<s>[INST] Show a polynomial is reducible over $F=\mathbb{Z}/(p)$ with $p$ prime and $p=a^2+b^2$ I have tried to solve the following question:
Recall that $\mathbb{Z}$ denotes the ring of integers, and $\mathbb{Q}$ is the rationals. We let $F[x]$ denote the polynomial ring in the variable $x$ over a field $F$, and we let $f(x)=x^2+1\in F[x]$
Assume that the integer $p$ is a prime of the form $p=a^2+b^2$. Let $F=\mathbb{Z} /(p)$. Show that $f(x)$ is reducible. (Fact: if $p \equiv 1$ modulo 4, then $p=a^2+b^2)$.
Solution :
If we take $a\in F^*=F\setminus \{0\}$ and $a^2+b^2=0$ in $F$, then $a^{-1}\in F^*$ exist. So :
\begin{align}
    a^2+b^2 &= 0\nonumber\\
    b^2 &=-a^2 \nonumber\\
    b^2 &=-\Big(\frac{1}{a^{-1}} \Big)^2 \nonumber  \\
    a^{-2}b^2 &= -1 \label{eq:2.3} \\
    a^{-2}b^2+1 &= 0  \nonumber  
\end{align}
Since $f(x)=x^2+1$ is in $F[x]$, we have to show that $f(x)$ is reducible. If we put $1=-a^{-2}b^2$ in $f(x)$ we get
$$ f(x)=x^2+1=x^2-(a^{-1}b)^2=(x-a^{-1}b)(x+a^{-1}b)$$
As such we have expressed $f(x)$ as factors of degree lower than that of said function, hence $f(x)$ is reducible.
Is this answer correct, and if so. Would this answer be sufficient?
 [/INST] $a\in F^*$ and $a^2+b^2=0$ in $F$, therefore $b^2a^{-2}+1=0$. Hence $x^2+1=(x-ba^{-1})(x+ba^{-1})$
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