# A question about subring of Rational Numbers involving prime and maximal ideals

Edited : I have this particular question in abstract algebra assignment given to me.

I have been studying algebra from Thomas Hungerford as a textbook.

Question : Let R be a subring of $$\mathbb{Q}$$ containing 1 . Then which 1 of following is nessesary true.

A. R is Principal ideal Domain.

B. R contains infinitely many prime ideals.

C. R contains a prime ideal which is a maximal ideal.

D. for every maximal ideal m in R, the residue field R/m is finite.

Attempt : I don't think ring is PID as it need not have a single element which will generate it.

Unfortunately , for B, C, D I am clueless on how can they be approached.

I understand one should give his attempt but I am unable to think anything about B,C and D.

• Hint: A field has only two ideals. Jul 31, 2020 at 8:57

B doesn't necessarily hold - take $$R = \mathbb{Q}$$.

D also doesn't necessarily hold - again, take $$R = \mathbb{Q}$$.

It seems like both A and C hold.

A - let $$w$$ be an ideal of $$R$$. Let $$k = \mathbb{Z} \cap w$$. Then we see that $$k$$ is an ideal of $$\mathbb{Z}$$. Take $$n \in k$$ s.t. $$k = (n)$$ since $$\mathbb{Z}$$ is a PID. Then $$n$$ generates $$w$$. For if we have fully simplified $$a/b$$ in $$w$$, then clearly $$a \in k$$ and therefore $$a$$ is a multiple of $$n$$. And since $$a$$ and $$b$$ are relatively prime, we can take $$x, y \in \mathbb{Z}$$ such that $$xa + yb = 1$$. Then $$x (a/b) + y = 1/b$$. Then $$1/b \in R$$. Then $$a (1/b)$$ is a multiple of $$n$$. Alternately, since $$n \in k$$, we have $$n \in w$$ and thus every multiple of $$n$$ is in $$w$$. Then $$R$$ is a PID.

C - every nonzero ring has a maximal ideal, and every maximal ideal is prime. So $$R$$ has a maximal, prime ideal.

• what Maximal ideal you took in case of D?
– user775699
Aug 4, 2020 at 17:10
• @User In a field, the only maximal ideal is $(0)$. And clearly, $\mathbb{Q} / (0) \simeq \mathbb{Q}$ is not finite. Aug 17, 2020 at 18:10