I need to construct such a polynomial, and more generally: given a group $G$, how can it be realized as a Galois group?

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    $\begingroup$ Does $D_8$ have order $8$ or $16$? $\endgroup$ Commented Aug 29, 2012 at 9:57
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    $\begingroup$ As for the second question. If you want it to be for an extension of $\mathbb{Q}$, it is still an unsolved problem whether that is possible. $\endgroup$ Commented Aug 29, 2012 at 9:59
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    $\begingroup$ It is a bit questionable to ask the same question simultaneously both here and at Math Overflow. The suggestion to newbies is to post it in one and wait for a couple of days. Otherwise people may waste their precious time thinking about it not knowing that an answer has already been given. You still haven't answered the question to the number of elements. $\endgroup$ Commented Aug 29, 2012 at 10:51
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    $\begingroup$ It is inexcusable to ask the same question simultaneously here and at MO without linking to each site from the other one. Why don't you just ask the whole world to work on your problem for you? $\endgroup$ Commented Aug 29, 2012 at 13:21
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    $\begingroup$ The polynomial $x^8-2x^4-4$ has Galois group $D_8$ of order 16. $\endgroup$ Commented Aug 30, 2012 at 1:39

2 Answers 2


I'm not sure how to make an extension for an arbitrary group $G$, but for finite groups there is a nice method.

So, assuming $|G|<\infty$

Embed $G$ into $S_{n}$ with $n=|G|$ and consider the ring $F=\mathbb{Q}[x_{1}, \dots, x_{n}]$. Let $E$ be the field of fractions of $F$.

If $\sigma\in G$, there is a $\mathbb{Q}$-automorphism (an automorphism fixing $\mathbb{Q}$) $\varphi_{\sigma}:E\rightarrow E$ given by $x_{i}\mapsto x_{\sigma(i)}$, where if $f_{1},f_{2}\in E$, $\varphi_{\sigma}(\frac{f_{1}}{f_{2}}) = \frac{\varphi_{\sigma}(f_{1})}{\varphi_{\sigma}(f_{2})}$. If $\sigma,\pi\in G$, it is clear that $\varphi_{\sigma}\circ\varphi_{\pi} = \varphi_{\sigma\circ\pi}$, and so these automorphisms form a group isomorphic to $G$. But then $E/E^{G}$, where $E^{G}$ is the fixed field of these automorphisms (of $G$), is a Galois extension with Galois group $G$.

There are some more details I left out, but this should give the general idea of how one can construct an extension for an arbitrary finite group.

  • $\begingroup$ Small typo: $F$-automorphism should be $\mathbb{Q}$-automorphism. As Dharam didn't specify the field (so I assume he doesn't care about it), I wonder why the solution on mathoverflow gets so many upvotes, but this generic solution didn't get any. Maybe you could mention what the fixed field for $G = S_n$ is, so that people recognize this solution from their algebra class... $\endgroup$
    – j.p.
    Commented Aug 30, 2012 at 13:31
  • $\begingroup$ Yes, it should be $\mathbb{Q}$-automorphism, thank you. Otherwise the definition $x_{i}\mapsto x_{\sigma(i)}$ doesn't make sense! $\endgroup$
    – Kirk Boyer
    Commented Aug 30, 2012 at 15:34

If you are interested in realization over $\mathbb{Q}$, then the book Inverse Galois Theory of Malle and Matzat tells us that $f=x^8-3x^5-x^4+3x^3+1$ has Galois group $D_8$ over $\mathbb{Q}$. In the end of this book there's a table of polynomials for all transitive groups up to degree $12$ over $\mathbb{Q}$. The body of the book explains in details the method to achieve realizations, and covers other topics in the area.


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