Let $R$ be a commutative ring with unity. Which of the following is true:
- If $R$ has finitely many prime ideals, then $R$ is a field.
- If $R$ has finitely many ideals, then $R$ is finite.
- If $R$ is a PID, then every subring of $R$ with unity is a PID.
- If $R$ is an integral domain which has finitely many ideals, then $R$ is a field.
The solution i tried-
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We take the example of $\mathbb Z_{2210}$ the prime ideals of this ring are prime divisors of 2210 which are 2,3,7,11,5, but the given ring is not a field, thus we can discard this option.
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If we take $R=\mathbb{Q}$, then this is a field (which is also a ring), the only ideals of $\mathbb{Q}$ are $(0)$ and $(1)$, but set of rational number is not finite, so we can discard 2nd option.
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In this option we can take $R=\mathbb{Q}[x]$ which is a P.I.D. and take its subring $\mathbb{Z}[x]$, and this subring is not P.I.D., so 3rd option is discarded.
The remained option is 4th which should be true (for this I can't find the example).
I am not satisfied by my approach to this question by only choosing particular examples.
Please suggest me a proper solution, and i am also confused with term "finitely many".
Please help.
Thank you.
Best Answer
Edited and corrected the answer based on Paul K's comment.
For the 4th case, let $a \neq 0$ be such that $a \in R$. Now consider the ideal $\langle a \rangle$. If this ideal is $R$, then $ar=1$ for some $r \in R$, in which case $a$ is invertible. If this ideal is not $R$, then we can create a chain of ideals
$$\ldots \langle a^3 \rangle \subset \langle a^2 \rangle \subset \langle a \rangle$$ But the number of ideals is finite, this means this chain will stabilize, i.e. $\langle a^i \rangle = \langle a^j \rangle$ for some $i<j$. Consequently, $a^i=ra^{j}$. In which case using the fact that $R$ is an integral domain and $a \neq 0$ we can infer that $ra^{j-i}=1$. Thus $a$ is invertible. This show that $R$ must be a field.
For (1), a simple example like $\Bbb{Z}_4$ works.
For (2), any infinite field like $\Bbb{R}$ works.