# Prove that the ring $(\{0\},+,\cdot)$ is a subring of any ring $(R,+,\cdot).$

I have to prove that the ring $(\{0\},+,\cdot)$ is a subring of any ring $(R,+,\cdot).$

Let $S = (\{0\},+,\cdot)$ and $R = (R,+,\cdot)$ then $S$ is a subring of $R$ iff $(R,+,\cdot)$ is a ring and $S \subseteq R$ and $S$ is a ring with the same operations.

As we know $S$ has an identity element of $0 \to 0+0=0$. It has an additive inverse of $0 \to 0-0=0$. It is commutative and associative, and it is distributive over addition.

So my only problem is that I am having difficulties cleaning this up and putting it into a proof. Maybe you can help me.

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THANKS :) I was assuming what properties needed to be proved since my textbook made no mention of subrings – math101 Aug 19 '12 at 18:10
If $R$ is a ring with unit, then $S$ might not be considered as a subring, since we usually assume $1_R\in S$ as a property of a subring. – Andrew Aug 19 '12 at 18:12
As it turns out, $0=1_R$ is possible if and only if $R$ is the zero ring, itself. Andrew's point (I suspect) is that some definitions of subring require that the multiplicative identities are the same. In this case, they clearly aren't requiring that. – Cameron Buie Aug 19 '12 at 18:22
I'm always surprised whenever anyone puts a lot of emphasis on requiring subrings to be "unital subrings" (to share the superring's identity). There are interesting subrings with identity which are not unital. If $e$ is any central idempotent of a ring $R$ with unity, then $eRe$ is a subring with identity $e$. This characterizes the factor rings in products of rings. – rschwieb Aug 19 '12 at 19:28
Some people define rings to have $1\neq0$, others don't. It is one of those things you have to check when you encounter a paper or a text book. On the whole it makes very little difference - a sentence to deal with an anomalous case occasionally. It can be material when Category Theory is involved - and different contexts can prefer different conventions. – Mark Bennet Aug 19 '12 at 19:48

Hint $\$ By the subring test it suffices to verify $\rm\:0-0,\, 0\cdot 0\,\in\, S,\:$ i.e. $\rm \,S\,$ is closed under subtraction and multiplication.