A recent question asks what makes degree 5 special when considering the roots of polynomials with integer coefficients etc. One answer is that the Galois Group of $S_5$ is not solvable. What I am looking for is the most straightforward example (with proof) of a polynomial with integer coefficients and Galois Group $A_5$.

Such an object ought to be standard ... if I ever knew one, I have forgotten it.

  • 1
    $\begingroup$ @Mark, I think you can use the theorem of Dedekind math.uconn.edu/~kconrad/blurbs/galoistheory/galoisSnAn.pdf to prove one has galois group An. Can maybe reverse engineer such a polynomial from the method too. I wish I had time to try.. $\endgroup$ – user58512 Jan 25 '13 at 21:51
  • 1
    $\begingroup$ @YACP has put the guts of a solution below, except for computing the discriminant - on a good day I can remember the formula for the discriminant of a cubic, but a quintic - well I need to be shown. What I mean by "straightforward" - thinking about it - is something I might be able to remember. Or maybe there is no way round this. $\endgroup$ – Mark Bennet Jan 27 '13 at 13:02

I'll borrow from DonAntonio's answer the part about the discriminant of the polynomial $f=X^5+20X+16\in\mathbb Q[X]$: this is $2^{16}5^6$ and thus is a perfect square. Then $G_f$, the Galois group of $f$ over $\mathbb Q$, is contained in $A_5$.

It remains to prove that $G_f=A_5$.

First reduce the polynomial modulo $3$: $\overline f=X^5+2X+1\in\mathbb Z_3[X]$, and prove that $\overline f$ is irreducible. Since $G_{\overline f}$ is a subgroup of $G_f$, it follows that $G_f$ contains a $5$-cycle.

Second, reduce the polynomial modulo $7$: $\hat f=X^5-X+2\in\mathbb Z_7[X]$, and note that $\hat f=(X+2)(X+3)(X^3+2X^2+5X+5)$. Furthermore, $X^3+2X^2+5X+5$ is irreducible over $\mathbb Z_7$ and for the same reason as above $G_f$ contains a $3$-cycle.

In particular, the order of $G_f$ is divisible by $15$ and then $[A_5:G_f]\leq 4$. On the other side, $A_5$ can't contain a proper subgroup of index less than $5$ (this is true for every nonabelian simple group). It follows that $G_f=A_5$.

Edit. In this case the simplest way to compute the discriminant is to use the determinant formula which involves the power sums $s_i=x_1^i+\cdots+x_5^i$, where $x_i$ are the roots of $X^5+aX+b$. (This formula can be found in Jacobson, Basic Algebra I, page 258.) After some easy calculations one finds the discriminant: $2^8a^5+5^5b^4$.

  • $\begingroup$ This is looking good, though there is yet the problem of computing the discriminant - and I am really looking for a "by hand" proof throughout. The question of non-existence of subgroups of $A_5$ of order 15 has been addressed before in various ways, but this is a neat way of looking at it. There are transitive subgroups of $A_5$ (indeed $S_5$) of order 5 and 10, but not 15 or 20. So why is it so difficult to find and prove a specific example for $A_5$? $\endgroup$ – Mark Bennet Jan 26 '13 at 21:56
  • $\begingroup$ NB - my computer was hanging on me - order 20 possible for $S_5$ not $A_5$ $\endgroup$ – Mark Bennet Jan 26 '13 at 22:25
  • $\begingroup$ How is $G_{\bar{f}}$ a subgroup of $G_f$? I assume you mean $G_{\bar{f}}$ is the Galois group of $\bar{f}$ over $\mathbb{Z}_7$? $\endgroup$ – Wyatt Kuehster Jun 16 at 3:26

You may want to read chapter $\,4\,$ here , page $\,84\,$ in the "ereaders" version.

After some rather messy and annoying calculations, the discriminant of for example the polynomial $\,x^5+20x+16\in\Bbb Q[x]\,$ , which appears in the other answer, is $\,5^2\cdot 80^4\,$ , which is a square in $\,\Bbb Q\,$ and thus its Galois Group is contained in $\,A_5\,$...

  • $\begingroup$ Thanks for the very good reference, which also shows that $S_5$ can be attained in a polynomial with five real roots. $\endgroup$ – Mark Bennet Jan 26 '13 at 7:23

Here are some found by computer search

$$x^5 - 55x - 88$$ $$x^5 - 55x + 88$$ $$x^5 + 20x - 16$$ $$x^5 + 20x + 16$$ $$x^5 + 95x - 76$$ $$x^5 + 95x + 76$$

$$x^5 + 3x^3 + 5x - 10$$ $$x^5 + 3x^3 + 5x + 10$$ $$x^5 + 6x^3 - 7x - 8$$ $$x^5 + 6x^3 - 7x + 8$$ $$x^5 + 10x^3 - 10x - 4$$ $$x^5 + 10x^3 - 10x + 4$$

$$x^5 - x^4 + x^3 + 2x^2 + x - 1$$ $$x^5 + x^4 + x^3 - 2x^2 + x + 1$$

  • $\begingroup$ @YACP: the post was created CW, and the author said that this was just a list. It is obviously not intended as a full answer. $\endgroup$ – robjohn Jan 26 '13 at 13:03
  • $\begingroup$ @user58512: there is nothing abusive in the tone of YACP's comments. Simply reiterate that this is just a list of examples and it was made CW for this reason. It is not intended as a full answer. $\endgroup$ – robjohn Jan 26 '13 at 13:05
  • 3
    $\begingroup$ What I was interested in, and still am, is a polynomial of degree 5 which has Galois Group $A_5$ together with a proof. I am also interested that it is apparently so difficult to obtain a proof. A list of examples known to have the requisite group is progress, but not a full answer. It is particularly interesting that this list includes a range of polynomials which look different, but have something in common. Thanks for the list. I will upvote it. Also "longer than comment" answers which explain the barriers to an easy proof. $\endgroup$ – Mark Bennet Jan 26 '13 at 19:05
  • $\begingroup$ @MarkBennet, I gave you a link showing how to do the proof. $\endgroup$ – user58512 Jan 26 '13 at 21:34
  • $\begingroup$ @user58512 I'm looking for an explicit proof in detail for a specific example. I am convinced that examples exist, but I want to be shown why I should believe in a particular one. $\endgroup$ – Mark Bennet Jan 26 '13 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.