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

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Hi, could anyone solve my confusion?

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

How could we conclude

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

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 1-x = 0, i.e. x = 1 and hence \mathbf{Z}[x]/(1-x)=\mathbf{Z}. So \frac{\mathbf{Z}[x]/(1-x)}{(x^{p-1}+\cdots+x+1)}=Z/(p) and therefore gives the result?
 
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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 ?
 
Eynstone said:
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 ?

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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