# Cyclotomic polynomial $\gamma_{p^n-1}(X)$ under ring homomorphism $\sigma: \mathbb C[X] \rightarrow F_p[X]$ and subring $R$ of ring $L$

Let $\gamma_{p^n-1}(X)$ denote a cyclotomic polynomial, where $p$ is a prime number.

I have a theorem stating that there exists an irreducible polynomial $f$ in $\mathbb F_p[X]$, such that $deg(f)=n$, dividing $\gamma_{p^n-1}(X)$ in $\mathbb F_p[X]$.

How can this be true ? I mean, if we use a ring homomorphisn $\sigma: \mathbb C[X] \rightarrow \mathbb F_p[X]$ on $\gamma_{p^n-1}(X)$ to view it as a polynomial in $\mathbb F_p[X]$, then we get a product of polynomials of degree $1$ - but these are irreducible contradicting that $f$ divides $\gamma_{p^n-1}(X)$ ? (polynomials of degree of $1$ are irreducible in a field and $\mathbb F[X]$ is a UFD?)

Also the theorem states that $R = \{\delta \in L | \delta^{p^n} = \delta \}$ is a subring of $L = \mathbb F_p[X]/\langle f \rangle$. However how can $-\delta$ be in this set if $p=2$ ?

Hope someone can give me some input. Thanks.

-
There is no "ring homomorphism $\sigma : \mathbb C[x] \to \mathbb F_p[x]$". – darij grinberg Dec 3 '13 at 19:54
If $p = 2$, then $-\delta = \delta$ is clearly in the set. – darij grinberg Dec 3 '13 at 19:55
Thanks, how can we view the cyclotomic polynomial in $\mathbb F_p[X]$ if no ring homomorphism exists ? – user111854 Dec 3 '13 at 21:20
There is a ring homomorphism $\mathbb Z[x] \to \mathbb F_p[x]$, and the cyclotomic polynomials lie in the former ring. – darij grinberg Dec 3 '13 at 21:41