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Does it have anything to do with the fact that when we divide $x^{p-1}-1$ by $x^d-1$, we get a remainder of zero? Would this mean that $x^{p-1}$ has exactly $p-1$ roots? Or does that even matter? This is the first course I've ever taken in abstract algebra and unfortunately the subject isn't quite clicking with me.

user3034400
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3 Answers3

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Here is another proof that if $x^n-1$ is a factor of $x^m-1$ then $n$ is a factor of $m$.

Write $x^m-1=(x^n-1)g(x)$. Note that $g$ must have integer coefficients because $x^n-1$ is monic.

Take derivatives: $mx^{m-1} = nx^{n-1}g(x) + (x^n-1)g'(x)$.

Evaluate at $x=1$ and get $m = n g(1)$. Since $g$ has integer coefficients, $g(1)$ is an integer and so $n$ divides $m$.

lhf
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the subject may click more if you think syntactically? here you wish to prove an assertion. since it comes from a field theoretic context the term $x^{p-1}$ occurs in its statement. however what is asserted is clarified and made easier to deal with if you drop what in this context is an irrelevant detail and change the statement of the assertion to: $$ x^m-1 | x^n-1 \Leftrightarrow m | n $$ this equivalence (if and only if) is thus actually two assertions, which must be verified separately. Marty has given the necessary clue for one direction. from the fact that $t^m-1$ vanishes for $t=1$ it follows that $t-1$ divides $t^m-1$. using the suggested substitution then shows that
$$ m | n \Rightarrow x^m-1 | x^n-1 $$ now to go in the other direction.

if $m$ is not a factor of $n$ then we may write $n$ as $km+r$ where $0 \lt r \lt m$. now, using a mini-telescoping trick, we rewrite: $$ x^n-1 = x^{km+r} - x^r + x^r - 1 = x^r(x^{km}-1) + (x^r -1) $$ if $x^n-1$ were divisible by $x^m-1$, then since we have just shown that $x^{km}-1$ is divisible by $x^m-1$ we must conclude that $x^r-1$ is also divisible by $x^m-1$ which ain't gonna happen.

David Holden
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just a footnote, which may show the result more intuitively (this isn't a proof, only an illustration)

you will notice that the quotient $\frac{x^n-1}{x-1}$ is $$ x^{n-1} +x^{n-2}+\dots+x^1 +x^0 = \sum_{k=0}^{n-1}x^k $$ i.e. it has exactly $n$ terms, the first $n$ powers of $x$ beginning with $x^0=1$

now check out: $$ (1+x^3+x^6)(1+x+x^2) = 1(1+x+x^2) +x^3(1+x+x^2) +x^6(1+x+x^2) \\ =1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8 $$ multiplying both sides by $(x-1)$ $$ (1+x^3+x^6)(x^3-1)=x^9-1 $$

David Holden
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