# Polynomials for the alternating group $A_n$

It is my understanding that the polynomial $$f_n(x)=x^n-1$$ has the Galois Group $$(\mathbb{Z}/n\mathbb{Z})^*$$, group of units of order $$\phi(n)$$. In some sense, these are the "simplest" polynomials with that Galois Group. Is there a formula for the polynomial, say $$g_n(x)$$, whose Galois group is $$A_n$$? And, I mean $$g_n(x)$$ in the same sense as $$f_n(x)$$: that is, the "simplest" polynomials with that Galois Group of $$A_n$$.

• The Galois group of $x^n-1$ (over $\mathbb{Q}$) is $(\mathbb{Z}/n \mathbb{Z})^{\times}$, the group of mulitiplicative units modulo $n$, not $C_n$. Aug 30, 2020 at 22:42
• See kconrad.math.uconn.edu/blurbs/galoistheory/cyclotomic.pdf , for example. Aug 30, 2020 at 22:51
• Note that there are distinct $n,m$ with $(Z/nZ)^\times$ and $(Z/mZ)^\times$ isomorphic (for instance, $m,n=1,2$, or $m,n=3,4$), so it's not a unique "simplest" polynomial.
– YCor
Aug 30, 2020 at 23:29
• The question was asked recently at math.stackexchange.com/questions/3808650/…. Generally speaking, when posting the same question on both sites this should be mentioned to avoid duplicate answers/work. Aug 31, 2020 at 0:30
• Schur proved that the truncated exponential polynomials $1 + x + x^2/2! + \cdots + x^n/n!$ are irreducible over $\mathbf Q$ for all $n \geq 1$ and their splitting field over $\mathbf Q$ has Galois group $A_n$ when $4 \mid n$. When $n$ is not divisible by $4$, the Galois group is $S_n$. The cases of $n$ being or not being divisible by $4$ corresponds to the discriminant of the polynomial being or not being a square, which is the minimial kind of information needed to know if a Galois group of an $n$th degree irreducible polynomial is or is not contained in $A_n$. Aug 31, 2020 at 1:18

For any integer $$n\ge7$$, we show how to explicitly build an infinite number of rational trinomals of degree $$n$$ whose Galois group over $$\bf Q$$ is isomorphic to $$A_n$$.