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

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.

• Where is the problem from? Are you sure that rings are not assumed to have an identity element? Jul 8 '14 at 5:09
• It's from Gallian. In Gallian, ring need not necessarily have a unity element Jul 8 '14 at 5:09
• Even with the zero product, a ring with 3 elements is simple. Jul 8 '14 at 5:12
• uhmm, .. $R/I = \{I,a+I,2a+I\}, a \notin I, 3a=0$. Why does that $I$ maximal? Jul 8 '14 at 5:19
• If $|R/I|=3$, then $R/I$ is simple because it has only $\{0\}$ and itself as additive subgroups. In general, $R/I$ is simple if and only if $I$ is maximal. Jul 8 '14 at 5:21

The ideal $I$ is maximal because $R/I$ is simple.
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.
• what does $\langle I,x\rangle$ means? Aug 23 '19 at 16:58
• @AnishRay It is the ideal generated by the set (ideal) $I$ and element $x$. If the ring is commutative and has unity (some assume unity as part of the definition and some don't) then $$\langle I,x \rangle=\{j+rx \, | \, r \in R, j \in I\}.$$ Aug 23 '19 at 18:50