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Z[x]/(1-x)/(x^{p-1}+ +x+1) = Z/pZ ?

  1. Aug 20, 2011 #1
    Hi, could anyone solve my confusion?

    Let [itex]p[/itex] be a prime and let [itex]x=\zeta_p[/itex] to be a primitive pth root of unity.

    How could we conclude

    [tex]\frac{\mathbf{Z}[x]/(1-x)}{(x^{p-1}+\cdots+x+1)}=\mathbf{Z}/p\mathbf{Z}=\mathbb{F}_p ?[/tex]

    This should be obvious but it seems that I missed something. Could anyone help?


    Am I correct to infer that if we want to force [itex]1-x = 0[/itex], i.e. [itex]x = 1[/itex] and hence [itex]\mathbf{Z}[x]/(1-x)=\mathbf{Z}[/itex]. So [itex]\frac{\mathbf{Z}[x]/(1-x)}{(x^{p-1}+\cdots+x+1)}=Z/(p)[/itex] and therefore gives the result?
     
    Last edited: Aug 20, 2011
  2. jcsd
  3. Sep 3, 2011 #2
    Are you sure the problem reads Z[x]/(1-x)(x^{p-1}+...+x+1) = Z/pZ and
    not Z[x]/(1-x)/(x^{p-2}+...+x+1) = Z/pZ ?
     
  4. Sep 3, 2011 #3
    I am pretty sure it is Z[x]/(1-x)(x^{p-1}+...+x+1) = Z/pZ

    it would be good that you write out why you think it is not the case.
     
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