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 $\Z$.

$\sqrt[m]{2}$ is an Irrational Number
Prove that $\sqrt[m]{2}$ is an irrational number for any integer $m \geq 2$.
Hint.
Use ring theory:
Consider the polynomial $f(x)=x^m-2$.
Apply Eisenstein's criterion, show that $f(x)$ is irreducible over $\Q$.
Proof.
Consider the monic polynomial […]

The Ring $\Z[\sqrt{2}]$ is a Euclidean Domain
Prove that the ring of integers
\[\Z[\sqrt{2}]=\{a+b\sqrt{2} \mid a, b \in \Z\}\]
of the field $\Q(\sqrt{2})$ is a Euclidean Domain.
Proof.
First of all, it is clear that $\Z[\sqrt{2}]$ is an integral domain since it is contained in $\R$.
We use the […]

There is Exactly One Ring Homomorphism From the Ring of Integers to Any Ring
Let $\Z$ be the ring of integers and let $R$ be a ring with unity.
Determine all the ring homomorphisms from $\Z$ to $R$.
Definition.
Recall that if $A, B$ are rings with unity then a ring homomorphism $f: A \to B$ is a map […]

Irreducible Polynomial Over the Ring of Polynomials Over Integral Domain
Let $R$ be an integral domain and let $S=R[t]$ be the polynomial ring in $t$ over $R$. Let $n$ be a positive integer.
Prove that the polynomial
\[f(x)=x^n-t\]
in the ring $S[x]$ is irreducible in $S[x]$.
Proof.
Consider the principal ideal $(t)$ generated by $t$ […]

Polynomial $x^4-2x-1$ is Irreducible Over the Field of Rational Numbers $\Q$
Show that the polynomial
\[f(x)=x^4-2x-1\]
is irreducible over the field of rational numbers $\Q$.
Proof.
We use the fact that $f(x)$ is irreducible over $\Q$ if and only if $f(x+a)$ is irreducible for any $a\in \Q$.
We prove that the polynomial $f(x+1)$ is […]

Prove that $\F_3[x]/(x^2+1)$ is a Field and Find the Inverse Elements
Let $\F_3=\Zmod{3}$ be the finite field of order $3$.
Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$.
(a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field. How many elements does the field have?
(b) […]

Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$
Let $R$ be a commutative ring. Consider the polynomial ring $R[x,y]$ in two variables $x, y$.
Let $(x)$ be the principal ideal of $R[x,y]$ generated by $x$.
Prove that $R[x, y]/(x)$ is isomorphic to $R[y]$ as a ring.
Proof.
Define the map $\psi: R[x,y] \to […]