# Does a homomorphism from a unital ring to an integral domain force a multiplicative identity?

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. – Lennart Apr 4 '12 at 10:27
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 $R'$. – Johannes Kloos Apr 4 '12 at 10:30
I had noticed that, but I can't see how it helps. – Matthew Kwan Apr 4 '12 at 10:36
@GeorgesElencwajg, see math.stackexchange.com/questions/16168/… and the links there. – lhf Apr 4 '12 at 11:34
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. – t.b. Apr 4 '12 at 12:21

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. – Marc van Leeuwen Apr 4 '12 at 13:12

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. – Marc van Leeuwen Apr 4 '12 at 13:00
@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. – user21436 Apr 4 '12 at 13:04
@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. – user21436 Apr 4 '12 at 13:13
You're welcome. – Marc van Leeuwen Apr 4 '12 at 13:15