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Let S be a non-empty subset of a ring R. Then S is a subring of R if and only if S is closed under $-$ and $\times$.

Proof: First, prove that S is a subgroup of R. Pick an arbitrary element $x$ from S. Since S is closed under subtraction, $x - x = 0 \in S$ and $0 - x = -x \in S$. Therefore, for all $x, y \in S, -x, -y \in S$ and hence $x-(-y) = x+y$ also belongs to $S$. Hence S is a subgroup.

Since R is a ring, distributive laws hold in R and hence also in S. Abelian property is also inherited from R.

S is closed under multiplication.

Hence, S is a subring.

Is my proof OK?

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    $\begingroup$ How do you define a subring? I'm asking since it can be defined as a subset closed under the two operations. $\endgroup$ – Asaf Karagila Jan 14 '15 at 13:50
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    $\begingroup$ Yes, the proof is correct. $\endgroup$ – darij grinberg Jan 14 '15 at 13:53
  • $\begingroup$ See also this thread on the subring test (there are others too iirc). $\endgroup$ – Bill Dubuque Jan 14 '15 at 14:59
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This community wiki solution is intended to clear the question from the unanswered queue.


This appears correct for one part of the proof: if $S$ is closed under subtraction and multiplication, then $S$ is a subring. The converse is fairly trivial, but it should at least be noted: Since $S$ is a subring, by definition it's closed under subtraction and multiplication.

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