end{array}$$ A concrete example is $f\left(x\right)=x^3-2$ which has roots $2^{1/3}, \omega 2^{1/3}, {\omega }^2{2}^{1/3},$ where $\omega$ is a primitive cube root of unity and $$\mathrm{Gal}(𝔼/\mathbb{Q})\cong S_3.$$ $$\begin{array}{cc}𝔼=\mathbb{Q}\left(2^{\frac{1}{3}},\omega \right)& \left\{1\right\}\\ |& |\\ \mathbb{Q}\left(2^{\frac{1}{3}}\right)& ⟨\sigma ⟩\cong \mathbb{Z}/3\mathbb{Z}\\ |& |\\ \mathbb{Q}& ⟨\sigma ,\tau |{\sigma }^3=1, {\tau }^2=1, \sigma \tau =\tau {\sigma }^{-1}⟩ \cong S_3\end{array}$$ Examples of the two cases are $$\begin{array}{rcl}f\left(x\right)=x^3-3x+1& which\; has& D^2=81,and\\ f\left(x\right)=x^3+3x+1& which\; has& D^2=-135,\end{array}$$ and roots of these polynomials are ????

Example 3. Let $f\left(x\right)=x^4-2$ which has roots $2^{\frac{1}{4}},i{2}^{\frac{1}{4}},-2^{\frac{1}{4}},-i{2}^{\frac{1}{4}} \in ℂ,$ and let $𝔼$ be the splitting field of $f\left(x\right)$. Then $$\mathrm{Gal}(𝔼/𝔽)= ⟨\sigma ,\tau |{\sigma }^3=1, {\tau }^2=1, \sigma \tau =\tau {\sigma }^{-1}⟩$$ is the dihedral group of order 8. $$\begin{array}{cccccccccc}& & \mathbb{Q}\left(2^{\frac{1}{4}},i\right)& & & & & & & \\ \mathbb{Q}\left((1-i)2^{\frac{1}{4}}\right)& \mathbb{Q}\left((1+i)2^{\frac{1}{4}}\right)& \mathbb{Q}\left(\sqrt{2},i\right)& \mathbb{Q}\left(i{2}^{\frac{1}{4}}\right)& \mathbb{Q}\left(2^{\frac{1}{4}}\right)& ⟨{\sigma }^3\tau ⟩ & ⟨\sigma \tau ⟩ & ⟨{\sigma }^2⟩ & ⟨{\sigma }^2\tau ⟩ & ⟨\tau ⟩ \\ & \mathbb{Q}\left(i\sqrt{2}\right)& \mathbb{Q}\left(i\right)& \mathbb{Q}\left(\sqrt{2}\right)& & & ⟨\sigma \tau ,{\sigma }^2⟩ & ⟨{\sigma }^2⟩ & ⟨\tau ,{\sigma }^2⟩ & \\ & & \mathbb{Q}& & & & & ⟨\sigma ,\tau ⟩ & & \end{array}$$

Example 4. Let $f\left(x\right)=x^4+1$ which has roots $\omega ,{\omega }^3,{\omega }^5,{\omega }^7$ where $\omega =e^{\frac{2\pi i}{8}}\in ℂ,$ and let $𝔼$ be the splitting field of $f\left(x\right)$. Then $$\mathrm{Gal}(𝔼/\mathbb{Q})= ⟨\sigma ,\tau |{\sigma }^2=1, {\tau }^2=1, \sigma \tau =\tau \sigma ⟩ \cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$$ is the Klein four group.
Let $a,b\in \mathbb{Q}$ and consider the extension $\mathbb{Q}\left(\sqrt{a},\sqrt{b}\right)$ which is the splitting field of $f\left(x\right)=(x^2-a)(x^2-b).$ Then $$\mathrm{Gal}(\mathbb{Q}\left(\sqrt{a},\sqrt{b}\right)/\mathbb{Q})= ⟨\sigma ,\tau |{\sigma }^2=1, {\tau }^2=1, \sigma \tau =\tau \sigma ⟩ \cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$$ is the Klein four group with $$\begin{array}{ccc}\sigma \left(\sqrt{a}\right)=-\sqrt{a},& & \tau \left(\sqrt{a}\right)=\sqrt{a},\\ & and& \\ \sigma \left(\sqrt{b}\right)=\sqrt{b},& & \tau \left(\sqrt{b}\right)=-\sqrt{b},\end{array}$$
$$\begin{array}{cccccc}& \mathbb{Q}\left(\sqrt{a},\sqrt{b}\right)& & & \left\{1\right\}& \\ \mathbb{Q}\left(\sqrt{b}\right)& \mathbb{Q}\left(\sqrt{a}\right)& \mathbb{Q}\left(\sqrt{a}\sqrt{b}\right)& ⟨\sigma \tau ⟩& ⟨\sigma ⟩& ⟨\tau ⟩\\ & \mathbb{Q}& & & ⟨\sigma ,\tau ⟩ & \end{array}$$ Note that $\mathbb{Q}\left(\sqrt{a},\sqrt{b}\right)$ since ${\left(\sqrt{a}+\sqrt{b}\right)}^2=a+2\sqrt{a}\sqrt{b}+b^2\in \mathbb{Q}(\sqrt{a}+\sqrt{b})$ and so $\left(\sqrt{a}\sqrt{b}\right)(\sqrt{a}+\sqrt{b})=(a\sqrt{b}+b\sqrt{a})\in \mathbb{Q}(\sqrt{a}+\sqrt{b}).$ So $(b-a)\sqrt{a}\in \mathbb{Q}(\sqrt{a}+\sqrt{b}).$

Example 5. Let $f\left(x\right)=x^n-1$ which has roots $\omega ,{\omega }^2,...,{\omega }^{n-1} \in ℂ$ where $\omega =e^{\frac{2\pi i}{n}},$ and let $𝔼$ be the splitting field of $f\left(x\right)$. Then $$\mathrm{Gal}(𝔼/\mathbb{Q})\cong {\left(\mathbb{Z}/n\mathbb{Z}\right)}^{*}and\left|{\left(\mathbb{Z}/n\mathbb{Z}\right)}^{*}\right|=\phi \left(n\right),$$ where $\phi \left(n\right)$ is Euler's phi function.

Example 6. Assume $𝔽$ contains a primitive $n^{th}$ root of unity and let $𝔼$ be a finite Galois extension of $𝔽$. Then $$\mathrm{Gal}(𝔼/𝔽)\cong \mathbb{Z}/n\mathbb{Z}if\; and\; only\; if𝔼 \; is\; the\; splitting\; field\; of\; an\; irreducible\; x^n-b \; for\; some\; b\in 𝔽.$$

Example 7. Assume $𝔼=𝔽\left(x_1,...,x_n\right)$ and $𝕂=𝔽\left(e_1,...,e_n\right),$ where $e_1,e_2,...,e_n$ are the elementary symmetric functions. Then $$𝔼 \; is\; the\; splitting\; field\; of\; f\left(x\right)=\prod _{i=1}^n(x-x_i)\in 𝕂\left[x\right]and\mathrm{Gal}(𝔼/𝕂)\cong S_n.$$

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