# Is the multiplicative identity unique in a unit ring?

In my coursebooks and on various websites online (wiki, proofwiki, etc.), among the first theorems which follow the definition of a ring are the uniqueness of the additive identity and the additive inverse. But I haven't found an answer to the following:

Question: If $R$ is a ring with multiplicative identity, is the multiplicative identity unique?

I suspect it is the case, since by definition, $1\cdot r=r\cdot 1=r$ for all $r\in R$; so if we assume that $1$ and $1'$ are two distinct multiplicative identities in $R$, we must have $1=1\cdot 1'=1'$, i.e. $1=1'$ — a contradiction. Is my line of reasoning correct? If so, why is this not included with all the simple theorems?

• Sorry, my previous comment was misleading and wrong (to an extent). The correct thing to say: The proof is done in elementary group theory books for any group. But if you look at that proof (as you just did yourself), you never use inverses. The proof works in fact for any monoid (not group, sorry, monoid is a group without inverses), especially the underlying multiplicative monoid of $R$. – Hamed May 29 '17 at 21:14
• @Hamed Thank you. We actually started with Groups straight away in our course, and in fact had proved it using inverses. – Luke Collins May 29 '17 at 21:16

You are correct, and $1 = 1 \cdot 1' = 1' \Rightarrow 1 = 1'$ is a great proof by contradiction.

As for "not included with all the simple theorems", it has probably already been presented in your book for far "simpler" groups and been treated as already-known facts by this point.

• It was presented for groups, but the proof we did in class used inverses (it's a one-step proof if you have multiplicative inverses). That's why I wasn't sure. Thanks for your reply. – Luke Collins May 29 '17 at 21:19

Is my line of reasoning correct?

Yes, it appears many places on this website, too. Actually you don't even need to frame it as a contradiction (the advice is usually to use direct proofs, where possible.) You simply say, "suppose $1$ and $1'$ are identities. Then $1=11'=1'$. Thus there is only a single identity. QED.

If so, why is this not included with all the simple theorems?

It typically is, or is included as an easy exercise.

If you think about it, a good general version of the theorem is just "the identity element of a monoid is unique." You only need the one operation, and the definition of an identity element. (It doesn't have anything to do with inverses. You could even relax the operation to be nonassociative.)

• Thank you. We actually started with groups straight away in our course, and in fact had proved a result for it using inverses. That's why I wasn't sure about whether or not the result is true. – Luke Collins May 29 '17 at 21:19
• @LukeCollins I can understand proving uniqueness of "inverses" in a group, but how would the proof for a group go any differently for the uniqueness of the identity? – rschwieb May 29 '17 at 21:20
• @rschwieb : at no point did you use associativity. If $(E,\star)$ is a set together with a binary operation, and if $e,e'\in E$ are identites ($\forall a\in E, e\star a = a\star e = a$, and the same for $e'$), then $e=e'$, whether or not $\star$ is associative. Even better, you can require $e$ to be a left identity, and $e'$ to be a right identity and it will suffice : no associativity whatsoever ;) – Max May 29 '17 at 21:46
• @Max you're right: I'm thinking of the uniqueness if inverses! – rschwieb May 29 '17 at 23:56