I'm reading Broker's paper "CONSTRUCTING SUPERSINGULAR ELLIPTIC CURVES", which gives an algorithm of constructing supersingular elliptic curves over $F_{p}$ (where $p$ is a prime number).
Theorem 2.1. [Deuring's lifting theorem] Let $E$ be an elliptic curve defined over a number field $L$ whose endomorphism ring is the maximal order $\mathcal{O}_K$ in an imaginary quadratic field $K$, and let $\mathfrak{p} \mid p$ be a prime of $L$ where $E$ has good reduction. Then $E$ mod $\mathfrak{p}$ is supersingular if and only if $p$ does not split in $K$ using Deuring's lifting theorem.
"Let $E$ be an elliptic curve as in the Theorem 2.1, and $P_K$ be a Hilbert class polynomial (so its degree equals the class number $h_K$ of $K$). If we now take $K$ such that $p$ remains inert in $\mathcal{O}_K$, then the roots of $P_K \in \overline{F}_p[x]$ are $j$-invariants of supersingular curves. Since the $j$-invariant of a supersingular curve is contained in $F_{p^2}$, the polynomial $P_K$ splits over $F_{p^2}$."
I understand up to this part. In the next paragraph he says
"If we take $K$ so that the class number $h_k$ is odd, then $P_K \in F_{p}[x]$ has a root in $F_{p}$."
Why does odd degree imply $P_K$ has a root in $F_{p}$? I know that polynomial of odd degree over real numbers must have a real root, but I'm not sure if a similar statement holds for general field, or at least for $F_{p}$.
Best Answer
$$P_K(x)=\prod_j f_j(x)^{e_j}\in \Bbb{F}_{p}[x]$$ Let $E$ be the splitting field. Because every finite extension of $\Bbb{F}_p$ is Galois we get that $$[E:\Bbb{F}_p]=lcm(f_1,\ldots,f_J)$$ Thus $E\subset \Bbb{F}_{p^2}$ tells us that $\deg(f_j)\le 2$.
That $\deg(P_K)$ is odd tells us that not all $\deg(f_j)$ is even, ie. some $\deg(f_i)=1$ and $P_K$ has a root.