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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.

math101
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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
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    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
  • Well I must prove this property that S is the trivial subring so in that case it must be – math101 Aug 19 '12 at 18:17
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    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
  • ohhh I was getting confused there – math101 Aug 19 '12 at 18:24
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    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
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    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
  • @rschwieb: I confess I'm in the opposite situation to you. Pretty much the only people I've encountered who use word "ring" to mean what I would call a "$\mathbb{Z}$-algebra" (or maybe a "rng") are students who aren't even aware that there is a different convention, so it's surprising to see someone who actually knows their stuff adopt that naming convention. Anyways, $1$ is part of the structure in the variety of (unital) rings. Homomorphisms preserve structure, and so map $1$ to $1$. The category of rngs with unit isn't as nice as a variety of universal algebras.... –  Aug 20 '12 at 05:19
  • @Hurkyl I'm not sure why you brought rings without identity into the conversation, because my comment was entirely about rings with identity. I'm speculating you just didn't read carefully. I just wanted to point out how awkward it was that assuming the factor rings of a product ring are no longer "subrings" of the product ring, if one assumes all subrings to be unital subrings. (PS Ah, I think you just read "unital subring" as "ring with identity". As I said, I just mean "shares identity with the superring") – rschwieb Aug 20 '12 at 12:04
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    @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. –  Aug 20 '12 at 12:36
  • @Hurkyl Ring theorists specializing in radical theory and related topics do often use the name "ring" for rngs (rings without 1), e.g. see the preface of Wiegandt's book. Perhaps someday there will be standard terminology for rings vs. rngs, but we are not there yet. For a related issue, see my MO post on the variety of ways to adjoin $1$ to a rng (vs. the "standard" Dorroh construction) - a point which, alas, is not well-appreciated by many. – Bill Dubuque Aug 20 '12 at 14:44
  • @rschwieb We have prior questions on importance of rngs (rings without $1$), e.g. see this answer and the MO thread linked there. Perhaps you and Hurkyl may wish to contribute answers there (alas, comments are not "first-class" SE objects, e.g. they are not searched by the SE search function). – Bill Dubuque Aug 20 '12 at 14:45
  • @BillDubuque I have nothing against rings without identity, and in more that one place I've commented on their usefulness. I only meant to clarify a point a fellow poster seemed confused about. – rschwieb Aug 20 '12 at 14:54
  • @Hurkyl It suffices to say that we do not all work in the same category with the same morphisms. I think I should end here, because it seems whenever I point out a natural point of view and you disagree, whatever I say will just result in a reiteration of your opinion stated as if it were a fact. This isn't a hard cycle for me to break, so I'll break it. – rschwieb Aug 20 '12 at 14:57
  • @rschwieb Yes, I inferred that. I merely wanted to emphasize that such remarks - if expanded to answers - may help many more of our readers than the few who may be reading these comments, esp. since comments won't appear in search results. As such, generally, if you write something that may be of value to the global community it is better to place it in an answer, than in a comment. – Bill Dubuque Aug 20 '12 at 15:05

1 Answers1

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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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