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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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    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 $R$! – 2017-02-13
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    Thank you! I feel stupid now haha. – 2017-02-13
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    Can I just say that because $ab$, $a$, and $b \in R_0$, if $ab=a\in R_0$ then $\exists 1$ such that $a \cdot 1 = a \in R_0$? – 2017-02-13
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    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 $1 \notin R_{0}$. – 2017-02-13
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    But isn't the multiplicative identity a necessary axiom? – 2017-02-13
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    @The Bosco: If a candidate property of $R_0$ holds _automatically_ due to the fact that it holds in $R$, the commonly used phrase is that the property is "inherited from $R$". – 2017-02-13
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    @The Bosco: Whether or not a ring is required to have a multiplicative indentity depends on the book or course, and always needs to be explicitly specified. Once specified, that requirement is assumed to hold from that point on. If a ring is not required to have a multiplicative identity, then two-sided ideals qualify as subrings. – 2017-02-13
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    @The Bosco: Most authors require a multiplicative identity, but some don't require it to be distinct from $0$. However that's atypical -- most authors require a multiplicative identity 1 and require $1 \ne 0$. – 2017-02-13
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    @The Bosco: For those authors that require $1 \ne 0$, it's standard to require that any subring must contain $1$ (the same element $1$ as in the main ring). To test if $R_0$ is a subring of $R$, you _would_ need to verify that $1 \in R_0$. – 2017-02-13

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