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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 $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.

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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.

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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
@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

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 $B$.)

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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
@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

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