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It is well-known that a degree 2 or 3 polynomial over a field is reducible if and only if it has a root. But what about integral domains? Can we have a reducible polynomial over an integral domain having no roots in the domain?

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    Yes, I meant reducible.2012-10-21

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For example take $\mathbb{Z}$ and the polynomial $3(x^{2}+1)\in\mathbb{Z}[x]$

This polynomial have no roots in $\mathbb{Z}$ but $3\cdot(x^{2}+1)$ is a factorization

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    @ Belgi Oh I see.2012-10-21