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

\( \left ( a + b \right ) ^2 = a^2 + ab + ba + b^2 = a^2 + ab + ab + b^2 \)

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

\(\Rightarrow   \left ( a + b \right ) ^2 = a^2 + (1+1)ab + b^2 = a^2 + (0)ab + b^2 = a^2 + b^2\).

So, \( \left ( a + b \right ) ^2 = a^2 + (1+1)ab + b^2 = a^2 + b^2 \) 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 \( \left ( r + s \right )^2 = 0 \) 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 

 \( f(X^2) = r_{0} + r_{1} X^2 = T(s_{0}) + T(s_{1}) X^2 = {s_{0}}^2 + {s_{1}}^2X^2\).

By using the property we have proven before,

\( f(X^2) =  {s_{0}}^2 + {s_{1}}^2X^2 = \left ( s_{0} + s_{1}X\right )^{2}\).

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  \( f(X^2) = \left(g(X)\right )^{2}\).

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 \(deg(f) \leq k\) then \( f(X^2) \) is reducible and there exists \( g \in  K \left [ x \right ]\) such that  \( f(X^2) = \left (g(X)\right )^{2}\).

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

\( h(X^2) = \left ( m(X) \right )^2 +  t_{k+1}X^{2(k+1)} \).

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

\( h(X^2) = \left ( m(X) \right )^2 +  u^{2}X^{2(k+1)}  =  \left( m(X) + uX^{k+1} \right)^2  \).

Hence, \( h(X^2) \) reducible to \( \left( m(X) + uX^{k+1} \right)^2 \). 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 \( r in K \), \(X^2 - r \) 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 \( 1 \neq -1 \) because \( K \) is finite. Thus, \(1 \) must be equal to \(-1 \). In other words, \( 1 + 1 = 0 \). \(\blacksquare \)