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 $A_n$.
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4$\begingroup$ 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 $C_n$. $\endgroup$– David E SpeyerCommented Aug 30, 2020 at 22:42
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1$\begingroup$ See kconrad.math.uconn.edu/blurbs/galoistheory/cyclotomic.pdf , for example. $\endgroup$– David E SpeyerCommented Aug 30, 2020 at 22:51
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1$\begingroup$ 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. $\endgroup$– YCorCommented Aug 30, 2020 at 23:29
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3$\begingroup$ 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. $\endgroup$– KConradCommented Aug 31, 2020 at 0:30
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1$\begingroup$ 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 $A_n$. $\endgroup$– KConradCommented Aug 31, 2020 at 1:18
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1 Answer
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Hermez and Salinier, Rational trinomials with the alternating group as Galois group, Journal of Number Theory, Volume 90, Issue 1, September 2001, Pages 113-129 has the abstract,
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 $A_n$.
