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Show that the polynomial $h(x)=(x-1)(x-2) \cdots (x-n)-1$ is irreducible in $\mathbb{Z}[x]$ for all $n \geq 1$.

This problem seems to be hard to solve. I thought I could use Eisenstein in developping this polynomial, but it is a bad idea. Another idea would be to suppose the existence of $f(x)$ and $g(x)$, and suppose that $f(x)g(x)=(x-1)(x-2) \cdots (x-n)-1$. In this direction, we could analyse the roots of $h(x)$ I guess.

Is anyone could help me to solve this problem?

user26857
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Taj Mohamed Bandalandabad
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    As others have observed, you are asking a great many questions about irreducibility. In principle, this isn't bad, but it seems like you are going through Dummit and Foote and asking for help on every irreducibility question. These questions aren't easy, so there's no shame in this, but if you are having this much difficulty answering these questions after receiving ample solutions to other similar questions, you might consider revisiting those questions to test your understanding as opposed to moving onto the next one. – Alex Wertheim Mar 10 '16 at 04:45
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    Not sure if this leads to a solution, but clearly you have $f(k) = \pm 1$ and $g(k) = \pm 1$ for quite a few integer values of $k$. And there's a limit to how many times a polynomial can take the same value without being constant. – David Mar 10 '16 at 05:25
  • And if you are going throught a wdiely used textbook you should most definitely search the site before asking :-( – Jyrki Lahtonen Mar 10 '16 at 07:28

2 Answers2

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Suppose you had a nontrivial factorization $h=fg$, where $f$ and $g$ are (monic) integral polynomials each of degree strictly less than $n$. Then for the integers $1\leq k\leq n$, $h(k)=f(k)g(k)=-1$, so $f(k)$ and $g(k)$ must be one of the values $\pm 1$, and necessarily $f(k)=-g(k)$ for each $1\leq k\leq n$.

Then the polynomial $p(x)=f(x)+g(x)$ has degree strictly less than $n$, but the integers $1\leq k\leq n$ are all roots of $p(x)$, so $p(x)\equiv 0$, so $f(x)=-g(x)$, a contradiction since their leading coefficients aren't equal.

Ben West
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  • why can't their leading coefficients be equal? – bmmcutet12 Jul 15 '17 at 18:19
  • @buchiplumanama $f(x)$ and $g(x)$ are monic, so $f(x)$ has leading coefficient $1$, but $-g(x)$ has leading coefficient $-1$. – Ben West Jul 15 '17 at 18:26
  • After getting $f(x)=-g(x)$, you can also argue by concluding that $h(x)=-f(x)^2$ , but this certainly not true as $h$ takes positive values too. – Mathronaut May 08 '18 at 18:58
  • Why can we conclude that $1 \le k < n$ are all roots of $p(x)$?. I don't see why you need this piece for the final conclusion either. Assuming the factorization exists is assuming $h$ is reducible. It follows that $f=-g$, but the leading coefficient of $h$ is 1, hence the contradiction. Not sure why we need to consider $p(x) = f(x) + g(x)$? – jeffery_the_wind May 12 '20 at 15:20
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    @jeffery_the_wind Introducing $p$ is a way to be explicit that $f(k)=-g(k)$ for some values $k$ actually implies $f=-g$. I don't think Bhauryal's comment is saying you don't need $p$, he's saying you can argue without considering leading coefficients. – Ben West May 12 '20 at 20:20
  • I finally got it, thank you. You are implicitly using the evaluation homomorphism (at $\alpha$) $\phi_\alpha: \mathbb{Z}[x] \to \mathbb{Z}$, $f(x) \mapsto f(\alpha)$, and using the properties $h(x)=f(x)g(x) \iff \phi(h(x)) = \phi(f(x)) \phi(g(x))$ and $\phi(a(x) + b(x)) = \phi(a(x)) + \phi(b(x))$. Perhaps it would be good to mention the existence of this homomorphism in the answer? – jeffery_the_wind May 13 '20 at 12:23
  • @jeffery_the_wind Not so much, the main point of introducing $p$ is to appeal to the fact that a nonzero polynomial in a single variable has at most $n$ roots, counting multiplicity, where $n$ is its degree. Since $p$ has more roots than its degree, it must be the zero polynomial, that is, $f+g=0$, so $f=-g$. – Ben West May 13 '20 at 15:37
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David's observation that if $f=gh$, then $g(k)=-h(k)=1$ or both $=-1$ for each of $k=1,2,\ldots , n$ is spot on. So both $g$ and $h$ take at least one of these values at least $n/2$ times. If we now take the polynomial of smaller degree (let's say it's $g$), so $\deg g=m\le n/2$, then the only way to avoid a constant $g$ would be $m=n/2$ (so we're done already if $n$ is odd) and a polynomial of the type $$ g = (x-k_1)(x-k_2) \ldots (x-k_m)+ 1 . $$ But we also have to have $g(j)=-1$ for $1\le j\le n$, $j\not= k_r$, and this clearly isn't working.