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

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The discussion centers on the equivalence of the ring quotient Z[x]/(1-x)/(x^{p-1}+\cdots+x+1) and the finite field Z/pZ, where p is a prime and x=\zeta_p represents a primitive pth root of unity. The participants clarify that setting 1-x=0 leads to Z[x]/(1-x)=Z, which simplifies the expression to Z/(p). The confusion arises from the interpretation of the polynomial degree in the denominator, specifically whether it is x^{p-1}+\cdots+x+1 or x^{p-2}+\cdots+x+1.

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This discussion is beneficial for mathematicians, algebraists, and students studying abstract algebra, particularly those interested in ring theory and finite fields.

rukawakaede
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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?
 
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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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