A commutative ring with unit element, which only has a finite number of ideals, is a field

Question

Let $R$ be a commutative ring with unit element. Suppose that it only has a finite number of ideals. Show that it is a field

My reasoning is as follows:

Let $a \in R-\{0\}$. Since $R$ is a commutative ring, it can be shown that $Ra$ is an ideal of $R$. It follows that $Ra^n$ is also an ideal of $R$, for every $n \in \mathbb N^*$, and $Ra \supset Ra^2 ... \supset Ra^n \supset ...$. But $R$ only has a finite number of ideals, so there must exists $m \in \mathbb N^*$ such that $Ra^n=Ra^{n+1}=(Ra)a^n$, for every $n \geq m$.

Now, $(Ra)a^n=\{(ba)a^n,b \in R\}$. If $ba \neq 1, \forall b \in R \Rightarrow 1a^n \notin (Ra)a^n = Ra^n \Rightarrow$ contradiction because $1 \in R$. In other words, there exists $b \in R$ such that $ba=1$, or $R$ is a field.

Is my reasoning correct? Can we conclude that: $R=Ra$, and thus $R=Ra^n$, for every $n \in \mathbb N^*$?

Answer 1

You must add that $R$ does not have zero divisor if not $Z/n\times Z/m$ is a counterexample.

Answer 2

The last implication in your reasoning supposes $a^n\ne 0$, which is not necessarily true. If it were true, it would imply any finite ring is a field.

Counter-example:

$\mathbf Z/6\mathbf Z$ is definitely not a field. It has $6$ elements, and a fortiori a finite number of ideals.