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Let $R$ be a ring and $R_0$ a non-empty subset of $R$. Show that $R_0$ is a subring iff, for any $a,b∈R_0$, we have $a-b$, and $ab$ in $R_0$.

In the previous question, when proving that if for any $a,b∈R_0$, we have $a-b$, and $ab$ in $R_0$ then $R_0$ is a subring, do I have to individually prove that all of the axioms of closure, associativity, commutativity, identity for addition and multiplication, additive inverses and distributivity hold?

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    $\begingroup$ The short answer is yes. The long answer is that most of these properties already hold (e.g. associativity, commutativity, distributivity) because they are true for elements of $R$ and $R_0$ consists of elements of $R$! $\endgroup$ – Eoin Feb 13 '17 at 7:42
  • $\begingroup$ Thank you! I feel stupid now haha. $\endgroup$ – The Bosco Feb 13 '17 at 7:46
  • $\begingroup$ Can I just say that because $ab$, $a$, and $b \in R_0$, if $ab=a\in R_0$ then $\exists 1$ such that $a \cdot 1 = a \in R_0$? $\endgroup$ – The Bosco Feb 13 '17 at 8:04
  • $\begingroup$ One thing you will have to prove is that $0 \in R_{0}$, but this is straightforward. But there is no way you can prove that if $R$ has a unity $1$, then $1 \in R_{0}$. For instance, if $R$ is the ring of integers, then the even integers $R_{0}$ satisfy the two conditions, but of course $1 \notin R_{0}$. $\endgroup$ – Andreas Caranti Feb 13 '17 at 8:17
  • $\begingroup$ But isn't the multiplicative identity a necessary axiom? $\endgroup$ – The Bosco Feb 13 '17 at 8:22

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