## The extension $𝔽\left(\alpha \right)$

Last update: 02 February 2012

## The extension $𝔽\left(\alpha \right)$

Let $𝔽$ be a field and let $𝔼$ be an extension of $𝔽$.

• The degree of $𝔼$ over $𝔽$ is $[𝔼:𝔽] = dim𝔽 (𝔼).$
• Let $\alpha \in 𝔼$. The minimal polynomial of $\alpha$ over $𝔽$ is the monic irreducible polynomial ${m}_{\alpha ,𝔽}\left(x\right)\in 𝔽\left[x\right]$ which has $\alpha$ as a root.
• An element $\alpha \in 𝔼$ is algebraic over $𝔽$ if ${m}_{\alpha ,𝔽}\left(x\right)$ exists.
• An element $\alpha \in 𝔼$ is separable if all roots of ${m}_{\alpha ,𝔽}\left(x\right)$ are multiplicity 1.
• An element $\alpha \in 𝔼$ is normal if all roots of ${m}_{\alpha ,𝔽}\left(x\right)$ are in $𝔼$.
• Let $\alpha \in 𝔼$. The field generated by $𝔽$ and $\alpha$ is the subfield $𝔽\left(\alpha \right)$ of $𝔼$ such that
1. $𝔽\left(\alpha \right)$ contains $𝔽$ and $\alpha$,
2. if $𝕂$ is a subfield of $𝔼$ which contains $\alpha$ and $𝔽$ the $𝕂\supseteq 𝔽\left(\alpha \right).$
• Let $\alpha \in 𝔼.$ The ring $𝔽[α] = im(evα :𝔽[x] →𝔼)$ is the image of the evaluation homomorphism ${\mathrm{ev}}_{\alpha }$.

Let $𝔼$ be an extension of $𝔽$ and let $\alpha \in 𝔼$. Then $𝔽(α) = 𝔽[α] ≅ 𝔽[x] ( mα,𝔽(x) ) .$

(Theorem of the primitive element) If $𝔼$ over $𝔽$ is a finite separable extension of $𝔽$ then there exists an element $\theta \in 𝔼$ such that $𝔼=𝔽\left(\theta \right)$.

Let $𝔼$ be a finite extension of $𝔽$. Then $𝔼=𝔽\left(\theta \right)$ for an element $\theta \in 𝔼$ if and only if there are only a finite number of intermediate fields $𝔼\supseteq 𝕂\supseteq 𝔽$.

1. All roots of ${m}_{\alpha ,𝔽}\left(x\right)$ have the same multiplicity.
2. If $\mathrm{char}\left(𝔽\right)=0$ or if $𝔽$ is finite then all elements of $\stackrel{_}{𝔽}$ are separable.

