Prove that $K\times K[X]/(X^7-1)\cong K\times \dots \times K$

Question

Given that $K$ is a finite field of order $q\equiv1\text{ mod } 7$, I have to prove that $$K\times K[X]/(X^7-1)\cong K\times \dots \times K\ (8 \text{ times } K).$$ It's the same to prove that $$K[X]/(X^7-1)\cong K\times \dots \times K\ (7 \text{ times } K).$$

I thought I could make an isomorphism $\phi$ by $f=a_0+a_1X+\dots +a_6X^6\mapsto (a_0,a_1,\dots,a_n)$, because $X^7=1$ in $K[X]/(X^7-1)$, so there are only polynomials of degree 6 and lower. Showing that $\phi(f+g)=\phi(f)+\phi(g)$ is easy, but I don't know how to show its multiplicative. Is it correct what I have done so far? I need help.

Answer 1

Extended hints:

1. The multiplicative group $K^*$ is cyclic of order $q-1$, so there is a primitive seventh root of unity $\omega\in K$. Thus $X^7-1$ has seven distinct roots in $K$, and it factors as $$ X^7-1=(X-1)(X-\omega)(X-\omega^2)\cdots(X-\omega^6). $$
2. Chinese remainder theorem
3. $K[X]/(X-a)\cong K$ for any $a\in K$.

Try $\phi:K[X]/(X^7-1)\to K^7$ defined by $\phi(f)=(f(1),f(\omega),f(\omega^2),\ldots,f(\omega^6))$ instead.