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.

Explicit Field Isomorphism of Finite Fields
(a) Let $f_1(x)$ and $f_2(x)$ be irreducible polynomials over a finite field $\F_p$, where $p$ is a prime number. Suppose that $f_1(x)$ and $f_2(x)$ have the same degrees. Then show that fields $\F_p[x]/(f_1(x))$ and $\F_p[x]/(f_2(x))$ are isomorphic.
(b) Show that the polynomials […]

The Number of Elements in a Finite Field is a Power of a Prime Number
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 […]

In a Field of Positive Characteristic, $A^p=I$ Does Not Imply that $A$ is Diagonalizable.
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 […]

Prove that $\F_3[x]/(x^2+1)$ is a Field and Find the Inverse Elements
Let $\F_3=\Zmod{3}$ be the finite field of order $3$.
Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$.
(a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field. How many elements does the field have?
(b) […]

Polynomial $x^p-x+a$ is Irreducible and Separable Over a Finite Field
Let $p\in \Z$ be a prime number and let $\F_p$ be the field of $p$ elements.
For any nonzero element $a\in \F_p$, prove that the polynomial
\[f(x)=x^p-x+a\]
is irreducible and separable over $F_p$.
(Dummit and Foote "Abstract Algebra" Section 13.5 Exercise #5 on […]

Prove that any Algebraic Closed Field is Infinite
Prove that any algebraic closed field is infinite.
Definition.
A field $F$ is said to be algebraically closed if each non-constant polynomial in $F[x]$ has a root in $F$.
Proof.
Let $F$ be a finite field and consider the polynomial
\[f(x)=1+\prod_{a\in […]

Two Quadratic Fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are Not Isomorphic
Prove that the quadratic fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are not isomorphic.
Hint.
Note that any homomorphism between fields over $\Q$ fixes $\Q$ pointwise.
Proof.
Assume that there is an isomorphism $\phi:\Q(\sqrt{2}) \to \Q(\sqrt{3})$.
Let […]

Any Automorphism of the Field of Real Numbers Must be the Identity Map
Prove that any field automorphism of the field of real numbers $\R$ must be the identity automorphism.
Proof.
We prove the problem by proving the following sequence of claims.
Let $\phi:\R \to \R$ be an automorphism of the field of real numbers […]

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