Luz Stokes

2022-07-19

Consider the polynomial ring $F\left[x\right]$ over a field F. Let d and n be two nonnegative integers.
Prove：${x}^{d}-1\mid {x}^{n}-1$

Sandra Randall

Expert

Suppose ${x}^{d}-1\mid {x}^{n}-1$. By division algorithm, we can write $n=qd+r$ for some $q,r\in {\mathbb{N}}_{0}$ with $0\le r. Now, observe that
${x}^{d}-1\mid \left({x}^{d}-1\right)\left({x}^{n-d}+{x}^{n-2d}+\cdots +{x}^{n-qd}+1\right)$
Expanding the above, and cancelling many terms, we get that
${x}^{d}-1\mid {x}^{n}+{x}^{d}-{x}^{n-qd}-1={x}^{n}-1+{x}^{d}-{x}^{r}$
Together with ${x}^{d}-1\mid {x}^{n}-1$, this implies:
${x}^{d}-1\mid {x}^{d}-{x}^{r}=\left({x}^{d}-1\right)+\left(1-{x}^{r}\right)$
which gives ${x}^{d}-1\mid {x}^{r}-1$. This is a contradiction, unless $r=0$, in which case $d\mid n$
The converse easily follows from the identity ${x}^{n}-1=\left(x-1\right)\left({x}^{n-1}+{x}^{n-2}+\cdots +x+1\right)$

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