$\textbf{ Lemma:}$ A non constant primitive polynomial $f(x) \in \mathbb{Z}[x]$ is irreducible in $\mathbb{Q}[x]$ if and only if $f(x)$ is irreducible in $\mathbb{Z}[x]$.
I am reading a book in which it is given that $f(x)=x^4+2x^2+1$ is primitive in $\mathbb{Z}[x]$ and it has no roots in $\mathbb{Q}$ but it is reducible over $\mathbb{Z} $ as $f(x)=(x^2+1)(x^2+1)$.
My question is doen't it conradict this lemma? since irreducible over $\mathbb{Z}$ and irreducible over $\mathbb{Q}$ is the same thing for primitive polynomial.
Best Answer
There is no contradiction here. The polynomial $f(x)=x^4+2x^2+1$ is reducible both in $\mathbb{Q}[x]$ and in $\mathbb{Z}[x]$, since it can be factored as $(x^2+1)(x^2+1)$ in either ring. All that's going on is that a polynomial can be reducible without having any roots.