Polynomial $(x-1)(x-2)\cdots (x-n)-1$ is Irreducible Over the Ring of Integers $\Z$

Problems and solutions of ring theory in abstract algebra

Problem 372

For each positive integer $n$, prove that the polynomial
\[(x-1)(x-2)\cdots (x-n)-1\] is irreducible over the ring of integers $\Z$.

 
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Proof.

Note that the given polynomial has degree $n$.
Suppose that the polynomial is reducible over $\Z$ and it decomposes as
\[(x-1)(x-2)\cdots (x-n)-1=f(x)g(x) \tag{*}\] for some polynomials $f(x)$ and $g(x)$ in $\Z[x]$ of degree less than $n$.

Evaluating at $x=1,2, \dots, n$, we obtain
\[f(k)g(k)=-1\] for $k=1, 2, \dots, n$.
Since $f(x)$ and $g(x)$ have integer coefficients, both $f(k)$ and $g(k)$ are integers for $k=1, 2, \dots, n$.
It follows that
\[f(k)=1=-g(k) \text{ or } f(k)=-1=-g(k). \tag{**}\]

Consider the polynomial $h(x):=f(x)+g(x)\in \Z[x]$.
Then we have
\begin{align*}
h(k)=f(k)+g(k)=0
\end{align*}
for $k=1, 2, \dots, n$ by (**).
Thus, the polynomial $h(x)$ has at least $n$ roots.

However, the degree of $h(x)$ is less than $n$ as so are the degrees of $f(x)$ and $g(x)$.
This forces that $h(x)=0$ for all $x$, and hence $f(x)=-g(x)$.

Then (*) becomes
\[(x-1)(x-2)\cdots (x-n)-1=-g(x)^2.\] The leading coefficient of the left hand side is $1$ but the leading coefficient of the right hand side is a negative number, hence it is a contradiction.

Therefore, the polynomial $(x-1)(x-2)\cdots (x-n)-1$ must be irreducible over $\Z$.


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