Section 4.9 Polynomial reducible over every \(\mathbb{F}_p\)
We show that \(x^4+1\) is reducible over every \(\mathbb{F}_p\text{,}\) where \(p\) is an odd prime number. The case when \(p=2\) is left to the reader.
Proof.
\begin{align*}
-1\amp\equiv (p-1)!\\
\amp\equiv \left(1\cdot 2\cdots\frac{p-1}{2}\right)\left((-1)\cdot (-2)\cdots\left(-\frac{p-1}{2}\right)\right)\\
\amp\equiv (-1)^{\frac{p-1}{2}}\left(1\cdot 2\cdots\frac{p-1}{2}\right)^2\\
\amp\equiv \left(1\cdot 2\cdots\frac{p-1}{2}\right)^2
\end{align*}
Lemma 4.9.2.
Suppose that \(p\) is an odd prime number. If \(a,b\in\mathbb{F}_p^{\times}\) are such that neither \(a\) nor \(b\) is a square in \(\mathbb{F}_p\) then \(ab\in\mathbb{F}_p^{\times2}\)Proof.
\begin{equation*}
\varphi\colon\mathbb{F}_p^\times\to\mathbb{F}_p^\times\quad\text{given by}\quad x\mapsto x^2.
\end{equation*}
The kernel is \(\{\pm 1\}\text{.}\) Therefore, \(|{\rm Im}(\varphi)|=\tfrac{p-1}{2}\text{,}\) i.e., \([\mathbb{F}_p^\times:{\rm Im}(\varphi)]=2\text{.}\) This implies that \(a\,{\rm Im}(\varphi)\cdot b\,{\rm Im}(\varphi)=ab\,{\rm Im}(\varphi)={\rm Im}(\varphi)\text{,}\) i.e., \(ab\in{\rm Im}(\varphi)\text{,}\) i.e., \(ab\in\mathbb{F}_p^{\times 2}\text{.}\)Proposition 4.9.3.
If \(p\equiv -1\mod 4\) then \(x^4+1\) is reducible over \(\mathbb{F}_p\text{.}\)Proof.
\begin{equation*}
x^4+1=x^4-a^2=(x^2-a)(x^2+a).
\end{equation*}
We make some computations.
\begin{align*}
x^4+1\amp=(x^2+ax+b)(x^2+cx+d)\\
\amp=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd
\end{align*}
Hence
\begin{equation*}
a=-c;\;ac+b+d=0;\;ad+bc=0;\;bd=1.
\end{equation*}
Therefore, \(b=d=\pm 1\text{,}\) \(c=-a\text{,}\) and \(a^2=\pm 2\text{.}\) So, if there is a factorization of \(x^4+1\) into quadratic polynomials then we necessarily have
\begin{equation}
x^4+1=(x^2+ax\pm 1)(x^2-ax\pm 1)\quad\text{with}\;b=d=\pm 1, c=-a,\text{ and }a^2=\pm2\tag{4.9.1}
\end{equation}
