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Splitting Fields. Prove Q(alpha^4)=Q(alpha)

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Let $f(x)\in \mathbb Q[x]$ be an irreducible polynomial of degree $n\geq 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Given that $[L:\mathbb Q]=n!$, prove that $\mathbb Q(\alpha^4)=\mathbb Q(\alpha)$.
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Attempt:

Lemma:
Let $F$ be any field and $f,g \in F[x]$. Let $K$ and $L$ be splitting fields for $f$ and $g$ over $F$ respectively. Write $h=fg\in F[x]$. Let $E$ be the splitting field for $h$ over $F$. Then $[E:F]\leq [K:F][L:F]$

Proof: Let $f=(x-\alpha_1)\cdots(x-\alpha_k)$ in $K[x]$ and $g=(x-\beta_1)\cdots(x-\beta_l)$ in $L[x]$. Its easy to see that $K=F(\alpha_1,\ldots,\alpha_k)$, $L=F(\beta_1,\ldots,\beta_l)$ and $E=F(\alpha_1,\ldots,\alpha_k,\beta_1,\ldots,\beta_l)$. From here its easy to conclude that \begin{equation*}[E:K]\leq [L:F]\tag{1}\end{equation*}Now, $[E:F]=[E:K][K:F]\leq [K:F][L:F]$. Hence the lemma is settled.$\blacksquare$

Corollary: Let $F$ be any field and $f\in F[x]$ with deg $f=n$. Let $E$ be the spliting field for $f$ over $F$. Assume that $[E:F]=n!$. Then $f$ is irreducible over $F$.

Assuming there are no mistakes in my argument..
In the light of the corollary I think the question redundantly confers irreducibility to $f$.

Now. Let $f=(x-\alpha_1)\cdots (x-\alpha_n)=(x-\alpha_1)g$ be the representation of $f$ in $L[x]$. Define $F_1=\mathbb Q(\alpha_1)$. Then $L$ is the splitting field for $g$ over $F_1$ satisfying $[L:F_1]=(n-1)!=\text{ deg }g$ and hence $g$ is irreducible over $F_1$. This piece of information suggests that induction might be helpful. I can settle the base case, that is, $n=3$. If you require then I can post the proof for the base case in my following posts.
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Can anybody see how to settle the question?