Consider a field $k$ of characteristic $0$ and a positive integer $n.$ In the proof of Theorem 4.19 of Polytopes, Rings, and K-Theory by Bruns and Gubeladze, it is stated that $k[x] / (x^n - 1)$ is a tensor product of integral domains; however, I fail to see why this is true. I have made the following observations but to no avail.
Considering that $k$ has characteristic $0,$ we have that $x^n - 1 = \prod_{d \,|\, n} \Phi_d(x),$ where $\Phi_d(x)$ is the $d$th cyclotomic polynomial. Consequently, the polynomial $x^n - 1$ splits into a product of pairwise relatively prime irreducible polynomials; however, I don't see how $k[x] / (x^n - 1) \cong \otimes_{d \,|\, n} k[x] / (\Phi_d(x)),$ if this is indeed what the authors assert. Even more, it seems to me that the tensor product of the quotients of $k[x]$ by two relatively prime irreducible ideals would be zero, as we would have that $$\frac{k[x]}{(p(x))} \otimes_{k[x]} \frac{k[x]}{(q(x))} \cong \frac{k[x]}{(p(x), q(x))} = \frac{k[x]}{\gcd(p(x), q(x))} = \frac{k[x]}{k[x]} \cong 0.$$
I would greatly appreciate any insight, comments, or suggestions. Thank you.
I think everything you say is correct. The authors are only arguing $k[x]/(x^n-1)$ is reduced, so they probably meant to observe that it is the direct product of integral domains, which is just the Chinese Remainder Theorem.
Edit: In fact, when I look at the Google Books preview version of the book, it says exactly that. So perhaps this error is fixed in later editions.