Let $K$ be a field. We consider the polynomial $f(x) = x^4 + 1$. It is known that $f(x)$ is irreducible over $\mathbb{Q}$ but reducible over any finite field. Thus $ f(x)$ is reducible over any field $ K $ having positive characteristic.


I am interested in determining for which fields $f(x)$ remains irreducible, based on the case $\text{char}(K) = 0$.



irreducibility of the polynomial $ x^4 +1 $
Among $-1$, $2$, and $-2$, there are $0$, $1$ or $3$ squares. This determines the irreducible factorization of $X^4+1$.


1. If none is a square, it is irreducible;
2. If only $-1=i^2$ is a square, the irreducible factorization is $(X^2-i)(X^2+i)$;
3. If only $-2=u^2$ is a square, the irreducible factorization is $(X^2+uX-1)(X^2-uX-1)$;
4. If only $2=v^2$ is a square, the irreducible factorization is $(X^2+vX+1)(X^2-vX+1)$;
5. If all are squares and char is $\neq 2$, the polynomial is split, with simple roots $i^k(1+i)/v$, $k=0,1,2,3$;
6. If the char is 2 (so all are squares), the polynomial is $(X-1)^4$.




---


Added: here's a proof. The char. 2 case (Case 6) being trivial, I assume $2\neq 0$. First, if there is a root $\theta$ in the field, then the polynomial splits with distinct roots $\theta^k$, $k=\pm 1,\pm 3$ (with $\theta^{-3}=-\theta$). In particular, if the polynomial is not irreducible, it is product of two degree 2 polynomials.


If $\theta$ is a root (possibly in a larger field) and the polynomial is not irreducible, the polynomial therefore splits with one of the following factors: $A=(X-\theta)(X-\theta^{-1})$, $B=(X-\theta)(X-\theta^3)$, $C=(X-\theta)(X+\theta)$. We observe that $A=X^2+vX+1$ with $v^2=2$, $B=X^2+uX-1$, $C=X^2-i$ with $v^2=2$, $u^2=-2$, $i^2=-1$. Hence if $2$ resp $1$ resp $-1$ is a square then $A$ resp $B$ resp $C$ yields a factorization in the ground field. If all are squares, $(1+i)/u$ is a root and the polynomial is totally split.


Conversely if $-2$ has no square root then $A$ is not a polynomial in the ground field and hence the polynomial is not totally split, etc, and if none of the three is a square, none of the possible degree two factors occurs and hence the polynomial is irreducible. Thus all of 1, 2, 3, 4, 5 are proved.


NB in char. $3$, we have $2=-1$ but this is no contradiction: just Cases 2 and 4 are excluded (and 1 is excluded as well). But actually in char $p>0$, for given $p$, at most one of 2, 3, 4 is possible and 1 is excluded, so char. $3$ is not really special.


