Why is it that $x^4+2x^2+1$ is reducible in $\mathbb{Z}[x]$ but has no roots in $\mathbb{Q}$?

Question

$\textbf{ Lemma:}$ 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.

Answer 1

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.

Answer 2

It is not necessary that a polynomial is reducible over a field iff it has a root in field. By meaning of a reducible polynomial is that we express the polynomial into product of non constant polynomials over the field. And a result that a polynomial having degree 2 or 3 is reducible iff it has root in field. In your example degree of polynomial is 4.