# True/False Statements about Commutative Rings with Unity. CSIR(2019)

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".

Thank you.

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

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 $$i. Consequently, $$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.
• 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. – Paul K Jul 19 '19 at 6:38