Let $\F$ be a finite field of characteristic $p$.

Prove that the number of elements of $\F$ is $p^n$ for some positive integer $n$.

Proof.

First note that since $\F$ is a finite field, the characteristic of $\F$ must be a prime number $p$. Then $\F$ contains the prime field $\F_p$ and $\F$ is a finite extension of $\F_p$, say, of degree $n$.

This means that we have a basis $\{v_1, \dots, v_n\}$ of $\F$ as a vector space over $\F_p$. Hence any element $x\in \F$ can be uniquely written as
\[x=a_1v_1+\cdots a_n v_n,\] where $a_i \in \F_p$ for $i=1, \dots, n$.

It follows that the fields $\F$ has $p^n$ elements.

The post The Number of Elements in a Finite Field is a Power of a Prime Number first appeared on Problems in Mathematics.

Let $F$ be a finite field.
Prove that each element in the field $F$ is the sum of two squares in $F$.

Proof.

Let $x$ be an element in $F$. We want to show that there exists $a, b\in F$ such that
\[x=a^2+b^2.\]

Since $F$ is a finite field, the characteristic $p$ of the field $F$ is a prime number.

If $p=2$, then the map $\phi:F\to F$ defined by $\phi(a)=a^2$ is a field homomorphism, hence it is an endomorphism since $F$ is finite.( The map $\phi$ is called the Frobenius endomorphism).

Thus, for any element $x\in F$, there exists $a\in F$ such that $\phi(a)=x$.
Hence $x$ can be written as the sum of two squares $x=a^2+0^2$.

Now consider the case $p > 2$.
We consider the map $\phi:F^{\times}\to F^{\times}$ defined by $\phi(a)=a^2$. The image of $\phi$ is the subset of $F$ that can be written as $a^2$ for some $a\in F$.

If $\phi(a)=\phi(b)$, then we have
\[0=a^2-b^2=(a-b)(a+b).\] Hence we have $a=b$ or $a=-b$.
Since $b \neq 0$ and $p > 2$, we know that $b\neq -b$.
Thus the map $\phi$ is a two-to-one map.

Thus, there are $\frac{|F^{\times}|}{2}=\frac{|F|-1}{2}$ square elements in $F^{\times}$.
Since $0$ is also a square in $F$, there are
\[\frac{|F|-1}{2}+1=\frac{|F|+1}{2}\] square elements in the field $F$.

Put
\[A:=\{a^2 \mid a\in F\}.\] We just observed that $|A|=\frac{|F|+1}{2}$.

Fix an element $x\in F$ and consider the subset
\[B:=\{x-b^2 \mid b\in F\}.\] Clearly $|B|=|A|=\frac{|F|+1}{2}$.

Observe that both $A$ and $B$ are subsets in $F$ and
\[|A|+|B|=|F|+1 > |F|,\] and hence $A$ and $B$ cannot be disjoint.

Therefore, there exists $a, b \in F$ such that $a^2=x-b^2$, or equivalently,
\[x=a^2+b^2.\]

Hence each element $x\in F$ is the sum of two squares.

The post Each Element in a Finite Field is the Sum of Two Squares first appeared on Problems in Mathematics.

Show that the matrix $A=\begin{bmatrix}
1 & \alpha\\
0& 1
\end{bmatrix}$, where $\alpha$ is an element of a field $F$ of characteristic $p>0$ satisfies $A^p=I$ and the matrix is not diagonalizable over $F$ if $\alpha \neq 0$.

Comment.

Remark that if $A$ is a square matrix over $\C$ with $A^k=I$, then $A$ is diagonalizable.
For a proof of this fact, see If a power of a matrix is the identity, then the matrix is diagonalizable

Thus, over a field of characteristic $p>0$ the condition $A^p=1$ dose not always imply that $A$ is diagonalizable.

Proof.

By induction, it is straightforward to see that
\[A^m=\begin{bmatrix}
1 & m\alpha\\
0& 1
\end{bmatrix}\] for any positive integer $m$.
Thus
\[A^p=\begin{bmatrix}
1 & p\alpha\\
0& 1
\end{bmatrix}=I\] since $p\alpha=0$ in the field $F$.

Since the eigenvalues of $A$ is $1$, if $A$ is diagonalizable, then there exists an invertible matrix $P$ such that $P^{-1}AP=I$, or $AP=P$.
Let $P=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$. Then we have
\[AP=\begin{bmatrix}
a+\alpha c & b+\alpha d\\
c& d
\end{bmatrix}\] and this is equal to $P$, hence
\begin{align*}
a+\alpha c=a \text{ and } b+\alpha d=b \\
\end{align*}
Thus
\begin{align*}
\alpha c=0 \text{ and } \alpha d=0 \\
\end{align*}
If $\alpha \neq 0$, then $c=d=0$ but this implies that the matrix $P$ is non-invertible, a contradiction.
Therefore we must have $\alpha=0$.

The post In a Field of Positive Characteristic, $A^p=I$ Does Not Imply that $A$ is Diagonalizable. first appeared on Problems in Mathematics.