# Roots of polynomial irreducible over the rationals

If a polynomial is "irreducible over the rationals", does it mean that it has no rational roots?

I would say yes because otherwise I could divide out the linear factors (i.e. rational roots) but maybe I'm wrong?

• If $f$ is irreducible over $\Bbb Q$ of degree $n\ge 2$, then it has no rational root, yes. The converse is not true in general. – Dietrich Burde Jan 15 at 14:37

For example, the polynomial $$X-\frac23$$ is irreducible and yet it has a rational root. However, this (i.e., all linear polynomials) is the only exception, as otherwise your correct reasoning applies.
Note however, that the converse is not true: A polynomial may be recucible even if it does not have rational roots; consider e.g., $$X^4-5X+6=(X^2-2)(X^2-2)$$
A polynomial over the rationals is a polynomial $$f(x)\in{\Bbb Q}[x]$$. If the polynomial is linear, $$f(x)=ax+b$$, $$a\ne 0$$, it is irreducible over the rationals. If the polynomial has degree $$\geq 2$$, it is irreducible over the rationals provided that it has no root in the rationals.