I need to construct such a polynomial, and more generally: given a group $G$, how can it be realized as a Galois group?
A polynomial whose Galois group is $D_8$
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11Does $D_8$ have order $8$ or $16$? – 2012-08-29
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2As 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. – 2012-08-29
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0Try playing around with quartic polynomials with even degree only – 2012-08-29
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0Start from a [presentation of $D_8$](http://en.wikipedia.org/wiki/Dihedral_group#Equivalent_definitions) in terms of generators and relations. – 2012-08-29
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4It is a bit questionable to ask the same question simultaneously both here and at [Math Overflow](http://mathoverflow.net/questions/105811/a-polynomial-whose-galois-group-is-d-8). 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. – 2012-08-29
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0But it seems to me that the polynomial $x^8-3$ over $\mathbb{Q}(\sqrt2)$ should work. The generating automorphisms amount to multiplying the eighth roots of $3$ by $(1+i)/sqrt2$ and the complex conjugation. All this assuming that $|D_8|=16.$ (People are asking you about that because there are two reasonably common conventions, and we want to be sure, which is used by your book/teacher). – 2012-08-29
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3It 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? – 2012-08-29
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0D8 is the dihedral group of symmetries of a regular polygon with 8 vertices. It is of order 16. – 2012-08-29
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1The polynomial $x^8-2x^4-4$ has Galois group $D_8$ of order 16. – 2012-08-30
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1The polynomial $x^8+2$ also has Galois group $D_8$. – 2012-08-30
2 Answers
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.
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0Small 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... – 2012-08-30
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0Yes, it should be $\mathbb{Q}$-automorphism, thank you. Otherwise the definition $x_{i}\mapsto x_{\sigma(i)}$ doesn't make sense! – 2012-08-30
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.