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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

3 Answers3

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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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Let $B$ denote arbitrary boolean ring. Recall that $B/I$ inherits boolean structure for any given ideal $I\subset B$. Since a boolean ring will have no zero divisors (i.e. will be an integral domain) if and only if it has cardinality 2, and since all integral domains of finite cardinality are necessarily fields, we deduce that $P\subset R$ is a prime ideal $\iff$ $P$ is a maximal ideal.

Statements Used.

  1. For any ideal $I\subset \text{boolean ring } B$, then $B/I$ inherits boolean structure.
  2. A boolean ring is an integral domain if and only if it has cardinality 2.
  3. A finite ring is an integral domain if and only if it is also a field.
  4. Given commutative $R$, then an ideal $P\subset R$ is prime if and only if $R/P$ inherits integral domain structure.
  5. Given commutative $R$, then an ideal $P\subset R$ is maximal if and only if $R/P$ inherits field structure.
  6. All boolean rings are commutative.
J.G.131
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