Let $F$ be a field. Show that there exist $a, b \in F$ with the property that $x^2 + x + 1$ divides $x^{43} + ax + b$.

Question

Original problem for formatting: Let $F$ be a field. Show that there exist $a, b \in F$ with the property that $x^2 + x + 1$ divides $x^{43} + ax + b$.

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This is a homework problem so you don't need to give me the complete answer. I just need a good hint.

I will be honest and say that I haven't tried anything yet because the only thing that comes to mind is just brute forcing the long division of polynomials. And that can get ugly rather quickly. I'm sure a much more elegant solution is possible. just not sure where to start.

This problem comes from a chapter that includes the division algorithm for polynomials so that theorem might apply here, but it's not clear to me where.

Polynomials are a big weakness for me in abstract algebra, so it would be helpful if people could highlight certain theorems or useful lemmas to help me in my journey in understanding this material. In particular, there are 2 other problems following this (Chapter 16 of the 8th edition of Contemporary Abstract Algebra) that involve $x^{25}$ and $x^{51}$ so I could use some help "breaking down" those large leading terms.

Answer 1

The division algorithm isn't a terrible idea, actually. We're not going to apply it to $x^{43} + ax + b$ and $x^2 + x + 1$, though. We're going to apply it to $x^{43}$ and $x^2 + x + 1$.

The division algorithm then tells us that there is a polynomial $q(x)$, and a polynomial $r(x)$ with degree at most $1$ such that $$ x^{43} = (x^2 + x + 1)q(x) + r(x). $$

We then have that $x^{43} - r(x)$ is divisible by $x^2 + x + 1$.

Answer 2

Hint: let $\bar{F}$ be the algebraic closure of $F$. Then the roots of $x^2 + x + 1$ (in $\bar{F}$) are $\omega$ and $\omega^2$, where $\omega$ is a cubic root of unity (do you see why?). Now, working in $\bar{F}$, what can you say for $x^2 + x + 1$ to be a factor of $x^{43} + ax + b$?