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I want to show that a prime ideal in a non-unital Boolean ring $B$ is maximal ideal.

If the ring contains unity then it is easy. As Boolean rings are commutative, for a prime ideal $P$ the ring $B/P$ is both integral domain aswell as Boolean ring. The only non trivial integral domain and Boolean ring is $\mathbb{Z}_2$ which is a field, so $P$ is maximal in $B$.

But there are Boolean rings without unity. For example consider all the finite subsets of real numbers with symmetric difference and intersection; this is a Boolean ring without unity. (This is also the subring of $\mathcal{P}(\mathbb{R})$, the power set of real numbers under the same operations).

Generally for rings with unity, there is a technique to show that an ideal $I$ is maximal ideal. We consider an ideal which strictly contains $I$ and somehow show that it contains unity, so that it is an improper ideal. In the above case this way doesn't work. I'm looking for a technique which can be applied in a wider settings for similar problems such as in non Commutative rings etc..

Infinity_hunter
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  • @NoahSchweber What exactly is Unital ring? I am not familiar with this terminology. – Infinity_hunter Apr 20 '22 at 18:35
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    Unital rings are required to have multiplicative identities (= units) and homomorphisms of unital rings are required to preserve these identities; non-unital rings don't have these features. The term "ring" used without qualification is a bit ambiguous; my understanding is that it usually but not always means "unital ring" these days (with the term "rng" being sometimes used for non-unital rings, see the second paragraph of the wiki page)). – Noah Schweber Apr 20 '22 at 18:36

2 Answers2

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Note that $P$ being prime is the same as $B/P$ being a ring without nontrivial zero divisors, so multiplication by a fixed nonzero element is injective.

We will show that a nonzero boolean ring $B$ without nontrivial zero divisors has a unit, so it is in fact isomorphic to $\mathbb Z/2\mathbb Z$:

Let $j\in B$ be some nonzero element, then $j$ is the multiplicative identity:
For all $x\in B$, we have $xj=xjj$ therefore $x=xj$ (since multiplication by $j$ is injective), similarly we have $jjx=jx$ so $jx=x$. (This also shows that there is only one nonzero element, since the multiplicative identity is unique)

The point is that in a ring with no nontrivial zero divisors, the only idempotent elements are $0$ and the multiplicative identity.

Mor A.
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  • How no non trivial zero divisors imply that multiplication is injective? For example take integers which have no non trivial zero divisor a but multiplication is not injective : $4\times 3 = 6 \times 2$ – Infinity_hunter Apr 21 '22 at 11:59
  • @quantum_spin I meant that the maps $\lambda_a,\rho_a :R \to R$ defined by $\lambda_a(x)=ax$, $\rho_a(x)=xa$ are injective for all nonzero elements $a\in R$ (this what I meant by "multiplication by nonzero elements"), I guess the correct term would be that multiplication is cancellative. – Mor A. Apr 21 '22 at 12:40
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    There is a confusion; When $P$ is prime the ring $B/P$ does not have non trivial zero divisors, so it is a field with two elements. In my opinion that does not necessarily imply that $P$ is maximal as $B$ may not have multiplicative identity. – Infinity_hunter Apr 21 '22 at 17:31
  • It implies that $P$ is maximal due to the correspondence theorem, namely ideals of $B$ containing $P$ correspond to ideals of $B/P$, but if $B/P$ is a field it is simple, so there are no ideals of $B$ containing $P$ (other than $B$ or $P$), hence $P$ is maximal – Mor A. Apr 21 '22 at 17:46
  • Thank you. Now I am getting clarity. – Infinity_hunter Apr 21 '22 at 18:24
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Here is the elementary proof:

First note that $$\forall x,y \in R:\ xy(x+y)=x^2y +xy^2=xy+xy=0$$

Suppose $I$ is a prime ideal:

$$\forall x,y \not \in I: xy \not \in I $$ but $$xy(x+y)=0 \in I$$ so $$\forall x,y \not \in I: x+y \in I.$$

Suppose $J$ is an ideal that strictly includes $I$. Pick $j \in J$ but $j \not \in I$. $$\forall r \not \in I: r+j \in I \Rightarrow r+j \in J \Rightarrow r \in J.$$ That implies that $J=R$, so $I$ is maximal.

khashayar
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