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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 $x^{43} + ax + b$.


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.

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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 $x^2 + x + 1$.

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  • $\begingroup$ But don't you have to say what $a$ and $b$ are in this case? $\endgroup$ – farleyknight May 16 '18 at 13:50
  • $\begingroup$ Come to think of it, since $deg(r) < 2$ then $r(x) = ax + b$, for some $a, b \in \mathbb{Z}$ correct? I'm not sure where you would get values for $a$ and $b$. $\endgroup$ – farleyknight May 16 '18 at 14:56
  • $\begingroup$ Yes, that is almost correct. We have that $r(x) = ax + b$ for some $a$ and $b$ in $F$, not in $\mathbb{Z}$. They don't ask you what the values of $a$ and $b$ are though, they just ask you to show that $a$ and $b$ exist. The division algorithm shows that $r(x)$ exists, and the $a$ and $b$ are just the additive inverses of the coefficients of $r(x)$. (So it might be better to say that $r(x) = -ax -b$ for some $a$ and $b$ in $F$.) $\endgroup$ – Dylan May 17 '18 at 14:30
  • $\begingroup$ No, that is technically correct. I guess I was hoping there would be some neat theorem, lemma, or other math trick that would get actual values for $a$ and $b$. But if it's not possible, that's fine. $\endgroup$ – farleyknight May 17 '18 at 15:32
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    $\begingroup$ You can use P. Senden's method to show that $a = -1$ and $b = 0$. An alternative way to solve the problem would be to show that $x^2 + x + 1$ divides $x^{43} - x$. In fact, we have that $x^{43} - x = x(x - 1)(x^2 + x + 1)(x^{39} + x^{36} + \dots + x^3 + 1)$. It's possible to factorise the last bracket further, but you don't need to to show that $x^{43} - x$ is divisible by $x^2 + x + 1$. $\endgroup$ – Dylan May 17 '18 at 20:14
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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 $x^{43} + ax + b$?

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  • $\begingroup$ Algebraic closures come later in the book, so while I feel like I could use that strategy, I don't think I'm supposed to use that strategy. $\endgroup$ – farleyknight May 16 '18 at 13:52
  • $\begingroup$ You're probably right. Nonetheless, this is (in my opinion) a rather elegant way to solve problems like this, instead of using the division algorithm. Final remark: it's not necessary to consider $\bar{F}$, the field $F(\omega)$ suffices. $\endgroup$ – P. Senden May 16 '18 at 14:04

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