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.

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    $\begingroup$ @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
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    $\begingroup$ @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
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    $\begingroup$ 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

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