 Proof. Let $\alpha$ and $\beta$ be roots of ${m}_{\alpha ,𝔽}\left(x\right).$ The isomorphism $σ: 𝔽(α) ≅ 𝔽[x] ( mα ,𝔽 (x)) ≅ 𝔽(β) α ↦ x ↦ β$ extends to an automorphism $σ: 𝔽_ → 𝔽_ α ↦ β$ which is the identity on $𝔽$. Thus $\sigma \left({m}_{\alpha ,𝔽}\left(x\right)\right)={m}_{\alpha ,𝔽}\left(x\right)$ and $\sigma \left({\left(x-\alpha \right)}^{k}\right)={\left(x-\beta \right)}^{k}.$ So if $\alpha$ is an root of ${m}_{\alpha ,𝔽}\left(x\right)$ of multiplicity $k$ then ${\left(x-\beta \right)}^{k}$ is a factor of ${m}_{\alpha ,𝔽}\left(x\right)$ in $\stackrel{_}{𝔽}\left[x\right].$ So $\beta$ has multiplicity $k$. Let ${m}_{\alpha ,𝔽}$ be the minimal polynomial of $\alpha \in \stackrel{_}{𝔽}$ over $𝔽$. Then ${m}_{\alpha ,𝔽}\left(x\right)=\left(x-\alpha \right)n\left(x\right)$ and $d dx mα,𝔽 (x) = (x-α) dn dx + n(x).$ So ${m}_{\alpha ,𝔽}\left(x\right)$ has $\alpha$ as a multiple root if and only if ${m}_{\alpha ,𝔽}\text{'}\left(\alpha \right)=0.$ Since ${m}_{\alpha ,𝔽}\left(x\right)$ is the minimal polynomial which has $\alpha$ as a root, ${m}_{\alpha ,𝔽}\left(x\right)$ has a multiple root if and only if ${m}_{\alpha ,𝔽}\text{'}\left(\alpha \right)=0.$ If $mα,𝔽 (x) = ∑ i=0 r aixi then mα,𝔽' (x) = ∑ i=0 r iai xi-1.$ So ${m}_{\alpha ,𝔽}^{\text{'}}\left(x\right)=0$ if and only if $p$ divides $i{a}_{i}$ for all $1\le i\le r$, where $p=\mathrm{char}\left(𝔽\right).$ Since $p$ divides $i{a}_{i}$ if and only if $p$ divides ${a}_{i}$ it follows that ${m}_{\alpha ,𝔽}\left(x\right)$ has no multiple root when $\mathrm{char}\left(𝔽\right)=0$. If $\mathrm{char}\left(𝔽\right)=p$ then ${m}_{\alpha ,𝔽}\left(x\right)$ has a multiple root if and only if ${m}_{\alpha ,𝔽}\left(x\right)=g\left({x}^{p}\right)$ for some polynomial $g\left(x\right)\in 𝔽\left[x\right].$ $\square$

(Theorem of the primitive element) If $𝔼$ over $𝔽$ is a finite separable extension of $𝔽$ then there exists an element $\theta \in 𝔼$ such that $𝔼=𝔽\left(\theta \right).$

 Proof. Case 2: $𝔽$ is infinite. Since $𝔼$ is a finite extension of $𝔽$, $𝔼=𝔽\left({\gamma }_{1},{\gamma }_{2},...,{\gamma }_{k}\right)$ for some ${\alpha }_{1},...,{\alpha }_{k}$ in $𝔼$. The proof is by induction on $k$. It is sufficient to show that if $𝔼=𝔽\left(\alpha ,\beta \right)$ then there is an element $\theta \in 𝔼$ such that $𝔼=𝔽\left(\theta \right)$. Let $\alpha = {\alpha }_{1},{\alpha }_{2},...,{\alpha }_{n} ,$ and $\beta = {\beta }_{1},{\beta }_{2},...,{\beta }_{m}$ be the roots of ${m}_{\alpha ,𝔽}\left(x\right)$ and ${m}_{\beta ,𝔽}\left(x\right),$ respectively. Since $𝔽$ is infinite there exists $a\in 𝔽$ such that Let $\theta =\alpha +a\beta$. Let $p\left(x\right)={m}_{\alpha ,𝔽}\left(\theta -ax\right)\in 𝔽\left(\theta \right)\left[x\right].$ Then since $\alpha +a\left(\beta -{\beta }_{j}\right)\ne {\alpha }_{i}$ for any $i$. Now ${m}_{\beta ,𝔽\left(\theta \right)}\left(x\right)$ divides ${m}_{\beta ,𝔽}\left(x\right)$ and ${m}_{\beta ,𝔽\left(\theta \right)}\left(x\right)$ divides $p\left(x\right)$. Since the only factor in common between $p\left(x\right)$ and ${m}_{\beta ,𝔽}\left(x\right)$ is $\left(x-\beta \right)$, $mβ,𝔽(θ) (x) = (x-β).$ So $\beta \in 𝔽\left(\theta \right)$ and $\alpha =\theta -a\beta \in 𝔽\left(\theta \right).$ So $𝔼=𝔽\left(\alpha ,\beta \right)=𝔽\left(\theta \right).$ $\square$

## Notes and References

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