5
$\begingroup$

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

  • 11
    Does $D_8$ have order $8$ or $16$?2012-08-29
  • 2
    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.2012-08-29
  • 0
    Try playing around with quartic polynomials with even degree only2012-08-29
  • 0
    Start from a [presentation of $D_8$](http://en.wikipedia.org/wiki/Dihedral_group#Equivalent_definitions) in terms of generators and relations.2012-08-29
  • 4
    It 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
  • 0
    But 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
  • 3
    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?2012-08-29
  • 0
    D8 is the dihedral group of symmetries of a regular polygon with 8 vertices. It is of order 16.2012-08-29
  • 1
    The polynomial $x^8-2x^4-4$ has Galois group $D_8$ of order 16.2012-08-30
  • 1
    The polynomial $x^8+2$ also has Galois group $D_8$.2012-08-30

2 Answers 2