What is the splitting field $\mathbb{F}_q$ of the polynomial $x^4+x+1$ over $\mathbb{F}_2$?
I already knew the polynomial $x^4+x+1$ is irreducible and its roots are distinct in some extension field of $\mathbb{F}_2$. However, I am not sure if the splitting field must be of the form $\mathbb{F}_{2^k}$ and if the polynomial $x^4+x+1$ must be a divisor of $x^{2^k}-x$.
Note: I am new to field extension and I haven't learnt about the degree of field extension, so please provide explanation without using it.


The splitting field of this polynomial is $$ K=\Bbb{F}_2[x]/\langle x^4+x+1\rangle. $$ This follows from your observation that $p(x)=x^4+x+1$ is irreducible, and from the fact that if $\gamma=x+\langle x^4+x+1\rangle$ is a zero of that polynomial, then

  • The other zeros of $p(x)$ are $\gamma^2$, $\gamma^4$, and $\gamma^8=\sqrt\gamma$, so $p(x)$ splits into linear factors over $K$ and
  • All the non-zero elements of $K$ are actually powers of $\gamma$, so no smaller field will do.

These facts can be seen from the first principles as follows:

  • We know $p(\gamma)=\gamma^4+\gamma+1=0$. Squaring this equation using the binomial formula and the fact that $2=0$ in $K$ we get $$0=\gamma^8+\gamma^2+1+2\cdot\text{something}=\gamma^8+\gamma^2+1=p(\gamma^2).$$ Repeating this shows that $\gamma^4$ and $\gamma^8$ are also zeros of $p(x)$. This trick is often facetiously called freshman's dream, because we have all met beginners who want to square binomials like $(a+b)^2=a^2+b^2$ - a formula that only works in a commutative ring of characteristic two.
  • The other fact I did as the mid section of this answer I prepared for referrals like this. You see that I denote the field $K$ by $\Bbb{F}_{16}$ there.

Remarks (and/or extras)

  • The second part could, indeed, be more easily deduced using basic facts about degrees of field extensions. Because $p(x)$ has degree four, we can deduce that $K$ is a four-dimensional space over $\Bbb{F}_2$.
  • That freshman's dream -trick works over all finite fields. It implies that adjoining a single root of an irreducible polynomial always automatically gives the other roots as well. This fact is special to finite fields. You have surely seen examples of irreducible polynomials over $\Bbb{Q}$ where this does not happen. $x^3-2$ is the standard example of this phenomenon. The same fact can be rephrased as All finite extensions of finite fields are Galois extensions. But it sounds like you may not have heard of the concept of a Galois extension yet.
  • We have, indeed, that $p(x)=x^{16}+x$. See this question for the other factors and a few more details.
  • $\begingroup$ Do you mean, $F(\gamma)\subseteq K$ because $F(\gamma)$ is a splitting field and $K\subseteq F(\gamma)$ since any element of K is a power of $\gamma$ so $F(\gamma)=K$? $\endgroup$ – Vinh Khang Jan 16 '16 at 7:12
  • $\begingroup$ More like: $K=\Bbb{F}_2(\gamma)$, so $K$ must be contained in the splitting field. And the splitting field is contained in $K$ because $K$ contains all four zeros of $p(x)$. $\endgroup$ – Jyrki Lahtonen Jan 16 '16 at 11:16

you are right about the form of the polynomial and the size of the field. let $a$ be a root of $x^4+x+1$ and find its multiplicative order. you get $a,a^2,a^3,a^4=a+1,a^5=a^2+a,\ldots,a^{15}=1.$ so theorder is 15, all these elements are distinct and thus $F_{2^4}$ is the splitting field.


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