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I encountered the following lemma about homomorphisms in Ring Theory.

I found the lemma stated in the book, "Topics in Algebra " by I.N Herstein (on Chapter-2, Ring Theory, Page no-131, 2nd edition ) as follows:

If $\phi$ is a homomorphism of $R$ into $R'$ , then

  1. $\phi(0)=0$,

2.$\phi(-a)=-\phi(a),$ for every $a\in R.$

I think, in this scenario, $0$ is being used to refer to both of the zero elements or equivalently, the additive identity of both the rings $R$ and $R'.$

I proceeded to prove the 1st part of the above lemma as:

If $0\in R$ and $0'\in R'$ then as $\phi$ is the homomorphism we can say, $\phi(0)=\phi(0)+\phi(0).$ Adding, the additive inverse of $\phi(0)$ say, $-\phi(0)$ on both sides of the qbove equation, we get, $\phi(0)=0'.$

But here's the problem, I ended up, with, $\phi(0)=0'$ and not $0$. Are they assuming that the iditive identity in $R'$ is also $0$ and not $0'$ as this seems to be the only possible explanation(?) If it so, then, how is this assumption really, justified. I mean, without any prior reasoning how can we be sure that the additive identity of $R'$ is $0$ as well.

This looks like an incorrect statement and NOT a notation abuse.

Thomas Finley
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    I am not familiar with the book, but to me it seems like they just use $0$ to notate the identities of both rings (although they are not necessarily the same of course). But what you wrote is correct - one should distinct between the two identities, so the statement should be $\phi(0)=0'$ (using your notations). – GSofer Jul 27 '23 at 07:34
  • @GSofer If you consider adding your comment as an answer it will be really be very helpful. – Thomas Finley Jul 27 '23 at 07:44
  • It's exactly notational abuse: reliance on the reader's willingness to use context when interpreting a symbol, rather than insisting they be interpreted in a vacuum. – rschwieb Jul 27 '23 at 11:33
  • Constants like $0$ and $1$ are (nullary) operations, and generally we use uniform notation for the operations of an algebraic structure (e.g. in rings $+$ and $,\cdot,).,$ This already occurs in your $(2)$ where the same symbol $,-,$ is used for the additive inverse operation in both rings. See here for some reasons why. – Bill Dubuque Jul 27 '23 at 12:35

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This is a purely notational thing (and it somewhat bugs me too sometimes).

The answer is simple: they assume $0$ to mean the element of the ring it is implicitly a part of. So, in the statement $\phi(0)=0$, we might rewrite this as $\phi(0_R) = 0_{R'}$. (This clarifies which zero is where. We know the argument of $\phi$ lies in $R$, since that is how $\phi$ is defined: its domain is $R$. We know $\phi(0)$ lies in $R'$ likewise).

This is not unlike a similar "simplification" of notation you might have seen in studying groups or rings: for a homomorphism $\phi : G \to H$ of groups (or rings), $\phi$ must satisfy $$ \forall a,b \in G, \;\; \phi(ab) = \phi(a) \phi(b) $$ but we didn't even state the operations on the groups! It looks like $ab$ and $\phi(a)\phi(b)$ represent the same sort of "multiplication" (or whatever the operation happens to be). Yet at the same time, writing things this way also compactifies things, and it is technically nonambiguous in some sense (since you can deduce which items lie where, and hence deduce what operation must be represented).

Consequently, you are correct in noting that the zeroes here may differ, and in fact generally can. This theorem is saying, phrased differently, that homomorphisms of rings preserve the additive identity, and preserve the additive inverse.

PrincessEev
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