## Extensions of $ℚ$

Last update: 02 February 2012

## Extensions of $ℚ$

• An algebraic number is an element of $\stackrel{_}{ℚ}$ the algebraic closure of $ℚ$.
• An algebraic number field is a finite extension of $ℚ$.
• Let $f\left(x\right)\in ℚ\left[x\right].$ The discriminant of $f\left(x\right)$ is $D2 where D= ∏ 1≤i where ${\alpha }_{1},...,{\alpha }_{k},\in ,\stackrel{_}{ℚ}$ are such that $f\left(x\right)=\left(x-{\alpha }_{1}\right)\cdots \left(x-{\alpha }_{k}\right)\in \stackrel{_}{ℚ}\left[x\right].$

Let $𝔼$ be the splitting field of $f\left(x\right)\in ℚ\left[x\right].$ Note that if $\sigma \in \mathrm{Gal}\left(𝔼/𝔽\right)$ 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 $ℚ\left(D\right)$ is a degree two extension of $ℚ$ since the minimal polynomial of $D$ is ${x}^{2}-{D}^{2}.$ If $D\in ℚ$ then $\mathrm{Gal}\left(𝔼/ℚ\right)\subseteq {A}_{n},$ the alternating group.

Example 1. If $f\left(x\right)={x}^{2}+bx+c\in ℚ\left[x\right]$ and $fx= x2+bx +c= (x-α1) (x-α2) ∈ℚ_ [x]$ then $ℚα1 = ℚα2 since α1=b -α1.$ If $\sigma \in \mathrm{Gal}\left(ℚ\left({\alpha }_{1}\right)/ℚ\right)$ is the element given by $σα1 =α2 then σα2 =α1,$ since ${\alpha }_{2}+\sigma \left({\alpha }_{2}\right)=\sigma \left({\alpha }_{1}+{\alpha }_{2}\right)=\sigma \left(b\right)=b={\alpha }_{1}+{\alpha }_{2}.$ So $Gal(ℚ α1/ ℚ) = 1σ ≅ ℤ/2ℤ = S2.$ Now the discriminant ${D}^{2}={b}^{2}-4c\in ℚ$ and $D=\sqrt{{b}^{2}-4c}\in ℚ\left({\alpha }_{1}\right).$ So $ℚα1 = ℚD and σD= -D.$

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 $fx = y3- a2y2+ 3ya22 9 - a23 27 + a2y2- 2a2x2 3 y+ a23 9 + a1y- a1a2 3 +a0 =0.$ 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 If $f\left(x\right)={x}^{3}+bx+c=\left(x-{\alpha }_{1}\right)\left(x-{\alpha }_{2}\right)\left(x-{\alpha }_{3}\right)$ then $e3 = α1 α2 α3 = -c, e2 = α1 α2 + α2 α3 + α1 α3 = b, e1 = α1 + α2 + α3 .$ $ℚ α1 α2 α3 {1} ℚ α1 ℚ α2 ℚ α3 1 1,3 1 23 1 23 ℚD A3 ℚ S3$ A concrete example is $f\left(x\right)={x}^{3}-2$ which has roots where $\omega$ is a primitive cube root of unity and $Gal(𝔼/ℚ) ≅ S3.$ Examples of the two cases are $fx = x3- 3x+1 which has D2=81, and fx = x3+ 3x+1 which has D2= -135,$ 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 is the dihedral group of order 8. $ℚ 2 1 4 i ℚ (1-i) 2 1 4 ℚ (1+i) 2 1 4 ℚ 2 i ℚ(i 2 1 4 ) ℚ 2 1 4 σ3τ στ σ2 σ2τ τ ℚ(i 2 ) ℚ(i) ℚ( 2 ) στ σ2 σ2 τ σ2 ℚ στ$

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 is the Klein four group.
Let $a,b\in ℚ$ and consider the extension $ℚ\left(\sqrt{a},\sqrt{b}\right)$ which is the splitting field of $f\left(x\right)=\left({x}^{2}-a\right)\left({x}^{2}-b\right).$ Then is the Klein four group with $σ a =- a , τ a = a , and σ b = b , τ b =- b ,$
$ℚ a b {1} ℚ b ℚ a ℚ a b ⟨στ⟩ ⟨σ⟩ ⟨τ⟩ ℚ στ$ Note that $ℚ\left(\sqrt{a},\sqrt{b}\right)$ since ${\left(\sqrt{a}+\sqrt{b}\right)}^{2}=a+2\sqrt{a}\sqrt{b}+{b}^{2}\in ℚ\left(\sqrt{a}+\sqrt{b}\right)$ and so $\left(\sqrt{a}\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=\left(a\sqrt{b}+b\sqrt{a}\right)\in ℚ\left(\sqrt{a}+\sqrt{b}\right).$ So $\left(b-a\right)\sqrt{a}\in ℚ\left(\sqrt{a}+\sqrt{b}\right).$

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 $Gal(𝔼/ℚ) ≅ ℤ/nℤ * and | ℤ/nℤ * |=φ(n),$ 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

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

## Notes and References

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