# A polynomial with integer coefficients is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$

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?

• The polynomial needs to be monic at least, as InsideOut's answer shows. Commented Apr 14, 2020 at 11:37

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 $$\Bbb Q$$.
• 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. Commented Apr 14, 2020 at 14:32
• 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. Commented Apr 14, 2020 at 15:33
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 $$1$$.