Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I made a mistake here. –  Lennart Apr 4 '12 at 10:27
2  
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
1  
@GeorgesElencwajg, see math.stackexchange.com/questions/16168/… and the links there. –  lhf Apr 4 '12 at 11:34
1  
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

2 Answers 2

up vote 2 down vote accepted

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 $R'$.

share|improve this answer
2  
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.)

share|improve this answer
    
+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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.