# $\displaystyle \mathbb KG\simeq \frac{\mathbb K[t]}{\langle t^n-1\rangle}$ as algebras?

Suppose $G$ is a cyclic group of order $n<\infty$. How can I show the isomorphism of algebras: $$\mathbb KG\simeq \frac{\mathbb K[t]}{\langle t^n-1\rangle}?$$

Above $\mathbb K$ is a field, $\mathbb KG$ is the group algebra of $G$ and $\mathbb K[t]$ the algebra of polynomials in $t$ with coefficients in $\mathbb K$.

I was trying to define an algebra morphism $f:\mathbb K[t]\longrightarrow \mathbb KG$ having kernel $\langle t^n-1\rangle$. The most natural choice would be $$\sum_{j=0}^n a_j t^n\longmapsto \sum_{j=0}^n a_j g^n,$$ where I'm supposing $G=\langle g\rangle$ (where $g$ has order $n$). Certainly $\langle t^n-1\rangle\subset \textrm{ker}(f)$ however I don't know how to prove the reverse inclusion.

Thanks

We will show that $\ker(f) \subseteq \def\g#1{\left\langle#1\right\rangle}\g{t^n - 1}$: Let $p(t) \in \ker(f)$. Dividing $p(t)$ by $t^n - 1$ gives polynomials $q,r\in \def\K{\mathbf K}\K[t]$ with $\def\deg{\mathord{\rm deg}}\deg(r) < n$, say $r(t) = \sum_{i=0}^{n-1} a_it^i$ and $$p(t) = (t^n - 1)\cdot q(t) + r(t)$$ Applying $f$ gives (as $f(t^n - 1) = f(p) = 0$) $$0 = \sum_{i=0}^{n-1} a_i g^i.$$ As $g^i$, $i=0,\ldots, n-1$ are different elements of $g$, we have $a_i =0$; giving $r = 0$. That is $$p(t) = (t^n - 1) \cdot q(t) \in \g{t^n - 1}.$$
$\newcommand{\K}{\mathbb K}$Just compare dimensions.
$B = \K G$ has dimension exactly $n$, while $A = K[t]/\langle t^{n} - 1 \rangle$ has dimension at most $n$ (as $1, t, \dots, t^{n-1}$ span it), and you have defined a surjective morphism (of algebras, so in particular of vector spaces) $f : A \to B$.
So $n = \dim(B) \le \dim(A) \le n$, thus $\dim(A) = \dim(B)$, and $f$ is bijective.
• I think it is fairly easy (in fact easier than the part you are using) to see that the images of $1,t,\ldots,t^{n-1}$ in $A$ are linearly independent over $K$. Indeed (over an integral domain) the ideal generated by a polynomial of degree $n$ never has nonzero elements of degree less than $n$. That every polynomial reduces modulo the ideal to one of degree less than $n$ on the other hand requires Euclidean division. Not really hard, but I would say slightly less obvious than the other direction. – Marc van Leeuwen Apr 9 '15 at 12:45