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From Berkeley Problems in Mathematics, Spring 1999, Problem 17.

Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n\ge 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Give that $[L:\mathbb{Q}]=n!$, prove that $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$.

(BELOW IS NONSENSE, SHOULD BE IGNORED)

Following Gerry Myerson's advice, since $L$ is the splitting field of $f$, $L$ must be Galois over $\mathbb{Q}$. Thus the Galois group for $[L:\mathbb{Q}]=S_{n}$. Gerry now asserts that either $\mathbb{Q}(\alpha^{4})$ is of degree 2 over rationals or it must be $\mathbb{Q}$. But assuming this one can rule out the case $\mathbb{Q}$, since this would imply $\alpha$ is the root of a 4th or 2nd degree polynomial over $\mathbb{Q}$. But we know $n\ge 3$. So $\alpha$ must be the root of a 4th degree polynomial. We know $S_{n}$ has normal subgroups $A_{n}$ when $n\not=4$ and $V,A_{4}$ when $n=4$. I do not know how to proceed any further.

Now assuming $\mathbb{Q}(\alpha^{4})$ is a degree 2 abelian extension over $\mathbb{Q}$. We should have a chain of normal extensions $\mathbb{Q}(\alpha)\supset \mathbb{Q}(\alpha^{2})\supset \mathbb{Q}$. This would imply $S_{n}$ has a normal subgroup which has a normal subgroup. But we know $A_{n}$ for $n\ge 5$ are simple. So the only possibility is $\mathbb{Q}(\alpha)$ has a Galois group isomorphic to $V$. In this case $n=4$ as well. I can only proceed to here.

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Each $a_i$ has degree $n$ over the rationals, but degree $i$ over the field you get by adjoining all the $a_j$ with $j\gt i$. –  Gerry Myerson Aug 13 '12 at 5:21
    
Also, over a field of characteristic zero, like $\mathbb Q$, every algebraic extension is separable, so you need not worry about that. –  Galois Group Aug 13 '12 at 5:29
    
I have never heard of it - why? –  Bombyx mori Aug 13 '12 at 5:31
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That's not what the primitive element theorem says! There's a primitive element, but it doesn't have to be $\alpha$ –  Cocopuffs Aug 13 '12 at 5:35
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Dear user, If you cannot show that $\mathbb Q(2^{1/3})$ equals $\mathbb Q(2^{4/3})$, then you should be studying more basic aspects of algebra than Galois theory. Regards, –  Matt E Aug 13 '12 at 5:47

2 Answers 2

up vote 8 down vote accepted

Look at an example. Let $f(x)=x^3-2$. You should be able to work out the splitting field, and see that it has degree 6 over the rationals. If you have any problem doing this, come back and let us know where you get stuck.

EDIT: much of what I wrote a few hours ago was not right. Let me try again.

$L$ is normal over the rationals, over $K={\bf Q}(\alpha^4)$, and over $E={\bf Q}(\alpha)$. The group of $L$ over the rationals is $S_n$, the group of $L$ over $E$ is $S_{n-1}$, so the group of $L$ over $K$ is a subgroup $H$ of $S_n$ containing $S_{n-1}$. We're trying to prove that $H=S_{n-1}$.

We can rule out $H=S_n$, as follows. If $H=S_n$, then $\alpha^4=q$ is rational, and the minimal polynomial for $\alpha$ over the rationals is $x^4-q$ (or some factor of that polynomial), and we're not talking about a polynomial with Galois group $S_n$.

So all we have to show is that there is no proper subgroup of $S_n$ properly containing $S_{n-1}$, and we're done. And, hey, there's a proof on m.se, so we're done.

I think what follows can safely be ignored.

Back to the ${\bf Q}(\alpha^4)$ question: if ${\bf Q}(\alpha^4)$ is a proper subfield of ${\bf Q}(\alpha)$, then it's either the rationals or degree 2 over the rationals. It's easy to rule out the first case (details left to you - I have to teach a class in a few minutes). In the second case, it's normal over the rationals, and ${\bf Q}(\alpha)$ is normal over it, and that says something about normal subgroups of the Galois group of $L$ which don't fit with that group being the symmetric group, $S_n$ (that group doesn't have a lot of normal subgroups).

I know I've left a lot out. I hope it's of some use. But if you haven't done the Galois Theory, my answer won't do much for you.

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Thanks Gerry, but I am still stuck. –  Bombyx mori Aug 13 '12 at 5:42
    
Where?${}{}{}{}$ –  Gerry Myerson Aug 13 '12 at 5:49
    
I editted - you can see. –  Bombyx mori Aug 13 '12 at 5:50
    
I have `done' Galois theory a few years ago by studying myself but my knowledge is not very solid. Let me try to fill in the details. –  Bombyx mori Aug 13 '12 at 6:07
    
I edited, still have a lot of blanks to fill out. –  Bombyx mori Aug 13 '12 at 8:08

I wish to give a solution that do not require to know anything about the normal subgroups of $S_n$:

First note $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{4})]\leq4$ since $\alpha$ is a root of $x^{4}-\alpha^{4}\in\mathbb{Q}(\alpha^{4})[x]$.

Now there is need to divide to cases: If the degree of the extension is $1$ then you are done, we need to show that it can't be $2,3,4$.

If the degree of the extension is $3$ then since $\mathbb{Q}(\alpha^{2})$ is a subextension we have $$3=[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{4})]=[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{2})][\mathbb{Q}(\alpha^{2}):\mathbb{Q}(\alpha^{4})]$$ hence one of $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{2})],[\mathbb{Q}(\alpha^{2}):\mathbb{Q}(\alpha^{4})]$ is of degree $3$ which is clearly a contradiction since, for example, $\alpha$ is a root of $x^{2}-\alpha^{2}\in\mathbb{Q}(\alpha^{2})$.

if $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{2})]=2$ then there is $g(x)=x^{2}+ax+b\in\mathbb{Q}(\alpha^{2})$ s.t. $g(\alpha)=0$, but then $g|f$ in $\mathbb{Q}(\alpha^{2})[x]\implies f=gh$ where $h\in\mathbb{Q}(\alpha^{2})[x],\deg(h)=\deg(f)-\deg(g)=n-2$.

Now, $f$ is irreducible (over $\mathbb{Q})$ hence $[\mathbb{Q}(\alpha):\mathbb{Q}]=\deg(f)=n$ so $[\mathbb{Q}(\alpha^{2}):\mathbb{Q}]=\frac{n}{2}$.

Denote $M$ as the splitting field of $h$ over $\mathbb{Q}(\alpha^{2})$, then $[M:\mathbb{Q}(\alpha^{2})]\leq(n-2)!$, also note that since $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^{2})]=2$ then $[M(\alpha):\mathbb{Q}(\alpha^{2})]\leq2(n-2)!$ and from the previews line $[M(\alpha):\mathbb{Q}]\leq2(n-2)!\frac{n}{2}=(n-2)!n<n!$

That is: $M(\alpha)$ is a field over $\mathbb{Q}$ containing all the roots of $f$ and is of degree $<n!$, and this is a contradiction.

In the same manner you get a contradictions to the rest of the cases which are similar, I leave it out to you to fill the rest of the details (I am in a hurry to get to class too).

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This =4 case can be similarly dismissed, and this is elementary enough for me to understand. Thanks. –  Bombyx mori Aug 13 '12 at 8:26

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