# Question about the solution of the polynomial $(x−1)(x−2)⋯(x−n)−1$ is irreducible in $\mathbb{Z}\left [ x \right ]$ for all $n≥1$

The solution of the polynomial $$(x−1)(x−2)⋯(x−n)−1$$ is irreducible in $$\mathbb{Z}\left [ x \right ]$$ for all $$n≥1$$ is in here.

I think it's no problem that do the same thing on $$\mathbb{Q\left [ x \right ] }$$ and $$\mathbb{R\left [ x \right ]}$$ and get the same result that the polynomial $$(x−1)(x−2)⋯(x−n)−1$$ is irreducible.

But my question is on $$\mathbb{R\left [ x \right ]}$$, the polynomial is continuous in $$\mathbb{R}$$, let $$f(x):=(x−1)(x−2)⋯(x−n)−1$$ Obviously,$$f(1)=-1<0;\;f(n+1)=n!-1>0$$, so there is a solution between $$0$$ and $$(n+1)$$ which means the polynomial is reducible. There's a contradiction, so which part is wrong. Thank you!

• The irreducible polynomials in $\mathbb{R}[x]$ are of degree $1$ or $2$. Once you have a real root $r$ there is a factor $x-r$. If there is a complex root $r$, then since the coefficients are real the complex conjugate $\overline{r}$ is also a root. Therefore, the polynomial is divisible by $x^2-(r+\overline{r})x+|r|^2$
– NDB
Commented Jul 15, 2023 at 3:23
• For the irreducibility in $\mathbb{Q}[x]$ you can deduce it from that in $\mathbb{Z}[x]$ using Gauss' lemma
– NDB
Commented Jul 15, 2023 at 3:26
• @NDB Thank you! I completely understand now :) Commented Jul 15, 2023 at 3:36

## 1 Answer

The logic doesn't generalize even to $$\mathbb{Q}[x]$$.

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 $$\pm 1$$

What if $$f(k) = \frac{1}{g(k)}$$? This would be entirely permissible in $$\mathbb{Q}$$ since $$k$$ is rational and hence $$g(k)$$ is too, if $$g \in \mathbb{Q}[x]$$.

Of course, $$1/n$$ is a non-integer for integers $$n$$ (except $$\pm 1$$). Since $$g \in \mathbb{Z}[x] \implies g(k) \in \mathbb{Z}$$ (or more broadly, the evaluation of a polynomial over a ring, at an element in that ring, gives an element in that ring) this where they draw their conclusion in the linked post: it is clear the product must evaluate to $$-1$$, and the factors must be integers since they are polynomials over $$\mathbb{Z}$$ too.

Of course, then, this need not hold for polynomials over $$\mathbb{Q}$$.

Some playing around in WolframAlpha does suggest irreducibility over $$\mathbb{Q}$$ regardless for $$n \ge 2$$, where nontrivial cases are concerned. My instinct is suggesting perhaps the rational root theorem may help in proving that irreducibility holds (i.e. prove that $$n!-1$$ is not a root for any $$n$$).

It does not hold over $$\mathbb{R}[x]$$, as you have seen.

• See Gauss' lemma
– NDB
Commented Jul 15, 2023 at 3:21
• I think your idea only holds over R[x], because in Gauss's lemma — A non-constant polynomial in Z[X] is irreducible in Z[X] if and only if it is both irreducible in Q[X] and primitive in Z[X]. So the polynomial is irreducible in Q[x] since it is irreducible in Z[x] Commented Jul 15, 2023 at 3:30
• @Zhiwei: Your statement is not quite right; you need the polynomial to be non-constant and primitive (the gcd of the coefficients is $1$); but your polynomial is monic, and therefore it is necessarily primitive. Commented Jul 15, 2023 at 5:58