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To check a subset of a given ring is a subring, is it enough to check that the subset is closed under induced operations(multiplication and addition) or do I also need to show that it contains 0 and additive inverses of each element?

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    $\begingroup$ Yes, you need to check that the additive inverse of an element in your set is in your set. The fact that $0$ is in the set follows, as long as the set is non-empty. $\endgroup$ Aug 5, 2013 at 2:56

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Let's denote this set by $S$. It's a subring if:

  • $S\ne \emptyset$ and in practice we prove that $0\in S$;
  • $\forall a,b\in S$, $a-b\in S$ that's $S$ is a subgoup;
  • $\forall a,b\in S$, $ab\in S$.
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Yes, you need to check that the additive inverse of an element in your set is in your set. The fact that $0$ is in the set follows from closure under addition, as long as the set is non-empty.

A technical detail: Some people make the existence of a multiplicative identity part of the definition of ring. And some people don't! Usually, for those people who insist on a multiplicative identity, that multiplicative identity must be in any subring.

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You do need to show that it contains an additive inverse for each of its elements. (For example, $\mathbb{N}$ is not a subring of $\mathbb{Z}$ though it is closed under addition and multiplication.) Provided that you know the subset is nonempty, this together with it being closed under addition will then imply that $0$ is in there.

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If you're interested in a single additive property to check, it suffices for the subset to be closed under subtraction and multiplication. Subtraction gives you $0$ by $a-a$, the additive inverse by $0-a$ and thus addition by $a-(-b)$. In practice this isn't really easier than just checking addition and negation, however.

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