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3 fixed result

I've stumbled across the family of polynomials $f_p(x) = x^{p-1} + 2 x^{p-2} + \cdots + (p-1) x + p$, where $p$ is an odd prime. It's not too hard to show that $f_p(x)$ is irreducible over $\mathbb{Q}$ -- look at the Newton polygon of $f_p(x+1)$ over $\mathbb{Q}_p$ and you see that it factors as the product of an irreducible polynomial of degree $p-2$ and a linear. Since $f_p(x)$ has no real roots (look at the derivative of $f_p(x) (x-1)^2$) it must be irreducible over $\mathbb{Q}$. It's also not hard to see that the only primes dividing the discriminant are $2, p$ and primes dividing $p+1$. I would expect that the Galois group of a randomish polynomial would be the full symmetric group. Indeed, according to Magma this is true for $f_p(x)$ for $p=3,5, \dots, 61$ with the exception of $p=7,17$. So my question is -- are these the only exceptions?

Added later: I've had Magma find the Galois group for primes through 101 and it found another exception: $p=97$. So my initial guess was wrong.

Another addition: If one looks at odd $p$ (not just prime) for $p < 100$ there is another exception, 49. Also 241 is not an exception (misread magma's output).

The ideas in the following two papers may be of help:

"On the Galois Groups of the exponential Taylor polynomials" by Robert Coleman, in L'Enseignement Mathematique, v 33 (1987) pp 183-189

and

"On the Galois Group of generalized Laguerre polynomials" by Farshid Hajir, J. Th´eor. Nombres Bordeaux 17 (2005), no. 2, 517–525 (also available on the author's web page).

2 added results of further computations

I've stumbled across the family of polynomials $f_p(x) = x^{p-1} + 2 x^{p-2} + \cdots + (p-1) x + p$, where $p$ is an odd prime. It's not too hard to show that $f_p(x)$ is irreducible over $\mathbb{Q}$ -- look at the Newton polygon of $f_p(x+1)$ over $\mathbb{Q}_p$ and you see that it factors as the product of an irreducible polynomial of degree $p-2$ and a linear. Since $f_p(x)$ has no real roots (look at the derivative of $f_p(x) (x-1)^2$) it must be irreducible over $\mathbb{Q}$. It's also not hard to see that the only primes dividing the discriminant are $2, p$ and primes dividing $p+1$. I would expect that the Galois group of a randomish polynomial would be the full symmetric group. Indeed, according to Magma this is true for $f_p(x)$ for $p=3,5, \dots, 61$ with the exception of $p=7,17$. So my question is -- are these the only exceptions?

Added later: I've had Magma find the Galois group for primes through 101 and it found another exception: $p=97$. So my initial guess was wrong.

1

# Galois Groups of a family of polynomials

I've stumbled across the family of polynomials $f_p(x) = x^{p-1} + 2 x^{p-2} + \cdots + (p-1) x + p$, where $p$ is an odd prime. It's not too hard to show that $f_p(x)$ is irreducible over $\mathbb{Q}$ -- look at the Newton polygon of $f_p(x+1)$ over $\mathbb{Q}_p$ and you see that it factors as the product of an irreducible polynomial of degree $p-2$ and a linear. Since $f_p(x)$ has no real roots (look at the derivative of $f_p(x) (x-1)^2$) it must be irreducible over $\mathbb{Q}$. It's also not hard to see that the only primes dividing the discriminant are $2, p$ and primes dividing $p+1$. I would expect that the Galois group of a randomish polynomial would be the full symmetric group. Indeed, according to Magma this is true for $f_p(x)$ for $p=3,5, \dots, 61$ with the exception of $p=7,17$. So my question is -- are these the only exceptions?