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Given that the degree of splitting field of a polynomial $f(x)$ over $\mathbb{Q}[x]$ is equal to $n!$ where $n$ is the degree of $f$, $n>2$. If $\alpha$ is a root of $f$ in the splitting field, then we have to show that $\mathbb{Q}(\alpha^4)=\mathbb{Q}(\alpha)$.

My attempt: $\mathbb{Q}(\alpha^4)\subseteq\mathbb{Q}(\alpha)$ is obvious.Also, $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^4)]=4$ as $x^4-\alpha^4$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}(\alpha^4)$. So, $n$ must be a multiple of $4$, by Tower Law. I have come to a dead end it seems Please guide me.

Shanghaikid
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  • Why do you think that $x^4-\alpha^4$ is the minimal polynomial? Consider for example the case of $\alpha=\root3\of2$ when $\alpha^4=2\alpha$. – Jyrki Lahtonen Apr 20 '19 at 05:51
  • I would attack this as follows. Let $\alpha_1=\alpha, \alpha_2,\ldots,\alpha_n$ be the zeros of $f(x)$ in the splitting field. Then the conjugates of $\alpha^4$ are exactly the numbers $\alpha_j^4,j=1,2,\ldots,n$. If you can show that they are all distinct, it follows that $[\Bbb{Q}(\alpha^4):\Bbb{Q}]=n$, and you are done. Now, if $\alpha_j^4=\alpha_k^4$, $j\neq k,$ then we must have $\alpha_k=i^t\alpha_j$ for some $t=1,2,3$. Hint: If $n>3$ show that $[\Bbb{Q}(\alpha_j,\alpha_k):\Bbb{Q}]<n(n-1)$, and arrive at a contradiction. – Jyrki Lahtonen Apr 20 '19 at 05:59
  • @JyrkiLahtonen Got it. Thanks a lot. – Shanghaikid Apr 20 '19 at 12:54
  • @JyrkiLahtonen By the way, Earlier I said $x^4-\alpha^4$ is the minimal polynomial over $\mathbb{Q}(\alpha^4) $ assuming of course $\alpha$ does not belong to that field. Is that wrong? What am i missing? The example you gave does not fit here I guess as $\mathbb{Q}(\alpha^4) $=$\mathbb{Q}(\alpha) $. – Shanghaikid Apr 20 '19 at 13:07
  • Good job completing the argument! Care to write it down as an answer? But, I don't see that assumption $\alpha\notin\Bbb{Q}(\alpha^4)$? It is hard to follow your reasoning, if you leave out pieces like that :-) There are also cases like $\alpha=\zeta_{12}$ when $\alpha\notin\Bbb{Q}(\alpha^4)$ but $\alpha^2\in\Bbb{Q}(\alpha^4)$. Doesn't come from $f(x)$ like here, but you have to show that if you make a claim about minimal polynomials like that :-) – Jyrki Lahtonen Apr 20 '19 at 13:33
  • @JyrkiLahtonen Thank you for your kind words. I admit I was being a little casual. I will keep that in mind. – Shanghaikid Apr 21 '19 at 06:38
  • @JyrkiLahtonen I would be grateful if you could kindly check the answer that I have posted. – Shanghaikid Apr 22 '19 at 09:17

1 Answers1

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I will simply try to join the dots mentioned in the comments.

First let us see why the conjugates of $\alpha^4$ are exactly $\alpha_i^4$, $ i=1,2,\cdots,n$.If $g(x)$ is the minimal polynomial of $\alpha^4$, then $\alpha$ is the root of $g(x^4)$ but $\alpha$ has $f(x)$ to be the minimal polynomial over $\mathbb{Q}$ because if $f(x)$ is reducible then the splitting field will have a degree strictly less than $n!$. Thus, $f(x)$ must divide $g(x^4)$, implying all the roots of $f(x)$, namely the $\alpha_i$'s must also be roots of $g(x^4)$. That is $\alpha_i^4$ is a root of $g(x)$. Hence the claim.

Now, all we have to show is that these conjugates are distinct.

$\alpha_j^4=\alpha_k^4,j\neq k$ implies four choices $\alpha_j=i^t\alpha_k$ , $t=1,2,3,4$ of which we claim none is possible. Let us first remove the obvious $t=4$ in which case $\alpha_j=\alpha_k$, not true.

For t=1, the minimal polynomial of $\alpha_j$ over $\mathbb{Q}(\alpha_k)$ is $x^2+\alpha_k^2$.Therefore, $[\mathbb{Q}(\alpha_j,\alpha_k):\mathbb{Q}(\alpha_k)]=2$ contradicting $[\Bbb{Q}(\alpha_j,\alpha_k):\Bbb{Q}]<n(n-1)$ , since $n>3$, so $n(n-1)>2n=[\Bbb{Q}(\alpha_j,\alpha_k):\Bbb{Q}]$.

And similarly the other cases.

Shanghaikid
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