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So I need to describe all subrings of Q with identity. My thought process so far is: Clearly $Z$ is contained in every $S$. Then, I considered adding $a/b$ (in lowest terms) to that subring, which makes $k(a/b)^n$ a member of $S$, for all integer $k$ and natural $n$.

So my answer would be: consider all rational numbers in lowest terms: $q_1, q_2, ...$ (the order does not matter, only the fact that they can be listed). Then picking an arbitrary number of different $q_i$ and considering integer linear combinations of these $q_i$ gives birth to a different subring of $Q$ with identity each time. So describing them in set notation, I would say $S =$ {${a +m_1q_1+m_2q_2+...} $}, for all integers $a$ and $m_i$ (I added the $a$ summand just to make sure that our selection of $q_i$ does not deprive $S$ of the multiplicative identity)

Is this close? It feels very simple and that there are things I have not considered. Thanks!

  • Please don't refer to the title for the statement of your question: include the full statement in the body of your question.If you are being asked to list the subrings of $Q$, then you have more work to do. (The subrings of $Q$ are in one-to-one correspondence with the sets of prime numbers). – Rob Arthan Feb 04 '20 at 23:32
  • @RobArthan yes, all subrings of Q that include the multiplicative identity 1. –  Feb 04 '20 at 23:37
  • My first comment contains a hint that should you solve the problem. – Rob Arthan Feb 05 '20 at 00:21
  • See also https://math.stackexchange.com/questions/104571/how-to-find-all-subgroups-of-mathbbq – lhf Feb 05 '20 at 00:59

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