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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 $$f(x)=1+x+\frac{x^2}2+\dots+\frac{x^n}{n!}$$ is irreducible.

Proof: We have to prove that the polynomial $$n!f(x)=x^n+nx^{n-1}+n(n-1)x^{n-2}+\dots+n!$$ 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 $p \le n - 1 < n$. $\hspace{20pt}\square$

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The irreducibility of these polynomials is due to Schur and uses prime ideal factorization in number fields. The proof in Prasolov's book is obviously bogus (for general $n$). I wrote up a correct proof at math.uconn.edu/~kconrad/blurbs/gradnumthy/schurtheorem.pdf. (By the way, the publisher name you give looks weird. The abbreviation they use in English is MCCME, not MCMO -- Moscow Center for Continuous Mathematical Education) –  KCd Apr 12 '12 at 11:13
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Thanks a lot @KCd, I'll have a look at the proof from your link. (I'll probably need some time to go through it.) Perhaps you could consider posting your comment as an answer? –  Martin Sleziak Apr 12 '12 at 11:22

1 Answer 1

"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. I AM NOT WORTHY.

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Yes, I certainly agree with what you wrote, we need $p\mid n!$ and $p^2\nmid n!$. This is fine. But we also need other coefficients to be divisible by $p$, i.e. $p$ should divide $n$, $n(n-1)$ etc. (At least if we apply Eisenstein's criterion directly to $n!f(x)$, without any tricks.) –  Martin Sleziak Apr 12 '12 at 18:50
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@MartinSleziak, sure. I'm just saying that Prasolov says he needs something about $n!$ and arranges that one thing. That is the excerpt you copied in from Prasolov. If his condition is still not good enough, that is another story. –  Will Jagy Apr 12 '12 at 19:00
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That proof from Prasolov's book is in a section on the Eisenstein irreducibility criterion. Taking $n = 4$, his $4!f(x)$ is $x^4 + 4x^3 + 12x^2 + 24x + 24$. The only prime dividing $4!$ once but not twice is 3 and $4!f(x)$ is not Eisenstein with respect to 3. Prasolov never does any kind of additive translate, like replacing $x$ with $x+1$, so it sure seems like his proof is based on using Eisenstein's criterion directly on $n!f(x)$. The case $n=4$ shows this approach has problems (for nonprime $n$). –  KCd Apr 12 '12 at 19:28
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@KCd, Hi Keith. Makes sense to me. I probably would not have posted anything if Martin had written "Now, it is true that Prasolov's $n! \equiv 0 \pmod p$ while $n! \neq 0 \pmod {p^2},$ however Eisenstein's criterion also requires $p$ to divide these other numbers..." in which case we all, including beginning undergraduates, would be up to speed. –  Will Jagy Apr 12 '12 at 20:31
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You may also note my reputation comes mainly from set theory, and in particular axiom of choice related questions (both here and on MO). I often say that forcing with large cardinals and no choice is a rich ground for puns. –  Asaf Karagila Apr 12 '12 at 21:48

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