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I was going through the book of Gallian, Contemporary Abstract Algebra, and got the following result:

The polynomial $f(x) = 2x^2 + 4$ is irreducible over $\mathbb{Q}$ but reducible over $\mathbb{Z}$, since $2x^2 + 4 = 2(x^2 + 2)$ and neither $2$ nor $x^2 + 2$ is a unit in $\mathbb{Z}[x]$.

I am not getting how the polynomial is reducible over $\mathbb Z$. Can anyone explain this point? Thanks for the help.

Mathmo123
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1 Answers1

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The element $2$ is a unit in $\mathbb Q[x]$ (it has inverse $\frac 12$) but not in $\mathbb Z[x]$ (since $\frac 12\notin \mathbb Z[x]$). We can write $$f(x) = 2(x^2+2).$$Since a polynomial $f$ is reducible iff it can be written as the product of two non-units, this means that $f$ is reducible in $\mathbb Z[x]$. However, this factorisation does not show that $f$ is reducible in $\mathbb Q[x]$ since $2$ is a unit. One can show in other ways (e.g. because $f$ has no rational roots) that $f$ is irreducible over $\mathbb Q$.

This condition is not arbitrary: it means that the ideal $(2x^2+4)$ is prime in $\mathbb Q[x]$ (and is equal to the maximal ideal ($x^2+2)$), but it is not prime in $\mathbb Z[x]$, since $2x^2+4\in(2x^2+4)$, but $$2,x^2+2\notin (2x^2+4).$$

Mathmo123
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  • This seems strange to me, as is stated here: http://mathonline.wikidot.com/reducible-and-irreducible-polynomials-over-a-field . The degree of the products should be less to the original function. Here only a constant is factored out, which leaves the other part with still a degree of 2. Why is it not irreducible over Z[x], because of the degree problem? –  Jul 14 '19 at 13:49
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    @user3892683 That result only holds for polynomials over a field! $2$ may be just another constant in $\mathbb Q[X]$, but that's because $2$ is invertible in $\mathbb Q[X]$. – Mathmo123 Jul 14 '19 at 16:24