The polynomials in \(\mathbb{F}_{2}[x]\) are infinite. We describe a way to make it finite and bounded by a certain degree through a ”polynomial relation”. For example fix the relations to be \(x^{3}=x+1\), or equivalently \(x^{3}+x+1=0\), as in the example above. We can limit the number of elements in \(\mathbb{F}_{2}[x]\) in this way: each time we find a monomial of degree greater than or equal to \(3\) we substitute \(x^{3}\) by \(x+1\). At the end of this process we obtain a new polynomial of degree strictly less than \(3\). For example
\begin{equation} x^{4}+x^{2}=x(x^{3})+x^{2}=x(x+1)+x^{2}=x^{2}+x+x^{2}=x.\nonumber \\ \end{equation}One can obtain the same result dividing the polynomial \(x^{4}+x^{2}\) by the polynomial \(x^{3}+x+1\). The remainder is \(x\). Hence the polynomials defined in the set
\begin{equation} \mathbb{F}_{2}[x],\mbox{ where }x^{3}+x+1=0\nonumber \\ \end{equation}are the polynomials
\begin{equation} 0,1,x,x+1,x^{2},x^{2}+x,x^{2}+1,x^{2}+x+1.\nonumber \\ \end{equation}These polynomials are \(8=2^{3}\). This number can be obtained directly if we consider that the generic polynomial of degree \(2\) is
\begin{equation} a_{2}x^{2}+a_{1}x+a_{0},\nonumber \\ \end{equation}where \(a_{2},a_{1},a_{0}\in\mathbb{F}_{2}\).
We define the set of polynomials \(A\) as
\begin{equation} \mathbb{F}_{2}[x],\mbox{ where }p=0\nonumber \\ \end{equation}The finite set \(A\) inherits the operations of sum and product defined in the polynomial ring \(\mathbb{F}_{2}[x]\). Moreover the following fact holds
The finite set \(A\) is a field if and only if \(p\) is an irreducible polynomial.
Nota (M): non capisco cosa intendi comunicare con l’esempio…
Let \(A\) defined by \(\mathbb{F}_{2}[x]\) where we fix the relation \(x^{2}+x+1=0\), that is \(x^{2}=x+1\). The finite set \(A\) is \(\{0,1,x,x+1\}\). We observed that \(x^{2}+x+1\) is irreducible. And in fact all the elements different from \(0\) in \(A\) have an inverse: the inverse of \(1\) is \(1\); the inverse of \(x\) is \(x+1\) (try it); the inverse of \(x+1\) is \(x\).
Now consider the reducible polynomial \(x^{2}\). Then \(A\) is defined by \(\mathbb{F}_{2}[x]\) where \(x^{2}=0\). Even in this case the set \(A\) is \(\{0,1,x,x+1\}\), but this time the inverse of \(x\) does not exist. In fact multiplying \(x\) by each element in \(\{0,1,x,x+1\}\) we have either \(0\) or \(x\). In fact
\begin{equation} x\cdot 0=0,x\cdot 1=x,x\cdot x=x^{2}=0,x\cdot(x+1)=x^{2}+x=x.\nonumber \\ \end{equation}That is \(x\) is not invertible.