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Let $R$ be a commutative ring with unity. Which of the following is true:

  1. If $R$ has finitely many prime ideals, then $R$ is a field.
  1. If $R$ has finitely many ideals, then $R$ is finite.
  1. If $R$ is a PID, then every subring of $R$ with unity is a PID.
  1. If $R$ is an integral domain which has finitely many ideals, then $R$ is a field.

The solution i tried-

  1. 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.

  2. 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.

  3. 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.

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    $\begingroup$ The solutions for 1) to 3) are fine. To disproof a statement it's sufficient to give a counterexample. $\endgroup$
    – user301452
    Jul 19, 2019 at 6:34

1 Answer 1

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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.

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  • $\begingroup$ If $(a) = R$, then you can only infere that there is some unit $\varepsilon \in R$ with $\varepsilon a = 1$, which is okay. How do you infere $a^k = 1$ for some $k$? Similarly in the latter case, $(a^i) = (a^j)$ implies in an integral domain that we have $a^i = \varepsilon a^j$ for some unit $\varepsilon \in R$. Now since $R$ is an integral domain we have $1 = \varepsilon a^{j - i}$ which shows that $a$ is a unit. $\endgroup$
    – user301452
    Jul 19, 2019 at 6:38
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    $\begingroup$ @PaulK you are right. I mistakenly was in the "group" mode. I will modify. $\endgroup$
    – Anurag A
    Jul 19, 2019 at 6:39
  • $\begingroup$ Your counter-example for $(1)$ doesn't seem to work properly because $\Bbb Z$ has infinitely many prime ideals. Please feel free to correct me if I am wrong. Thanks! $\endgroup$
    – Anacardium
    Nov 15, 2020 at 6:18
  • $\begingroup$ @Anacardium You are right. I have edited it. Thanks for pointing it out. $\endgroup$
    – Anurag A
    Nov 18, 2020 at 23:42

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