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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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    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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    @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