0
$\begingroup$

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.

Any hints please!!

$\endgroup$
1
  • 1
    $\begingroup$ Hint: A field has only two ideals. $\endgroup$
    – Anurag A
    Jul 31, 2020 at 8:57

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ what Maximal ideal you took in case of D? $\endgroup$
    – user775699
    Aug 4, 2020 at 17:10
  • $\begingroup$ can you please reply to my question asked in comments above? $\endgroup$
    – user775699
    Aug 17, 2020 at 10:02
  • $\begingroup$ @User In a field, the only maximal ideal is $(0)$. And clearly, $\mathbb{Q} / (0) \simeq \mathbb{Q}$ is not finite. $\endgroup$
    – Doctor Who
    Aug 17, 2020 at 18:10

You must log in to answer this question.