**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.



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


