Problem: Let $F$ be a finite field
- Show that characteristic of $F$ is a prime number $p>0$ and $|F| = p^n, n \in \mathbb{Z}^+$.
- Show that $F$ is a Galois extension on $\mathbb{Z}_p$.
- Show that for each $a \in F$ then $a = x^2 + y^2$ with $x,y \in F$.
My attempt:
- $F$ is a finite field implies $\text{char} (F)$ and $|F| = p^n$. If $\text{char} (F) = 0$ then $\mathbb{Q}$ is the prime subfield of $F$. Then $F$ infinite. Hence $\text{char} (F) = p>0$ and $\mathbb{Z}_p$ is the prime subfield of $F$. Consider the vector space $F$ on $\mathbb{Z}_p$. We have $\dim_{\mathbb{Z}_p} F = n = \dim_{\mathbb{Z}_p} \mathbb{Z}_p^n$. Hence, $F \cong \mathbb{Z}_p \times \dots \times \mathbb{Z}_p \Rightarrow |F| = p^n$
I need help for the question number $2$ and $3$. Thank all!
If $\mathbb{F}$ is a field with $p^n$, then try to show it is the splitting field of $x^{p^{n}}-x$ over $\mathbb{F}_{p}$.(You might want to recall the fact that $\mathbb{F}-\lbrace 0 \rbrace$ is a multiplicative group, and use Lagrange's theorem!). As for part 3, if $p=2$, consider the map $\tau: \mathbb{F}^{\star} \to \mathbb{F}^{\star}$, $\tau(x)=x^2$. Notice that $\tau$ is a homomorphism of groups, that is injective (check this!), and hence surjective, and hence bijective. Every element in $\mathbb{F}$ is hence a square. Now, if $p$ is odd, consider the same map, $\tau: \mathbb{F}^{\star} \to \mathbb{F}^{\star}$. It is also a group homomorphism, whose kernel has two distinct elements. This means that the image of $\tau$, has $\frac{p^{n}-1}{2}$ elements, which means that there are $\frac{p^{n}-1}{2}$ distinct non zero squares, and hence $\frac{p^{n}+1}{2}$ squares in $\mathbb{F}$, including $0 \\$.
Let $a \in \mathbb{F}$. Consider the sets $S= \lbrace s^2, s \in \mathbb{F} \rbrace$ and the set $H= \lbrace a-y^2, y \in \mathbb{F} \rbrace$. These are subsets of $F$ with $\frac{p^{n}+1}{2}$ elements each. They must intersect (if they did not, then what would the cardinality of the union be?). And hence... finish it!