(R, +, $\ast$) ( $\ast$ := Multiplication) is a ring. A subset S $\subseteq$ R is called subring when (S, +, $\ast$) is a ring. Show the following: S is a subring from R if the following properties apply:

1.) S $\subseteq$ R

2.) S $\neq$ $\varnothing$

3.) $\forall$ r, s $\in$ S : r + s $\in$ S $\land$ r $\ast$ s $\in$ S

4.) $\forall$ r $\in$ S : - r $\in$ S

My solution:

3.) Let S be S := {$\pi$ $\ast$ t : t $\in$ $\mathbb{R}$} ( Let t $\in$ $\mathbb{R}$ be arbitrary). Furthermore we say that r, s $\in$ S and we define r = $\pi$ $\ast$ $\Omega$ and s = $\pi$ $\ast$ $\lambda$ s.t $\Omega$, $\lambda$ $\in$ $\mathbb{R}$.

S.t r + s = $\pi$ $\ast$ $\Omega$ + $\pi$ $\ast$ $\lambda$ = $\pi$ ($\Omega$ + $\lambda$) $\in$ S (Closure under +). And r $\ast$ s = ($\pi$ $\ast$ $\Omega$) $\ast$ ($\pi$ $\ast$ $\lambda$) = $\pi$ $\ast$ ($\Omega$ $\ast$ $\pi$ $\ast$ $\lambda$) $\in$ S (Closure under $\ast$).

4.) Let r $\in$ S s.t r = $\pi$ $\ast$ $\Omega$, $\Omega$ $\in$ $\mathbb{R}$. We conclude that r $\in$ $\mathbb{R}$ $\Rightarrow$ $\exists$ (- r) $\in$ $\mathbb{R}$ s.t r + (- r) = (- r) + r = 0. It follows that ($\pi$ $\ast$ $\Omega$) + (- r) = 0 s.t (- r) = - ($\pi$ $\ast$ $\Omega$) s.t (- r) = $\pi$ $\ast$ (- $\Omega$) $\in$ S.

I am not sure if this proofs these properties, because it is only one example and I did not know how to show "the general proof" for that which is why I made up an example. But this proofs it only for this particular example right? I also does not know how to proof 1.) and 2.) since I don't know what I even have to show here. Doesn't the definition of the subring already proofs 1.) and 2.)? And are my solutions even correct?


Suppose first that $S$ is a ring with the operations $+$ and $\cdot$ inherited from the arbitrary ring $R\ $ (which is not the same as $\Bbb R$ here).
Then verify that all 4 conditions apply.

Next, suppose those conditions hold for a set $S$ and a ring $R$.
Then show that it's a ring with the inherited operations, i.e. the operations don't lead out from $S$, and the ring axioms hold.

  • $\begingroup$ Could you explain that a little bit further or give some kind of an example? I don't quite get what you mean. $\endgroup$ – Sibelephant May 25 '18 at 0:23
  • 1
    $\begingroup$ For example $\Bbb Z$ is a subring of $\Bbb Q$ or $\Bbb Z[x]$. $\endgroup$ – Berci May 25 '18 at 0:25
  • $\begingroup$ If I would prove that $\mathbb{Z}$ is a subring of $\mathbb{Q}$ would that be a sufficient proof for the properties or is there a good way to show a general proof? I would appreciate it a lot if you could also answer my other questions. $\endgroup$ – Sibelephant May 25 '18 at 0:44
  • $\begingroup$ I didn't really understand your thoughts in the question, either. One way to prove it is showing that it's a ring on its own within $\Bbb Q$, with the same operations. Another way is to verify the 4 conditions. Then the exercise is about to show that these two ways lead to the same conclusion. $\endgroup$ – Berci May 25 '18 at 0:49

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