4

I checked the Galois groups of the polynomials $$f(m,n) := mx^{n-m}+(m+1)x^{n-m-1}+\dots+(n-1)x+n$$ for $0 < m < n$, and I only found one polynomial whose Galois group is NOT the symmetric group, namely $x^{6} + 2x^{5} + 3x^{4} + 4x^{3} + 5x^{2} + 6x + 7$.

I have two questions :

  1. Is this the only example of a polynomial of the form $f(m,n)$ having not the symmetric group as the Galois group ?

  2. If a polynomial $f$ with integer coefficients has the symmetric group as the Galois group, must $f$ be irreducible over $\mathbb Q$?

Perhaps the Galois groups help to show the irreducibility of the polynomials $f(m,n)$ for all $0 < m < n$!

user26857
  • 53,190
Peter
  • 41

1 Answers1

3

Although I cannot answer the first question (though it seems unlikely to me), the answer to the second question is that the polynomial is always irreducible. You didn't specify which symmetric group, but I assume you mean $S_n$ where $n=\deg(f)$

Assume that $f=gh$. Let $\mathbb{Q}_f$ denote the splitting field of $f(x)$ over $\mathbb{Q}$. We have

\begin{align*} [\mathbb{Q}_f:\mathbb{Q}]&=[\mathbb{Q}_f:\mathbb{Q}_g][\mathbb{Q}_g:\mathbb{Q}]\\ &\leq [\mathbb{Q}_h:\mathbb{Q}][\mathbb{Q}_g:\mathbb{Q}]\\ &\leq(\deg(h))!(\deg(g))!\\ &<(\deg(g)\deg(h))!\\ &=(\deg(f))! \end{align*}

Thus the Galois group can't be $S_n$ where $n=\deg(f)$

Stella Biderman
  • 31,475