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Let K be a field and a,b $\in \Bbb N$ Show that $$a\text{ divides } b \text{ in } \Bbb Z \iff t^a-1\text{ divides } t^b-1 \text{ in } K[t] $$

My attempt at this:

I found a formula (speculation) for the polynomial division. It states as follows

$$(t^b-1):(t^a-1)=\sum_{i=1}^{\frac{b}{a}}t^{b-ia}$$ of course this is only defined as long as $\frac{b}{a}$ is a natural number, which is only the case when a divides b. So (I think) it's sufficient to show that this formula is true.

I try to show this by induction over $\frac{b}{a}$ as follows.

$\frac{b}{a}=1$

$\Rightarrow a = b$, so $(t^a-1):(t^a-1) = 1$ and $\sum_{1}^{1}t^{b-a} = 1$

$\frac{b}{a} \to \frac{b}{a}+1$ We have $\frac{b}{a}+1=\frac{b+a}{a}$ So

$(t^{b+a}-1):(t^a-1) = 1$ and $\sum_{i=1}^{\frac{b+a}{a}}t^{b-ia+a}$ (we have to prove this equality holds)

$\iff t^b(t^a-1):(t^a-1)=\sum_{i=1}^{\frac{b}{a}}t^{b-ia+a} + 1 $ Now susing the induction hypothesis we get

$\iff t^b\sum_{i=1}^{\frac{b}{a}}t^{b-ia} =\sum_{i=1}^{\frac{b}{a}}t^{b-ia}t^a + 1 $

I am stuck here. Thanks in advance

Thomas Andrews
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Travis
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1 Answers1

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Use long division. Clearly $t^a-1 | t^b -1 $ requires $a\le b$. But then the division at each step gives $t^{b-a}+t^{b-2a}+\ldots$ and terminates with $0$ remainder iff $b-da=0$ for some $d\in\Bbb N$. By definition $a|b$. Conversely, if $a|b$, multiply $t^a-1$ by the polynomial $(t^{b-a}+t^{b-2a}+\ldots + t^a+1)$ and verify that it gives $t^b-1$.

Adam Hughes
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