# Why is $p_{3}(x) = 9x^{2}-3=3(x\sqrt{3}-1)(x\sqrt{3}+1)$ reducible over integers?

I am learning Irreducibility of polynomials, and I am reading this entry of Wiki. However, I am confused why the polynomial $$p_{3}(x) = 9x^{2}-3=3(x\sqrt{3}-1)(x\sqrt{3}+1)$$ is reducible over integers. I though it falls right into the defintion of irreducibility(over integer), because it could only be written into two polynomials of strictly lower degree with coefficients not being integers. The entry explains that it is reducible because the factor 3 is not invertible in integers, but I just can't see how this makes the polynomial reducible. I am very thankful if someone could help explain it.

• Recall when is a polynomial in $\mathbb Z[X]$ reducible in $\mathbb Z [X]$ – Soumik Mar 10 '19 at 7:54

Well, the polynomial $$f(x)=3(x^2-1)$$ is reducible over the integers, since it factors into a product of non-units in $${\Bbb Z}[x]$$. The units in $${\Bbb Z}[x]$$ coincide with the units in $${\Bbb Z}$$ and are $$\pm 1$$.
The definition says that a polynomial $$f\in R[x]$$ is irreducible if it can only be factored as $$f=gh$$ with $$g,h\in R[x]$$ such that one of $$g,h$$ is a unit in $$R[x]$$. The units in $$R[X]$$ are the units in $$R$$.
• Your argument is correct, but you should have applied it to $p_3(x) = 3(3x^2-1)$. – Paul Frost Mar 10 '19 at 9:03