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This is a question in Herstein's Topics in Algebra ("unit element" refers to multiplicative identity):

If $R$ is a ring with unit element $1$, and $\phi$ is a homomorphism of $R$ into an integral domain $R'$ such that $\ker\phi\ne R$, prove that $\phi(1)$ is the unit element of $R'$.

Now, Herstein does not require that integral domains have a unit element. It seems like the question is suggesting that the existence of such a homomorphism forces $R'$ to have a unit element. Of course, if $R'$ is assumed to have a unit element then the proof is trivial.

I am having trouble finding a proof or a counterxample for the first interpretation. I'm even having trouble thinking of integral domains without unit elements.

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    I made a mistake here.2012-04-04
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    Hint: Since $\phi$ is a homomorphism, $\phi(1x) = \phi(1)\phi(x)$. From this, it is easy to conclude that $\phi(1)$ must be an idempotent element of $R'$.2012-04-04
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    I had noticed that, but I can't see how it helps.2012-04-04
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    I'm always baffled by questions on whether a ring has a unit element and whether morphisms of rings preserve that element. Are there really textbooks or mathematicians out there believing this might not be the case? Anyway, in commutative algebra or algebraic geometry I have never seen a written document or heard a mathematician using rings without units or morphisms not preserving it. And I challenge anyone here to name a book on these two subjects not requiring this.2012-04-04
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    @GeorgesElencwajg, see http://math.stackexchange.com/questions/16168/applications-of-rings-without-identity and the links there.2012-04-04
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    Just to expand on @JohannesKloos ' hint: If $\phi(1_R)$ is an idempotent, it cannot be the zero element (since $\ker \neq R$). But then, you can "cancel out" $\phi(1_R)$ every times it appears on both sides of an equality... just have to find the right equality...2012-04-04
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    Firstly are you sure your question is right? I have a copy of Herstein's book: which question are you talking about?2012-04-04
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    section 3.4, question 21. I think the question is correct, M Turgeon's hint was very helpful.2012-04-04
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    @MatthewKwan, if you have solved it, then please post an answer and accept it.2012-04-04
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    as a new user, I can't answer my own question immediately. I'll do so when possible.2012-04-04
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    I'll post my comment as an answer.2012-04-04
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    Dear @Georges: I happily confess that I strongly believe in non-discrete groups, their convolution algebras, and the usefulness of distributions and unitary group representations. However, I also confess that most of the time I want more than just *any* ring homomorphism in order to be able say anything sensible.2012-04-04
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    Dear @lhf, thanks for the links. I have moved this comment to meta. My comment was probably not very clear: of course the ideal of functions ( of some type of regularity)with compact support is tremendously important. My point of view is that we shouldn't call it a ring.2012-04-04
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    Dear @t.b., my comment has actually no mathematical content! I know that what you mention is important, interesting, etc. I just suggest not to call "rings" structures with no unit element. If my suggestion is rejected, so be it! Wouldn't you like to continue this friendlydiscussion on meta, where I have opened a questionon the subject ?2012-04-04
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    @Georges: if memory serves, I think t.b. was the one who created the (rngs) tag. So I am inclined to say that he will agree with you on the naming of such things.2012-04-04
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    Dear @Willie, I'm not surprised: count on t.b. for good initiatives!2012-04-04

2 Answers 2

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This solution is based on the hint given by Johannes Kloos in the comments above.

Let $R$ be a unital ring, with unit element $1_R$, and let $R'$ be an integral domain (which is not a priori assumed to be unital). Suppose we have a nonzero homomorphism of (nonunital) rings $\phi:R\to R'$. We want to prove that $R'$ is actually unital, with unit element $\phi(1_R)$.

First, we have the following equality: $$\phi(1_R)=\phi(1_R1_R)=\phi(1_R)\phi(1_R).$$ Hence, we see that $\phi(1_R)$ is an idempotent of $R'$. Now, since $\phi$ is nonzero, $\phi(1_R)\neq 0$. Therefore, for any element $x\in R'$, we have $$\phi(1_R)x=(\phi(1_R))^2x \Longrightarrow x=\phi(1_R)x,$$ and similarly, we have $x=x\phi(1_R)$. Therefore, we conclude that $\phi(1_R)$ is a unit element of the ring $R'$.

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    Since this doesn't actually use that $1_R$ is a unit in $R$, it proves a somewhat more general statement: a morphism to an integral domain $R'$ that sends some idempotent to a nonzero element $e$ of $R'$ makes $R'$ unital with unit $e$. Or: a nonzero idempotent of an integral domain has to be its unit element.2012-04-04
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Now here is my go:

Let $r' \in R$ be a fixed but arbitrary element of $R'$. Let $r \in R \setminus \ker \phi$.

$$\begin{align}\phi(r)r'&=\phi(r)\phi(1)r'\\r'&\overset{\dagger}{=}\phi(1)r' \tag{1}\end{align}$$

Note that $\dagger$ follows from the fact that $\phi(r) \neq 0$ and that $R'$ is an integral domain. Note that, cancellation law holds for non-zero elements in an integral domain. Further, since by definition, an integral domain is a commutative ring, $(1)$ gives us that, $$r'=r'\phi(1) \tag{2}$$

Now $(1)$ and $(2)$ force that $\phi(1)$ is the unit element in $R'$ from the definition of unit element.

Thanks are due to Prof. Marc van Leeuwen whose relevant observations made the solution nicer and shorter. (See Comments below.)

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    +1 Nice. You could just say that in an integral domain one can simplify by the nonzero element $\phi(r)$, to avoid the messy second line. By the way I thought that 'integral' means 'commutative' in this context, but maybe that's because I've spent too much time on wikipedia.2012-04-04
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    @MarcvanLeeuwen You're infact right about the definition of an integral domain. So, the whole second paragraph is superfluous. I shall edit that in. And, your observation about that cancellation is right and will make it much nicer. I will correct that as well. Thank you.2012-04-04
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    @MarcvanLeeuwen I hope it looks OK now. I really like the fact that you make your remarks in a very mild tone; That makes me feel at times I should have reacted nicely on the site, well, at least on one occasion when I had that exchange with Prof. Jyrki.2012-04-04