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I am a bit confused about the behavior of polynomials in finite fields. Why in a splitting field $\mathbb F2[x]/x^3+x+1$,$x^3=x+1$? I have problem in understanding it intuitively, if α is the root, should it be $α^3=-(α+1)$? In the case of deg 2，i.e. $\mathbb F2[x]/x^2+x+1$ I can sort of understand it using group theory, but not so well in higher degree. Can anyone help on it? (I apologize if this question seems basic, still new to abstract algebra.)

Yi Shen
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  • In every field $K$ containg $\Bbb{F}_2$ we have $1+1=0$ and consequently $$z+z=(1+1)z=0z=0$$ for all $z\in K$. Subtracting $z$ from both sides then gives $z=-z$. I collected a few examples of how the fields of $4,8$ and $16$ elements work into this Q&A pair. The focus there is on the use of discrete logarithms, but you may still benefit from taking a look. – Jyrki Lahtonen May 22 '24 at 19:36

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Denote $K := \mathbb F_2[x]/(x^3+x+1)$. Formally speaking, $K$ is a quotient ring whose elements are subsets $[f]\subseteq\mathbb F_2[x]$. For a given polynomial $f$, the equivalence class $[f]$ contains precisely those polynomials that yield the same remainder when dividing by $x^3+x+1$.

What corresponds to $x^3$ in $K$? It is the remainder of $x^3$ divided by $x^3+x+1$. In this instance one could work it out as $$ x^3 = (x^3+x+1) {\color{red}{-(x+1)}}. $$ Hence, the remainder is $-(x+1) = x+1$, since we calculate over $\mathbb F_2$. In other words, $[x^3] = [x+1]$ in $K$.

What about $x^5$? Again, we perform synthetic division of $x^5$ by $x^3+x+1$ yielding $x^2+x+1$ as the remainder. Therefore, $[x^5] = [x^2+x+1]$ in $K$.

AlvinL
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