(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.