# Does irreducible over $\mathbb{R}$ imply irreducible over $\mathbb{Z}$?

The question is True or False: "If a polynomial $$f$$ has integer coefficients and is irreducible over $$\mathbb{R}$$, then it is irreducible over $$\mathbb{Z}$$."

The solution manual claims this is True but I think that has to be a mistake. $$2x^2+2$$ is irreducible over $$\mathbb{R}$$ but 2 is not a unit in $$\mathbb{Z}$$ so $$2(x^2+1)$$ would be consider a non trivial factorization over $$\mathbb{Z}$$ right?

• Reducibility of a polynomial in this case refers to breaking it down into a product of polynomials of lower degrees. Aug 10 at 11:43
• It's a bit sad that "Irreducible in the ring $\Bbb Z[x]$" and "Irreducible polynomial over $\Bbb Z$" don't necessarily mean the same thing. But such is life, unfortunately. Aug 10 at 12:11
• Recall that an element $r$ of a unique factorization domain (UFD) is irreducible if and only if the equation $r = st$ implies that $s$ is a unit or $t$ is a unit. But the polynomial $2x^2 + 2 = 2(x^2 + 1)$ is not irreducible in $\mathbb Z[x]$ because neither $2$ nor $x^2 + 1$ is a unit in $\mathbb Z[x].$ Aug 10 at 17:43
• @Arthur Wait, what? Aug 11 at 0:39
• @Dylan r should be non-zero and a non-unit Aug 11 at 0:41