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Can $D_{8}$ be the Galois group of $f$ where $f$ is a degree 8 polynomial? I was thinking that the Galois group must act transitively on the roots of $f$ (if $f$ is irreducible). In that case $D_{8}$ cannot be the Galois group of $f$. Is that correct? Is there a better way to approach this problem?

What if $f$ is not irreducible?

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  • $\begingroup$ Is your polynomial irreducible? $\endgroup$ – Mark Bennet Jan 26 '13 at 18:27
  • $\begingroup$ not necessarilly. but what i said about the Galois group acting transitively assumes irreducibility. $\endgroup$ – Digital Gal Jan 26 '13 at 18:31
  • $\begingroup$ Indeed, but you could take a polynomial of degree 4, say, and multiply it by $x^4$ to get a polynomial of degree 8. So you have to specify irreducible. $\endgroup$ – Mark Bennet Jan 26 '13 at 18:36
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    $\begingroup$ Sorry, there is always some ambiguity with the notation for dihedral groups. Your $D_8$ has order 8 or (degree 8 and) order 16? $\endgroup$ – Andreas Caranti Jan 26 '13 at 19:14
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    $\begingroup$ I was thinking that since $D_{8}$ embeds in $S_{4}$ it cannot act transitively on a set with more than 4 elements. I guess after thinking about it some more $D_{8}$ also embeds in $S_{8}$ so I was incorrect. $\endgroup$ – Digital Gal Jan 26 '13 at 22:43
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Best to work backwards. Start with an extension $K\supset\mathbb Q$ whose G-group you know to be $D_8$. Let $\xi$ be a primitive element of $K$, then the minimal polynomial for $\xi$ over $\mathbb Q$ is octic, irreducible, and all of its roots, just as well as any one of them, generate(s) $K$ over $\mathbb Q$.

You expect that for squarefree $n$, the Galois closure of $\mathbb Q(\root4\of n)$, which you get by adjoining $i$ as well, will have $D_8$ for the G-group. Calling $\rho=\root4\of n$, almost any linear combination of $\rho$ and $i$ will be a primitive element. For instance, for $n=2$, you get the minimal polynomial of $\xi=\rho+i$ to be $X^8 +4X^6 +2X^4+28X^2+1$. I confess that I have not checked the irreducibility of this directly.

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