A Ring of Polynomial over a Field of Characteristic 2 Must Have This Property

 

Let $K$ be a finite field. $1+1=0$ in K if and only if for any $f\in K\left[x\right]$ such that the degree of $f$ is more or equal to 1, the polynomial $f(X^2)$ is a reducible polynomial.

Proof

From Left to Right ($\Rightarrow$)

Let $1+1=0$ with $0$ is the additive identity of field $K$ and 1 is the unity or the multiplicative identity of the field $K$.

Now, observe that for any  $a,b\in K\left[x\right]$, we have

${(a+b)}^2=a^2+ab+ba+b^2=a^2+ab+ab+b^2$

(by using the distributive and commutative property of the polynomial ring $K\left[x\right]$ )

$\Rightarrow {(a+b)}^2=a^2+(1+1)ab+b^2=a^2+(0)ab+b^2=a^2+b^2$.

So, ${(a+b)}^2=a^2+(1+1)ab+b^2=a^2+b^2$ for any $a,b\in K\left[x\right]$.

Next, let's define a function $T:K\to K$ such that $T(r)=r^2$ for all $r\in K$. We will show that $T$ is a surjective function.

Take any $r,s\in K$ such that $T(r)=T(s)$, then $r^2=s^2$ or $r^2-s^2$ = 0. Since $1=-1$, then $r^2+s^2=0$. Again, since $r,s\in K\subseteq K\left[x\right]$ then ${(r+s)}^2=0$ or $r+s=0$. Thus, $r=-s=s$. 

Hence, $T$ is injective. Since $K$ is finite, then we can conclude $T$ is also surjective.

Take any polynomial $f\in K\left[x\right]$ such that the degree of $f$ is $deg(f)=1$. We can write

$f(X)=r_0+r_{1}X$ for some $r_0,r_1\in K$. 

Then, we obtain

 $f(X^2)=r_0+r_1{X}^2$.

Since $T$ is a surjective function, there exist $s_0,s_1\in K$ such that 

 $f(X^2)=r_0+r_1{X}^2=T(s_0)+T(s_1)X^2={s_0}^2+{s_1}^2{X}^2$.

By using the property we have proven before,

$f(X^2)={s_0}^2+{s_1}^2{X}^2={(s_0+s_{1}X)}^2$.

Therefore, we conclude for any polynomial $f\in K\left[x\right]$ with degree 1, $f(X^2)$ is reducible and also there exists $g\in K\left[x\right]$ such that  $f(X^2)={(g(X))}^2$.

Furthermore, by using mathematical induction, this is also true for any $f\in K\left[x\right]$ of any degree.

Assume for some $k\in \mathbb{N}$, if $f\in K\left[x\right]$ has degree $deg(f)\le k$ then $f(X^2)$ is reducible and there exists $g\in K\left[x\right]$ such that  $f(X^2)={(g(X))}^2$.

Now, take any  $h\in K\left[x\right]$ such that its degree is $deg(h)=k+1$, we can write

$h(X)=t_0+t_{1}X+t_2{X}^2+...+t_k{X}^k+t_k{X}^{k+1}=\sum _{i=0}^k{t}_i{X}^i+t_{k+1}{X}^{k+1}$

with $t_0,t_1,t_2,...,t_k,t_{k+1}\in K$.

Let $j(X)=t_0+t_{1}X+t_2{X}^2+...+t_k{X}^k=\sum _{i=0}^k{t}_i{X}^i$, then

$h(X)=j(X)+t_{k+1}{X}^{k+1}$

$\Rightarrow h(X^2)=j(X^2)+t_{k+1}{X}^{2(k+1)}$

Since $deg(j)\le k$, by induction hypothesis then there exists $m\in K\left[x\right]$ such that

$h(X^2)={(m(X))}^2+t_{k+1}{X}^{2(k+1)}$.

Next, since $T$ is surjective, there exists $u\in K$ such that

$h(X^2)={(m(X))}^2+u^2{X}^{2(k+1)}={(m(X)+u{X}^{k+1})}^2$.

Hence, $h(X^2)$ reducible to ${(m(X)+u{X}^{k+1})}^2$. Therefore, we have proven the forward implication.

From Right to Left ($\Leftarrow$)

Assume for every $f\in K\left[x\right]$ if $f$ has degree at least 1 then $f(X^2)$ is reducible. 

Since for every $rinK$, $X^2-r$ is reducible, then $X^2=r$ has at least one solution in $K$. Therefore, as a result of this, for every $f\in K\left[x\right]$, $f(X)=0$ has some solution(s) in $K$ if and only if $f(X^2)=0$ also has some solution(s) in $K$. (*)

Now, since $x^2=-1$ has some solution(s), then by (*), $x^4=-1$, $x^8=-1$, $x^{16}=-1$, and so on, for every $n\in \mathbb{N}$, $x^{2^n}=-1$ has some solution(s) in $K$. This is impossible to be happenend if $1\ne -1$ because $K$ is finite. Thus, $1$ must be equal to $-1$. In other words, $1+1=0$. $◼$