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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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    $\begingroup$ 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. $\endgroup$ – Andrew Aug 19 '12 at 18:12
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    $\begingroup$ 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. $\endgroup$ – Cameron Buie Aug 19 '12 at 18:22
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    $\begingroup$ 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. $\endgroup$ – rschwieb Aug 19 '12 at 19:28
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    $\begingroup$ 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. $\endgroup$ – Mark Bennet Aug 19 '12 at 19:48
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    $\begingroup$ @rschweib: If you don't require structure-preserving maps to preserve $1$, that means you're not viewing the existence of a multiplicative unit as part of the algebraic structure, but instead as a property of the structure -- it means you're viewing it as "a $\mathbb{Z}$-algebra that has a unit", rather as a ring. As for products, normally the factors of a product are supposed to be quotient objects, not subobjects. That the fact $R$ (viewed as an $R$-algebra) happens to be a product of $R$-subalgebras is a property not shared by the ring structure doesn't seem awkward to me. $\endgroup$ – Hurkyl Aug 20 '12 at 12:36
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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.

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