Consider the polynomial ring F [ x ] over a field F. Let d and n be two nonnegative integers.Prove： x d − 1 ∣ x n − 1

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Consider the polynomial ring   F      [    x    ]   over a field F. Let d and n be two nonnegative integers.Prove：      x    d    −  1  ∣      x    n    −  1

Sandra Randall · Accepted Answer

Suppose       x          d        −  1  ∣      x          n        −  1. By division algorithm, we can write   n  =  q  d  +  r for some   q  ,  r  ∈            N        0   with   0  ≤  r  &amp;lt;  d. Now, observe that      x          d        −  1  ∣  (      x          d        −  1  )  (      x          n      −      d        +      x          n      −      2      d        +  ⋯  +      x          n      −      q      d        +  1  )Expanding the above, and cancelling many terms, we get that      x          d        −  1  ∣      x          n        +      x          d        −      x          n      −      q      d        −  1  =      x          n        −  1  +      x          d        −      x          r      Together with       x          d        −  1  ∣      x          n        −  1, this implies:      x          d        −  1  ∣      x          d        −      x          r        =  (      x          d        −  1  )  +  (  1  −      x          r        )which gives       x          d        −  1  ∣      x          r        −  1. This is a contradiction, unless   r  =  0, in which case   d  ∣  nThe converse easily follows from the identity       x          n        −  1  =  (  x  −  1  )  (      x          n      −      1        +      x          n      −      2        +  ⋯  +  x  +  1  )

Consider the polynomial ring F[x] over a field F. Let d and n be two nonnegative integers. Prove：x^d−1∣x^n−1 iff d∣n.

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