How to show that the ring $R= \mathbb Z_3[x]/\langle x^6-1\rangle$ is finite without using the concepts of vector space? I have understood that since $\mathbb Z_3$ is a finite field so $R$ has to be a finite dimensional vector space but is there any way of proving this from purely ring theoretic concepts ?

$R= Z_3[x]/\langle x^6-1\rangle$ is $Z_3[x]$, the set of polynomials over $Z_3$, reduced by the equivalence relation that says that $x^6 - 1 \equiv 0$, or equivalently that $x^6 \equiv 1$.
If we have any polynomial we can reduce its higher-order terms via this equivalence, until the result has degree at most 5. For example, $x^7 + 2x^6 + 2x^2$ reduces to $x + 2 + 2x^2$, because $x^7\equiv x$ and $2x^6\equiv 2$.
Since a polynomial of degree at most 5 is completely determined by its 6 coefficients, and these coefficients are chosen from $Z_3$, there are at most $3^6$ such polynomials, and we are done.
