## Reducible Polynomial for every prime p

In an algebra course, whenever we are asked to prove that a certain monic polynomial is irreducible in $\mathbb{Q}[x]$, it is a useful to trick to reduce the coefficients of this polynomial modulo $p$ for some prime $p$ and show that the resulting polynomial is irreducible in $\mathbb{F}_{p}[x]$ (the unique field with $p$ with elements). Proof of this result can be found in any standard introductory algebra course. Here is an application. If we are asked to show the irreducibility of $f(x)=x^3+3x^2+8x+1225$ in $\mathbb{Q}[x]$, we could reduce the coefficients mod $p$ to get the polynomial $\overline{f}(x)=x^3-x+1$ in $\mathbb{F}_{3}[x]$. Now we can easily show that $\overline{f}(x)$ is irreducible in $\mathbb{F}_{3}[x]$. Since $\overline{f}(x)$ is cubic, it suffices to show that $\overline{f}(x)$ has no root in $\mathbb{F}_{3}[x]$. This is clear, because $\overline{f}(x)=1$ for $x=0,1,2$. This shows that $\overline{f}(x)$ is irreducible in $\mathbb{F}_{3}[x]$, and as a result $f(x)$ is irreducible in $\mathbb{Q}[x]$, as desired.

What if we reduced coefficients of $f(x)$ mod 5? Then our reduced polynomial would be $\overline{f}(x)=f(x)=x^3+3x^2+3x$ in $\mathbb{F}_{5}[x]$. But wait! This polynomial is now reducible, namely $\overline{f}(x)=f(x)=x(x^2+3x+3)$ in $\mathbb{F}_{5}[x]$. Does this mean that our original function $f(x)$ is also reducible in $\mathbb{Q}[x]$? Well, of course not, for we just proved above that $f(x)$ is irreducible in $\mathbb{Q}[x]$. Moral of the story: if, for some prime $p$, reducing coefficients of a polynomial $f(x)$ modulo $p$ renders it reducible in $\mathbb{F}_{p}[x]$, we cannot conclude that polynomial $f(x)$ is reducible in $\mathbb{Q}[x]$.

Things are about to get more interesting. There exist some polynomials (and I will give an example) that are irreducible in $\mathbb{Q}[x]$ but reducible in $\mathbb{F}_p[x]$ for every prime $p$. Basically, these special polynomials fully destroy any possible converse for the theorem “if $f(x)$ is irreducible in $\mathbb{F}_p[x]$, then it is irreducible in $\mathbb{Q}[x]$.” One such polynomial is $g(x)=x^4+1$. First, let’s show that $g(x)$ is irreducible in $\mathbb{Q}[x]$. It suffices to show $g(x+1)$ is irreducible in $\mathbb{Q}[x]$. But we have

$\displaystyle g(x+1)=x^4+4x^3+6x^2+4x+2$

which is irreducible in $\mathbb{Q}[x]$ by Eisenstein’s Criterion for prime $2$. I will hopefully write about Eisenstein’s Criterion in one of the future blog posts. So we have established that $g(x)$ is irreducible in $\mathbb{Q}[x]$.

It remains to show that $g(x)=x^4+1$ is reducible for every prime $p$. Rotman’s Advanced Modern Algebra has two different proofs of this fact. First one relies on the following argument. For $p=2$ it is clear that $g(x)=x^4+1$ is reducible, since it has a root in $\mathbb{F}_p$. Now, for odd prime $p$, one can show that $x^4+1$ can be written as a product of two quadratic polynomials in $\mathbb{F}_p[x]$. To find coefficients of these polynomials, one simply writes down two arbitrary monic quadratics, and multiplies it out, and the result must equal to $x^4+1$. The coefficients can then be analyzed, and some very special cases of quadratic reciprocity allows one to conclude that such two quadratics exist.

Second approach given in the book is the use of field extensions, and elementary property of cyclic groups. Here is the full proof:

Proposition. The polynomial $x^4+1$ is reducible in $\mathbb{F}_p[x]$ for every prime $p$.
Proof. If $p=2$, then we have $x^4+1=(x+1)^4$, and we are done. Let $p$ be an odd prime. Then, it is easy to check that $p^2\equiv 1 \ (\textrm{mod} \ 8)$. Thus, $8\mid p^2-1$. If $\mathbb{F}_{p^2}^\times$ denotes the group of units for the unique finite field containing $p^2$ elements, then it is easy to see that $|\mathbb{F}_{p^2}^\times|=p^2-1$, and so $8\mid |\mathbb{F}_{p^2}^\times|$. By a well known theorem $\mathbb{F}_{p^2}^\times$ is cyclic. Thus, the group $\mathbb{F}_{p^2}^\times$ contains a cyclic group of order $8$, and so it contains all $8$-th roots of unity; in particular, all the roots of $x^4+1$ are contained in $\mathbb{F}_{p^2}$ (Here we use the fact that 8-th cyclotomic polynomial is $x^4+1$). Thus, the splitting field of $x^4+1$ over $\mathbb{F}_{p}$ is
$\mathbb{F}_{p^2}$. Now, $[\mathbb{F}_{p^2}:\mathbb{F}_{p}]=2$. So if $x^4+1$ were irreducible in $\mathbb{F}_{p}$, then would get $4\mid [\mathbb{F}_{p^2}:\mathbb{F}_{p}]$, contradiction.