2
$\begingroup$

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.

2) If $R$ has finitely many ideals, then $R$ is finite.

3) If $R$ is a PID, then every subring of $R$ with unity is a PID.

4) If $R$ is an integral domain which has finitely many ideals , then $R$ is a field.

The solution i tried-

for 1). we take the example of $Z_{2210}$ 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.

for 2). if we take $R$=$\mathbb{Q}$, then this is 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

for 3rd). in this option we can take $R$=$\mathbb{Q(x)}$ which is P.I.D. and take its subring $\mathbb{Z(x)}$, and this subring is not P.I.D., so 3rd option is discarded.

and 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.

$\endgroup$
  • $\begingroup$ Counterexamples don't help. $\endgroup$ – Wuestenfux Jul 19 '19 at 6:31
  • 2
    $\begingroup$ The solutions for 1) to 3) are fine. To disproof a statement it's sufficient to give a counterexample. $\endgroup$ – Paul K Jul 19 '19 at 6:34
4
$\begingroup$

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 $\Bbb{Z}$ works.

For (2), any infinite field like $\Bbb{R}$ works.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Paul K Jul 19 '19 at 6:38
  • 1
    $\begingroup$ @PaulK you are right. I mistakenly was in the "group" mode. I will modify. $\endgroup$ – Anurag A Jul 19 '19 at 6:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.