Suppose that $R$ is a commutative ring and $|R|=30$. If $I$ is an ideal of $R$ and $|I|=10$, prove that $I$ is maximal ideal

Question

Suppose that $R$ is a commutative ring and $|R|=30$. If $I$ is an ideal of $R$ and $|I|=10$, prove that $I$ is maximal ideal

Solution: $|R/I|=3 \implies R/I \approx Z_3$ which is a field.

If $R$ is a commutative ring with unity and $I$ is an ideal, then $R/I$ is a field if and only if $I$ is a maximal ideal. Hence, in this problem, $I$ is a maximal ideal iff $R$ contains a unity.

How do we prove that $R$ must contain a unity?

I know that a finite commutative ring with no zero divisors definitely contains a unity. But, then $R$ has not been stated to not contain zero divisors either.

Thank you for your help..

Answer 1

Let $x \in R-I$. Consider the ideal $J=\langle I, x \rangle$. Observe that $|J|>10$. Moreover this must be a subgroup of $R$ but based on Lagrange its order should divide $30$. But $I \leq J$ as well, therefore $10$ divides $|J|$. Thus $J=R$, hence $I$ is maximal.

Answer 2

The ideal $I$ is maximal because $R/I$ is simple.