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 $D_8$
5
$\begingroup$
galois-theory
-
11Does $D_8$ have order $8$ or $16$? – 2012-08-29
-
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
-
0Try playing around with quartic polynomials with even degree only – 2012-08-29
-
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
-
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
-
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
-
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
-
0D8 is the dihedral group of symmetries of a regular polygon with 8 vertices. It is of order 16. – 2012-08-29
-
1The polynomial $x^8-2x^4-4$ has Galois group $D_8$ of order 16. – 2012-08-30
-
1The polynomial $x^8+2$ also has Galois group $D_8$. – 2012-08-30