# Families of quintics in $\mathbb{Q}[x]$ with Galois group $A_5$

Theorem. The Galois group of a quintic polynomial $f\in\mathbb{Q}[x]$ is $A_5$ if and only if its discriminant is a rational square and its Weber sextic resolvent has no rational root.

Question. What are known infinite families of of quintic polynomials in $\mathbb{Q}[x]$, each with Galois group $A_5$?

I am aware of two explicit such families as follows.

Example 1. $f(x)=x^5+(5t^2-1)(5x+4)$ for $t\in\mathbb{Z}$ such that $t\equiv\pm 1\pmod{21}$. This is Exercise 3.7.2 in the book Generic Polynomials by Jensen, Ledet and Yui.

Example 2. $f(x)=x^5+(t^2-5^5)(x-4)$ for non-zero $t\in\mathbb{Q}$ such that $\forall u\in\mathbb{Q}: t\ne g(u)$ where $g(u)=\frac{(u^3-18u^2+8u-16)(u^3+2u^2+18u+4)}{2u^2(u^2+4)}$. This is from the paper Reducibility and the Galois group of a parametric family of quintic polynomials by Lavallee, Spearman and Williams.

Note. Both are all trinomials, presumably because it is easier to construct polynomials with square discriminant when some coefficients vanish. It would be nice to see otherwise.

• Some discussion at math.stackexchange.com/questions/286944/… – Gerry Myerson Jul 18 '14 at 0:08
• There may be something in Jensen, C. U.; Yui, N.; Polynomials with $D_5$ (resp. $A_5$) as Galois group, C. R. Math. Rep. Acad. Sci. Canada 2 (1980), no. 6, 297–302, MR0600565 (82c:12009). – Gerry Myerson Jul 18 '14 at 0:19
• Is there an electronic copy of the paper? I have been unsuccessful in finding it online. – Favst Jul 18 '14 at 0:57
• I don't know. That journal never circulated widely, and has always been hard to find. comptesrendus.math.ca has issues back to 1997; maybe you could contact someone through that website. – Gerry Myerson Jul 18 '14 at 2:04

A result of Mestre (Extensions regulieres de $\mathbb{Q}(T)$ de groupe Galois $\overline{A}_{n}$, Journal of Algebra, 1990) allows you to take "almost all" quintic $A_{5}$ polynomials $P(x)$ and extend them into an infinite family of polynomials giving $A_{5}$ extensions. (Mestre's argument works for $A_{2n+1}$ without any modification.)

More specifically, for an irreducible polynomial $P(x)$ of degree $5$ with square discriminant, there is usually a solution to the congruence $P'' R - 2 P' R' \equiv 0 \pmod{P}$. Solving this boils down to a system of linear equations with rational coefficients (and as shown by Mestre, this congruence will be solvable unless the coefficients of your original polynomial satisfy some specific polynomial equation). If $R$ is a solution with $\deg R < \deg P$, let $Q$ be the unique polynomial of degree less than $\deg P$ so that $-P' Q \equiv R^{2} \pmod{P}$. Then $\gcd(P,Q) = 1$ and $PQ' - P' Q = R^2$. From this, we get that $f(x,t) = P(x) - t Q(x)$ has square discriminant for all $t$. Moreover, since $P(x)$ has Galois group $A_{5}$ there are primes $p_{1}$, $p_{2}$ that do not divide a denominator of a coefficient in $Q(x)$ and for which $P(x)$ is irreducible modulo $p_{1}$, and for which $P(x)$ factors as an irreducible cubic multiplied by two (distinct) linear factors modulo $p_{2}$. This implies that if $t \equiv 0 \pmod{p_{1} p_{2}}$, then $f(x,t) \equiv P(x) \pmod{p_{i}}$ for $1 \leq i \leq 2$ and this proves that $f(x,t)$ has Galois group $A_{5}$.

For example, if $P(x) = x^{5} - x^{4} - 11x^{3} + x^{2} + 12x - 4$, which produces a totally real $A_{5}$ extension, then $R(x) = 7x^{4} + 13x^{3} - 20x + 16$ and $Q(x) = -49x^{4} - 91x^{3} + 93x^{2} + 92x - 52$. Then, $f(x,t) = P(x) - t Q(x)$ will have Galois group $A_{5}$ for "most" values of $t$.

• Would you mind checking your answer for typos? For example, $P'Q-PQ'=R^2$ gives $P'Q\equiv R^2 \pmod{P}$, whereas before that you wrote $-P'Q\equiv R^2 \pmod{P}$. – Favst Jul 21 '14 at 16:32
• I've fixed the typo. The correct equation is $PQ' - P'Q = R^{2}$. – Jeremy Rouse Jul 21 '14 at 16:37

You might be interested in a paper of Hashimoto and Tsunogai on generic polynomials for quintics (https://projecteuclid.org/euclid.pja/1116443730), including one for $A_5$. It is not a trinomial as a matter of fact.