Is it true that a polynomial with integer coefficients is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$? I read this in "Polynomials" by Victor Prasolov and they also give a proof of this, but I had previously read in a book from my country that if a polnomial $f\in \mathbb{Z}[X]$ is irreducible over $\mathbb{Z}$ then it is irreducible over $\mathbb{Q}$(this is Gauss' lemma), but they went on to say that if $f\in \mathbb{Z}[X]$ is a primitive polynomial which is irreducible over $\mathbb{Q}$ only then is it also irreducible over $\mathbb{Z}$.
So, my question is : in the second case, does the polynomial really need to be primitive?
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$\begingroup$ The polynomial needs to be monic at least, as InsideOut's answer shows. $\endgroup$– TonyKCommented Apr 14, 2020 at 11:37
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2 Answers
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I am going to give you a counterexample of the statement in the title.
It is false.
For instance, look at the polynomial $p(x)=2x+2\in\Bbb Z[x]\subset \Bbb Q[x]$. If you consider $p(x)$ as a polynomial with integer coefficient, then it is reducible as $2(x+1)$ because $2$ is not a unit in $\Bbb Z$ - it is not invertible in $\Bbb Z$. If you look at $p(x)$ as a polynomial in $\Bbb Q[x]$ instead, then it is irreducible. Indeed, even if you write $2(x+1)$, $2$ is a unit in $\Bbb Q$.
answered Apr 14, 2020 at 11:28
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$\begingroup$ I don't think that the counterexample is valid, $2$ is a constant polynomial and irreducible means not being able to be factored into a product of two non-constant polynomials. $\endgroup$ Commented Apr 14, 2020 at 14:32
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1$\begingroup$ I suggest you to look at the problem from a general point of view. Given an integral domain $D$, what does it mean that an element is irreducible? An element is said to be irreducible if it is not a product of two non-units - look here en.wikipedia.org/wiki/Irreducible_element. In $\Bbb Z[x]$, both $2$ and $x+1$ are not unit, right? Then $2x+2$ is reducible. Your statement in the comment above applies when you consider polynomial with coefficients in some fields, but it fails when you consider the coefficients in some integral domain, like $\Bbb Z$ for instance. $\endgroup$ Commented Apr 14, 2020 at 15:33
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Prasolov uses a nonstandard definition of irreducibility over $\mathbb Z$ (page 48):
He goes on to discuss the content of a polynomial and then to state that "a polynomial with integer coefficients is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$". In the proof he assumes that the polynomial has content $1$.
