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.

Let $(B, +, \cdot)$ be a ring (not necessarily unital!) with the property that every $x \in B$ satisfies $x \cdot x = x$.

How does one show that the kernel of any non-zero homomorphism of rings $h:B\rightarrow \mathbb{Z}_2$ is a maximal ideal of $(B, +, \cdot)$?

I'm looking for an elementary proof, not requiring anything more than the definitions of a ring, of an ideal, and of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.

share|improve this question
add comment

2 Answers 2

up vote 3 down vote accepted

Since $h:B \to \mathbb{Z}_2$ is a non-zero homomorphism, $\ker h$ is a proper ideal of $h.$ If $xy\in \ker h$ then $h(xy)=h(x)h(y)=0$ in $\mathbb{Z}_2$ so either $x$ or $y$ is in $\ker h,$ hence $\ker h$ is a prime ideal of $B.$

Now suppose $I$ is an ideal of $B$ that strictly contains $\ker h.$ Pick $x\in I\setminus \ker h.$ For any $y\in B$ we have $x(y-xy)=0\in \ker h$ and since $\ker h$ is a prime ideal we have $y-xy \in \ker h.$ Thus $y = (y-xy) +xy \in I$ and we conclude that $\ker h$ is maximal.

share|improve this answer
    
Nice proof. Thanks. My only quibble with it (and it's tiny) is that its reference to "prime ideals" was unnecessary, since it already contains the proof of $xy \in \ker h \Rightarrow x \in \ker h$ or $y \in \ker h$, which is all the rest of the proof needs. –  kjo May 14 '13 at 10:15
1  
@kjo The reference was a) to sum up both the implication $xy\in \ker h \implies x\in \ker h$ or $y\in \ker h$ and $\ker h \neq B$ (both properties being crucial) and b) to make it easier to see the natural generalization of the next paragraph (that any prime ideal of a Boolean rng is maximal). –  Ragib Zaman May 14 '13 at 10:20
    
Thanks, I had not noticed (b). –  kjo May 14 '13 at 11:20
add comment

Let $f:B\to\mathbb{Z}_2$ be a non-zero homomorphism, and let $I=\ker(f)$. Suppose $I\subsetneq J\subseteq B$. There exists an element $x\in J\setminus I$ (because $I\subsetneq J$), and we must have $f(x)=1$ (because there are only two elements of $\mathbb{Z}_2$ to go to), so that $$f(x-1_B)=f(x)-f(1_B)=1-1=0,$$ and hence $x-1_B\in \ker(f)=I\subset J$. Thus $$1_B=x-(x-1_B)\in J,$$ and therefore $J=B$. Thus $I$ is maximal.


The standard proof:

Any non-zero homomorphism of rings to $\mathbb{Z}_2$ is surjective; apply the first isomorphism theorem. Then use that an ideal $I$ of a ring $R$ is maximal $\iff$ $R/I$ is a field.

(None of this depended on any properties of $B$.)

share|improve this answer
    
Aw @Zev, you beat me to it. +1 –  Stahl May 14 '13 at 0:22
    
Thanks, I'm looking for more elementary proofs... (I've clarified my question) –  kjo May 14 '13 at 0:23
    
Thanks again, but I'm embarrassed to admit that there are several steps in your proof I can't follow. For one, I don't know how you know that $B$ has a unit, $1_B$. Second, I don't see why $x - (x - 1_B)$ has to belong to $J$, even if I assume that $J$ is an ideal. –  kjo May 14 '13 at 1:27
1  
@kjo I'm not sure about the unity part of your question, although it's usually a safe assumption: most rings we want to consider have unity. As for why $x - (x - 1_B)\in J$: $J$ is by assumption an ideal, and we assumed $x\in J$. However, the calculation Zev did in the proof showed that $x - 1_B\in\ker f$. Since $\ker f\subseteq J$, $x - 1_B\in J$ as well. Now, as ideals are abelian groups, we must have $x - (x - 1_B)\in J$, since $x, x - 1_B\in J$. –  Stahl May 14 '13 at 1:42
    
@Stahl: thanks for the explanation. I did not realize that some definitions of a ring assume that the ring has a unit element. I have modified my question to explicitly deny this assumption. –  kjo May 14 '13 at 9:28
add comment

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.