Indeterminacy of Boolean Ring

Definition 2.1.

¹¹ Consider S as a non-empty set. an Indeterminacy set $A_I$ on X is known as:

were

$T_{A\left(\alpha \right)}\subseteq [0,1],$ (Truth-membership),

$I_{A\left(\alpha \right)}\subseteq [0,1],$ (Indeterminacy membership),

$F_{A\left(\alpha \right)}\subseteq [0,1]$ (Falsity membership).

of an element x in the set $A_I$ .

Remark 2.2.

The fundamental feature of an Indeterminacy set is that it generalizes classical, fuzzy, and intuitionistic fuzzy sets by explicitly incorporating indeterminacy.

Example 2.3.

Let $S=\{\alpha ,\beta \}$ . Define an Indeterminacy set $A$ as:

Definition 2.4.

¹² For $(G,\cdot )$ , We have that it is a group. an Indeterminacy group is defined as:

Example 2.5.

For $G=(Z_3,+)$ the additive group of integers modulo 3. So, Indeterminacy group is:

For example, $(1+I)+(2+I)=(1+2)+(I+I)=0+I$ .

Definition 2.6.

⁸ For $R$ to be a ring. $R_I=\{\alpha +\mathit{\beta I}:\alpha ,\beta \in R\}$ is referred to as Indeterminacy ring generated by $R$ and $I$ .

Remark 2.7.

The indeterminate element $I$ satisfies the condition $I^{²}=I$ , which is essential in defining Indeterminacy rings.

Definition 2.8.

³ Let $R$ be a ring. The Indeterminacy ring $R_I$ is a ring generated by $R$ and $I$ .

Remark 2.9.

The angle bracket notation $R_I$ is sometimes used to emphasize the closure under ring operations.

Example 2.10.

⁶ We denote by $Z$ the ring of integers, $Z_I=\{\alpha +\mathit{\beta I}:\alpha ,\beta \in Z\}.$

This is a ring termed the Indeterminacy ring of integers. Also, $Z⊊Z_I$ .

Remark 2.11.

The enlargement from $Z$ to $Z_I$ highlights how Indeterminacy extensions generalize classical rings.

Example 2.12.

⁴ We denote by $Q$ the ring of rationales. $Q_I=\{\lambda +\mathit{\beta I}:\lambda ,\beta \in Q\}$ . This is the Indeterminacy ring of rationales.

Definition 2.13.

¹³ We denote by $R$ a ring. A subset $J_I\subseteq R_I$ is referred to as Indeterminacy ideal of the Indeterminacy ring $R_I$ if for all $r\in R$ and $j\in J$ , we have:

Remark 2.14.

The presence of the indeterminate $I$ ensures that classical ideals extend naturally into the Indeterminacy framework.

Example 2.15.

Let $R=Z_6$ and $J=\{0,2,4\}$ . Then $J$ is an ideal in $R$ . The Indeterminacy ideal is:

Definition 2.16.

¹⁴ An Indeterminacy extension of a field is an Indeterminacy field is An Indeterminacy algebraic structure $(F_I,+,\cdot )$ where $F$ is a classical field and $I$ is the Indeterminacy indeterminate with $I^{²}=I$ . It satisfies all field axioms extended with the Indeterminacy component.

Example 2.17.

Let $F=Q$ , the field of rational numbers. Then the Indeterminacy field is:

For example, $(1+I)\cdot (2+I)=2+3I+I^{²}=2+3I+I=2+4I$ .

Remark 2.18.

Although $Q$ is a field, $Q_I$ is not a field since $I^{²}=I$ and $I$ has no multiplicative inverse. Still, it is sometimes loosely designated the Indeterminacy field of rationales.

Example 2.19.

⁵ Let $R$ be fixed as the ring of real numbers. $R_I=\{\lambda +\mathit{sI}:\lambda ,s\in R\}$ . This is the Indeterminacy ring of real’s.

