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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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"Can we have an irreducible polynomial over an integral domain having no roots in the domain?" do you mean reducible ? – Belgi Oct 21 '12 at 13:28
Yes, I meant reducible. – Reader Oct 21 '12 at 13:35
up vote 1 down vote accepted

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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Thanks. But shouldnot factorization be in lower degree than 2? – Reader Oct 21 '12 at 13:44
@Reader - why ? I recall that the definiton only requires that both polynimials are $\neq 0$ and that they are not invertible – Belgi Oct 21 '12 at 13:45
A polynomial $f (x) \in F [x]$ is called irreducible over $F$ if $deg(f) > 0$ and if its only factors are $c$ and $cf(x)$, where $c \in F$, $c\neq 0$, is any non-zero constant. – Reader Oct 21 '12 at 13:52
@Reader - What is $F$ ? – Belgi Oct 21 '12 at 13:53
@Reader: If you don't like the example, look at $4x^2-4x+1=(2x-1)(2x-1)$. Certainly reducible over $\mathbb{Z}$, but no integer zeros. – André Nicolas Oct 21 '12 at 18:52

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