# Example such that $f_1(x)$ is reducible but $f(x)$ is irreducible.

(The (mod p) Irreducibility Test)

Let $p$ be a prime an suppose that $f(x) \in \mathbb Z[x]$ with $\deg f(x) \geq 1$. Let $f_1(x)$ be the polynomial in $\mathbb Z_p[x]$ obtained from $f(x)$ by reducing all the coefficients of $f(x)$ modulo $p$. If $f_1(x)$ is irreducible over $\mathbb Z_p$ and $\deg f_1(x)=\deg f(x)$, then $f(x)$ is irreducible over $\mathbb Q$.

I know that if $f_1(x)$ is reducible , then $f(x)$ can be reducible or irreducible. I want example such that $f_1(x)$ is reducible but $f(x)$ is irreducible.

You have $(x+1)^2=x^2+2x+3$ over $\mathbb F_2$, but $x^2+2x+3$ is irreducible over $\mathbb Q$: It has no Zero.
The polynomial $x^4+1$ is irreducible in $\Bbb Q$ but it's reducible in every $\Bbb Z_p$ for any prime $p$, for instance in $\Bbb Z_2$ we have: $$x^4+1=x^4-2x^2+1=(x^2-1)^2$$