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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 $n \in \mathbb N^*$?

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    $\begingroup$ But what about $\mathbb Z_6$? wouldn't the ring have to be a domain or something like that? $\endgroup$ – Jorge Fernández Hidalgo Dec 6 '15 at 22:01
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    $\begingroup$ This could not possibly be true, as it would say all finite commutative rings with units are fields, which is patently false $\endgroup$ – Alan Dec 6 '15 at 22:01
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    $\begingroup$ I would advise you to consider the chain $\mathbb{Z}/6\mathbb{Z} \supseteq 3 \cdot \mathbb{Z}/6\mathbb{Z} \supseteq 3^2 \cdot \mathbb{Z}/6\mathbb{Z} \dots$. Then actually $3 \mathbb{Z}/6\mathbb{Z} = 3^2\mathbb{Z}/6\mathbb{Z}$. Now it is true that $b \cdot 3 \neq 1$ for all $b \in \mathbb{Z}/6\mathbb{Z}$, but still $3 \in 3 \cdot\mathbb{Z}/6\mathbb{Z}$. You seem to be assuming you are working over an integral domain. $\endgroup$ – Improve Dec 6 '15 at 22:07
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    $\begingroup$ Do you mean infinite as in an infite number of elements? Then consider the commutative ring $\mathbb{C} \times \mathbb{C}$. $\endgroup$ – Improve Dec 6 '15 at 22:13
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    $\begingroup$ In fact every artinian integral domain is a field via an argument very similar to yours. $\endgroup$ – Jorge Fernández Hidalgo Dec 6 '15 at 22:24
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You must add that $R$ does not have zero divisor if not $Z/n\times Z/m$ is a counterexample.

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  • $\begingroup$ do you mean $Z/mZ \times Z/nZ$? $\endgroup$ – SiXUlm Dec 6 '15 at 22:01
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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.

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  • $\begingroup$ Ah I see. What if I impose $R$ to be infinite? $\endgroup$ – SiXUlm Dec 6 '15 at 22:06
  • $\begingroup$ I think if $ R$ is infinite then $\mathbb Z_2\oplus \mathbb Q $ is a counterexample. $\endgroup$ – Jorge Fernández Hidalgo Dec 6 '15 at 22:13
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    $\begingroup$ Or $K[X]/(X^2)$, $K$ any field. $\endgroup$ – Bernard Dec 6 '15 at 22:16

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