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I'm having a bit of trouble with a problem from Hungerford's Algebra concerning ring homomorphisms.

Let $f\colon R\to S$ be a homomorphism of rings such that $f(r)\neq 0$ for some nonzero $r\in R$. If $R$ has an identity and $S$ has no zero divisors, then $S$ is a ring with identity $f(1_R)$.

I've worked out a bit of an argument. Take $r$ to be as in the problem, so $$ f(r)=f(1_R)f(r)=f(r)f(1_R)\implies f(r)-f(1_R)f(r)=0_S. $$ This shows that $f(1_R)$ is the identity of $f(R)$, but I'm not sure if that's much use. However, if $1_S$ exists, then from $f(r)-f(1_R)f(r)=1_Sf(r)-f(1_R)f(r)=0_S$ and the distributive law, I would have $$ \bigl(1_S-f(1_R)\bigr)f(r)=0_S\implies 1_S=f(1_R) $$ since $S$ has no zero divisors, and $f(r)\neq 0_S$. But I don't see a way to prove $1_S$ exists, so if it's no bother, I'm hoping to get a hint on how to show that, or perhaps a prod in the right direction if I'm off track. Thank you.

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Just a remark: you are aware, I suppose, that nowadays when most people say "ring" and "ring homomorphism" they mean that multiplicative identities exist and are preserved? These "rngs" (i.e., not necessarily unital rings) do show up in nature sometimes...but not nearly as much as rings do. If you worked through just about any other contemporary algebra text, you wouldn't find problems like this. It would be nice to think that you have some specific application of all these rng theory exercises in mind... – Pete L. Clark May 23 '11 at 7:13
@Pete, oh, I'm afraid I wasn't aware. Thanks for letting me know. I tried my hand at Atiyah and Macdonald a week ago, which started out with rings, but a lot of the early results the authors seemed to take for granted went over my head, so I thought it would be wiser to go back and start from the beginning with Hungerford. Are you implying that I should perhaps get another book? My impression was that Hungerford was considered a good one, but I may be wrong. – yunone May 23 '11 at 7:18
@yunone: I'm not suggesting that you put Hungerford back on the shelf. Most of these standard graduate algebra texts have some good features and some bad features (we may not all agree on which are which!) which make it desirable -- at least in the long run -- to read more than one of them. My own commutative algebra notes reference Hungerford at certain points, so I know there's good stuff in there. Rather I'm saying that you might want to just assume that rings have multiplicative identities and ignore problems that trivialize under that assumption..for now. – Pete L. Clark May 23 '11 at 7:23
You should also get a second opinion... – Pete L. Clark May 23 '11 at 7:24
@yunone: personally I found Hungerford dry and impossible to read... as far as alternate sources, I liked Ash's Abstract Algebra, which has the merit of being freely available online: – Qiaochu Yuan May 23 '11 at 7:47
up vote 9 down vote accepted

Hint: If $s$ is in $S$, then $f(1_R)s=f(1_R)^2s$.

I left the above brief hint based on the request at the end of your question, not wanting to "give too much away," but now I'm happy to elaborate a bit.

You want to show that $f(1_R)$ is an identity for $S$. As you mentioned in a comment, it follows from the fact that $f$ is a homomorphism that $f(1_R)=f(1_R)^2$. You also know that $f(1_R)\neq 0$, because $f(1_R)f(r)=f(r)\neq 0$, and thus it is tempting to want to "cancel" $f(1_R)$ in the equation of the previous sentence to obtain $1_S=f(1_R)$. But of course, we still have to show that $1_S$ exists. In order to cancel $f(1_R)$, we have to be multiplying it with something, and this motivates multiplying both sides of the equation by some $s\in S$ to obtain the equation at the top of this answer. Now you can do some cancelling.

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Thanks Jonas. From that equation, I see $(f(1_R)-f(1_r)^2)s=0_S$, and if $s\neq 0$, then $f(1_R)=f(1_R)^2$, but I'm not sure if this is what you're hinting at, since that seems to follow by the homomorphism properties anyway. Is there someway to cancel $f(1_R)$? I also tried adjusting my original work to get $f(r)-f(1_R)f(r)=f(1_R)f(r)-f(1_R)^2f(r)=0_S$, but that didn't seem to lead anywhere either. Do you mind maybe expanding if it's no trouble? Thanks. – yunone May 23 '11 at 7:10
@yunone: wrong factorization. You want to show that $s=f(1_R)s$, so you want $s-f(1_R)s$ as one factor and something nonzero as the other. – Chris Eagle May 23 '11 at 7:13
Thank you for the addition Jonas, and your comment @Chris. Both very helpful for my trivial little problem. – yunone May 23 '11 at 7:40

Since the answer was not obvious to me immediately, I just want to post an answer so that the question has a full answer.

Putting it together, $f(1_R)s=f(1_R)^2s\implies f(1_R)[s-f(1_R)s]=0_S$. But $f(1_R)\neq 0$ since $f(1_R)f(r)=f(r)\neq 0$, so $s=f(1_R)s$ since $S$ has no zero divisors. Since $sf(1_R)=sf(1_R)^2$ as well, the same argument shows $s=sf(1_R)$, so $f(1_R)=1_S$.

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Consider $x \in S$ and consider $$x \cdot \Bigl[ f(1_{R}) - f(1_{R})^{2}\Bigr]$$

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Thanks Chandru, as I mentioned in my other comment, I suppose this implies $f(1_R)=f(1_R)^2$? But I'm not quite sure if this is what is intended, or where to go from there. – yunone May 23 '11 at 7:12
@Yunone: I hope Jonas's last comments makes the doubt clear. – user9413 May 23 '11 at 7:19

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