Exercise 11.3

Proof. Suppose that $E$ and $E^{\prime }$ are two fields containing $F$ both with $q^s$ elements. We first show that there is a isomorphism $\sigma :E\to E^{\prime }$ which fixes the elements of $F$, by showing that that both $E$ and $E^{\prime }$ are isomorphic over $F$ to $F[x]∕(f(x))$ for some irreducible polynomial $f(x)\in F(x)$.

There is a primitive element ${\alpha }^{\prime }\in E^{\prime }$, i.e. such that $E^{\prime }=F({\alpha }^{\prime })$. For example, take ${\alpha }^{\prime }$ to be a primitive $q^s-1$ root of unity : since $\alpha$ is a generator of ${E{}^{\prime }}^{\text{∗}}$, every element $\gamma \in {E{}^{\prime }}^{\text{∗}}$ is equal to ${\alpha {}^{\prime }}^k$ for some integer $k$, thus $\gamma \in F({\alpha }^{\prime })$ (and $0\in F({\alpha }^{\prime })$). This proves $E^{\prime }\subset F({\alpha }^{\prime })$, and since ${\alpha }^{\prime }\in E^{\prime }$ and $F\subset E^{\prime }$, $F({\alpha }^{\prime })\subset E^{\prime }$, so $E^{\prime }=F({\alpha }^{\prime })$.

Let $f(x)\in F[x]$ be the minimal polynomial of ${\alpha }^{\prime }$ over $F$. Then

where the isomorphism ${\sigma }_1:F({\alpha }^{\prime })\to F(x)∕(f(x))$ maps ${\alpha }^{\prime }$ to $\bar{x}=x+(f(x))$, and maps $a\in F$ on $\bar{a}=a+(f(a))$. Since ${\alpha }^{\prime }$ is a root of $x^{q^s}-x$, $f(x)\mid x^{q^s}-x$.

$E$ is a field with $q^s$ elements, so we have $x^{q^s}-x={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in E}(x-\alpha )$. Thus $f(x)\mid {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in E}(x-\alpha )$, where $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f(x))=s\ge 1$, so $f(\alpha )=0$ for some $\alpha \in E$. The polynomial $f$ being irreducible over $F$, $f$ is the minimal polynomial of $\alpha$ over $F$, thus $F(\alpha )\simeq F[x]∕(f(x))$ is a field with $q^s$ elements. Since $F(\alpha )\subset E$, and $|F(\alpha )|=|E|$, we conclude $E=F(\alpha )$, therefore

where the isomorphism ${\sigma }_2:F(\alpha )\to F(x)∕(f(x))$ maps $\alpha$ to $\bar{x}=x+(f(x))$, and maps $a\in F$ on $\bar{a}=a+(f(a))$.

Then $\sigma ={\sigma }_1^{-1}\circ {\sigma }_2:E\to E^{\prime }$ is an isomorphism, and $\sigma (a)=a$ for all $a\in F$.

We can now use the isomorphism $\sigma$ to induce a map

Then $\bar{\sigma }$ is injective: if $[\sigma ({\alpha }_0),\dots ,\sigma ({\alpha }_n)]=[\sigma ({\beta }_0),\dots ,\sigma ({\beta }_n)]$, then there is $\lambda \in F^{\text{∗}}$ such that $\sigma ({\beta }_i)=\mathit{\lambda \sigma }({\alpha }_i)=\sigma (\lambda )\sigma ({\alpha }_i)=\sigma (\lambda {\alpha }_i),i=0,\dots ,n$, thus ${\beta }_i=\lambda {\alpha }_i$, which proves $[{\alpha }_0,\dots ,{\alpha }_n]=[{\beta }_0,\dots ,{\beta }_n]$.

If $[{\gamma }_0,\dots ,{\gamma }_n]$ is any projective point of $P^n(E^{\prime })$, then

This proves that $\bar{\sigma }$ is surjective. So $\bar{\sigma }$ is a bijection.

Now take $f(y_0,\dots ,y_n)\in F[y_0,\dots ,y_n]$ an homogeneous polynomial, ${\bar{H}}_f(E)$ the corresponding projective hypersurface in $P^n(E)$, and ${\bar{H}}_f(E^{\prime })$ the corresponding projective hypersurface in $P^n(E^{\prime })$. We show that $\bar{\sigma }({\bar{H}}_f(E))={\bar{H}}_f(E^{\prime })$.

Since $\sigma$ is a $F$-isomorphism, $\sigma (f({\alpha }_0,\dots ,{\alpha }_n))=f(\sigma ({\alpha }_0),\dots ,\sigma ({\alpha }_n))({\alpha }_i\in E)$, and similarly ${\sigma }^{-1}(f({\beta }_0,\dots ,{\beta }_n))=f({\sigma }^{-1}({\beta }_0),\dots ,{\sigma }^{-1}({\beta }_n))({\beta }_i\in E^{\prime })$, thus

This shows $\bar{\sigma }({\bar{H}}_f(E))\subset {\bar{H}}_f(E^{\prime })$.

Conversely,

If we define ${\alpha }_i={\sigma }^{-1}({\beta }_i),i=0,\dots ,n$, then $[{\alpha }_0,\dots ,{\alpha }_n]\in {\bar{H}}_f(E)$, and $[{\beta }_0,\dots ,{\beta }_n]=\bar{\sigma }([{\alpha }_0,\dots ,{\alpha }_n])\in \bar{\sigma }({\bar{H}}_f(E))$. This shows ${\bar{H}}_f(E^{\prime })\subset \bar{\sigma }({\bar{H}}_f(E))$, and so

Since $\bar{\sigma }$ is a bijection,

So $N_s$ is independent of the choice of the extension $F_s={𝔽}_{q^s}$ of $F={𝔽}_q$. □