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

1. Aug 20, 2011

### rukawakaede

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?

Last edited: Aug 20, 2011
2. Sep 3, 2011

### Eynstone

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 ?

3. Sep 3, 2011

### rukawakaede

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.