Is 1+x+\frac{x^2}2+\dots+\frac{x^n}{n!}1+x+\frac{x^2}2+\dots+\frac{x^n}{n!} irreducible?

The polynomial f(x)=1+x+\frac{x^2}2+\dots+\frac{x^n}{n!} often appears in algebra textbooks as an illustration for using derivative to test for multiple roots.

Recently, I stumbled upon Example 2.1.6 in Prasolov’s book Polynomials (Springer, 2004), where it is shown that this polynomial is irreducible using Eisenstein’s criterion and Bertrand’s postulate. However, I do not think the argument is correct. Below you can find the argument presented in the book — I do not see how Eisenstein is applicable here, since we do not know p\mid n. And if we are using Eisenstein’s criterion directly to the polynomial n!f(x), this is one of the coefficients that would have to be divisible by p. (However, the argument works at least if n is prime.)

So my main question is about the irreducibility of the original polynomial, but I also wonder whether Prasolov’s proof can be corrected somehow. To summarize:

  • Is the polynomial f(x)=1+x+\frac{x^2}2+\dots+\frac{x^n}{n!} irreducible over \mathbb Q?
  • Is the Prasolov’s proof correct or can it be easily corrected? (Did I miss something there?)

Here is the (whole) Example 2.1.6 from Prasolov’s book. The same example is given in прасолов: многочлены(Prasolov: Mnogochleny; 2001,MCCME).

Example 2.1.6. For any positive integer n, the polynomial
is irreducible.

Proof: We have to prove that the polynomial
is irreducible over \mathbb Z. To this end, it suffices to find the prime p such that n!
is divisible by p but is not divisible by p^2, i.e., p \le n < 2p.

Let n = 2m or n = 2m + 1. Bertrand's postulate states that
there exists a prime p such that m < p \le 2m.

For n = 2m, the inequalities p \le n < 2p are obvious. For n = 2m + 1, we
obtain the inequalities p \le n-1 and n-1 < 2p. But in this case the number
n-1 is even, and hence the inequality n-1 < 2p implies n < 2p. It is also
clear that p \le n - 1 < n. \hspace{20pt}\square


"since we do not know p∣n."

I will take Keith's word for it that the argument, overall, is bogus.

However, please note that the condition Prasolov identifies as desired is

p | n! but not
p^2 | n!

If p \leq n, then
n! = 1 \cdot 2 \cdot \cdots (p-1) \cdot p \cdot (p+1) \cdots n
so indeed p | n!

If 2p \leq n, then
n! = 1 \cdot 2 \cdots (p-1) \cdot p \cdot (p+1) \cdots (2p-1) \cdot (2p) \cdot (2p+1) \cdots n
so p^2 | n!

If p \leq n < 2 p then n! is divisible by p but not by p^2

EDIT: it is clear from comments that Martin was considering the entire Eisenstein argument, meaning that it is not enough to know the behavior of the last coefficient n!. I did not have the Prasolov book in front of me and was late for an appointment, so I reacted strictly to the excerpt I saw on this site.

Source : Link , Question Author : Martin Sleziak , Answer Author : Cooperation

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