The group algebra $KG$ If $G$ is a cyclic group of order $m$. Then $KG\cong K[t]/(t^m-1)$. Where $K$ is a field.
I define
\begin{align*}
\varphi:K[t]&\longrightarrow KG\\
\sum_ia_it^i&\longmapsto\sum_ia_ig^i
\end{align*}
 where $\left\langle g\right\rangle=G$ and $a_i\in K$ and $\varphi$ is surjective and homomorphism. Let $I=\left\langle t^m-1\right\rangle$. So I want to prove that $ker\varphi\subseteq I$. Thus
Let $p(t)\in ker\varphi$, where $p(t)=\sum_ia_it^i$, and $e$ the identity of $G$. Then
\begin{align*}
\varphi(p(t))&=\varphi\left(\sum_ia_it^i\right)\\
&=0\\
&=\sum_ib_ig^i-\sum_ib_ig^i\\
&=\sum_ib_ieg^i-\sum_ib_ig^i\\
&=\sum_ib_ig^mg^i-\sum_ib_ig^i\\
&=\sum_ib_ig^{m+i}-\sum_ib_ig^i\\
&=\varphi\left(\sum_ib_it^{m+i}\right)-\varphi\left(\sum_ib_it^i\right)\\
&=\varphi\left(\sum_ib_it^{m+i}-\sum_ib_it^i\right)\\
&=\varphi\left(\sum_ib_it^{i}(t^m-1)\right)\\
&=\varphi\left(q(t)(t^m-1)\right)
\end{align*}
 Where $q(t)=\sum_ib_it^i$ and $b_i\in K$. How I can guarantee that $p(t)=q(t)(t^m-1)$?
 A: Hint: Use the division algorithm to write $p(t)=q(t)(t^m-1)+r(t)$ where $\deg(r)<m$. 

 If $p(t)\in\ker\varphi$, then $\varphi(p)=r(g)=0$. But since $\deg(r)<m$, necessarily $r\equiv 0$, so $t^m-1\mid p(t)$. 

A: It does not answer your question, but I would propose a different strategy in proving that
$$
K[\mathbf Z/n]\cong K[t]/(t^n-1).
$$
Indeed, one has 
$$
K[\mathbf Z]\cong K[t,t^{-1}]
$$
as $K$-algebras, mapping $1$ to $t$. Moreover,
if $H$ is a normal subgroup of a group $G$ then the kernel of the natural morphism
$$
K[G]\rightarrow K[G/H]
$$
is the two-sided ideal $I=(h-1|h\in H)$ of $K[G]$. The latter statement is clear since the quotient $K[G]/I$ satisfies the universal property of the group algebra $K[G/H]$.  Recall that, for $G$ a group, $K[G]$ is the universal $K$-vector space equipped with linear $G$-action and with a specified element $1$. Universal means here that for all $K$-vector spaces $V$ equipped with linear $G$-action and with a specified element $v$, there is one and only one $G$-equivariant $K$-linear map $f\colon K[G]\rightarrow V$ with $f(1)=v$.
It follows that
$$
K[\mathbf Z/n]\cong K[t,t^{-1}]/(t^n-1)\cong K[t]/(t^n-1).
$$
A: Since clearly $I \subseteq \ker \phi$, we have a unique homomorphism $\overline{\phi} \colon K[t] / \langle t^m - 1 \rangle \to KG$ such that $\overline{\phi} (p(t) + \langle t^m - 1 \rangle) = \phi(p(t))$. To see that $\overline{\phi}$ is injective, verify that $\overline{\psi} \colon KG \to K[t] / \langle t^m -1 \rangle, \sum_{i = 0}^{m-1} a_i g^i \mapsto \sum_{i=0}^{m-1} a_i t^i + \langle t^m -1 \rangle$ is its inverse. (Provided that you know that every element of $KG$ can be represented in the form above, it is clearly well-defined. Otherwise you'd have to verify that first.)
