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I need to prove that the truncated expression of the Taylor development of the exponential function is irreducible.

$ 1 + \frac{1}{1!}x + \frac{1}{2!}x^2+... + \frac{1}{p!}x^p \in \mathbb{Q}[x]$

where p is prime. Searching for answers I have found some proofs that use Galois theory, but that is very complex for me to understand. I know about Eisenstein's and reduction criterion.

Eisenstein's criterion

Suppose we have the following polynomial with integer coefficients. $ {\displaystyle Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}} $

$ {\displaystyle Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}} $

If there exists a prime number p such that the following three conditions all apply:

  • $p$ divides each $a_i$ for $0 ≤ i < n$,
  • $p$ does not divide $a_n$, and
  • $p^2$ does not divide $a_0$,

Reduction criterion (converse of reduction criterion?)

Any help would be appreciated. Thank you :).

shaymin shaymin
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1 Answers1

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Let us denote $$P(x)=1 + \frac{1}{1!}x + \frac{1}{2!}x^2+ \dots + \frac{1}{p!}x^p.$$ Then we have that

$$p!P(x)= p!+ p!x + \frac{p!}{2!}x^2+ \dots + p(p-1)x^{p-2}+px^{p-1}+x^p$$ satisfies the Eisenstein's criterion, from where it follows that $p!P(x)$ is irreducible. And thus $P(x)$ is irreducible.

mfl
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