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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?

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  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Jan 15 at 14:37
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Almost.

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., $X^4-5X+6=(X^2-2)(X^2-2)$

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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.

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