Remark 2.20.

Similarly to the rationals, it is only a ring and not a true field, but in literature it is sometimes termed the Indeterminacy field of real’s.

Example 2.21.

⁸ $C_I=\{z+\mathit{wI}:z,w\in C.$ This is the Indeterminacy ring of complex numbers.

Remark 2.22.

Even though $C$ is algebraically closed and a field, its Indeterminacy extension $C_I$ is not a field because of the special Indeterminacy element I.

Definition 2.23.

Let \(R_I\) be an Indeterminacy ring. The ring \(R_I\) is called commutative if, for all \(\alpha I,\beta I\in R_I\), we have

\[

(\alpha I)(\beta I)=(\beta I)(\alpha I).

\]

If there exists an element \(1_I\in R_I\) such that

\[

1_I\cdot \lambda I=\lambda I\cdot 1_I=\lambda I

\]

for every \(\lambda I\in R_I\), then \(R_I\) is called an Indeterminacy ring with identity.

Remarks 2.24.

1.⁷ Unity here generalizes the multiplicative identity of the base ring $R$ .

2.³ Again, although $Q$ is a field, $Q_I$ is only a ring since I lack an inverse.

3. It is not a field, but in many Indeterminacy studies it is referred to as the Indeterminacy field of complex numbers.

Definition 3.1.

¹⁰ A ring \(R\) is called a Boolean ring if.

\[

\alpha^2=\alpha

\]

for every \(\alpha\in R\).

Definition 3.2.

Let \(R_I\) be an Indeterminacy ring. Then \(R_I\) is called an Indeterminacy Boolean ring if

\[

x^2=x

\]

for every \(x\in R_I\). In particular, if \(x=\alpha I\), then the idempotent condition becomes

\[

(\alpha I)^2=\alpha I.

\]

Example 3.3.

An Indeterminacy ring ( ${Z_2}_I,$ ${+}_2,$ ${.}_2$ ) is an Indeterminacy B-Ring.

Example 3.4.

An Indeterminacy ring $(P\left(X\right),∆,\cap ),$ Where $P\left(X\right)=\{\mathit{AI}:\mathit{AI}\subseteq X_I\}$ is an Indeterminacy B-Ring Since $(P\left(X\right),∆,\cap )$ is an Indeterminacy ring with identity and $\forall \mathit{AI}\in P\left(X\right)⟹A^2{I}^2=\mathit{AI}\cap \mathit{AI}=\mathit{AI}.$

Example 3.5.

An Indeterminacy ring ( ${Z_3}_I,$ ${+}_3,$ ${.}_3$ ) is not Indeterminacy B-Ring. Since $\exists \overline{2I}\in {Z_3}_I$ is not idempotent element, s.t $2^2{I}^2$ = $2I{.}_{3}2I=I\ne 2I.$

Example 3.3–3.5.

clarify the role of idempotency in distinguishing Indeterminacy Boolean rings from general Indeterminacy rings. The ring \(\mathbb {Z}_{2I}\) satisfies the Boolean condition because each element is idempotent, whereas \(\mathbb {Z}_{3I}\) fails to be an Indeterminacy Boolean ring because it contains non-idempotent elements. These examples justify the need for the formal results that follow.

Example 3.6.

For $(R_I,+,\bullet )$ comm. Indeterminacy ring with unity

$R_I=\{\psi ;\psi :X_I\to {Z_2}_I\}$ and $\forall \mathit{\alpha I}\in X_I$ we have:

So, $(R_I,+,\bullet )$ be a comm. Indeterminacy ring with unity. It achieves the following and $\forall \psi \in R_I$ either $\psi \left(\mathit{\alpha I}\right)=0$ or $\psi \left(\mathit{\alpha I}\right)=1$ . Then ${\psi }^2=\psi$ . If $\psi \left(\mathit{\alpha I}\right)=0.$

Then,

Or

Hence,

And

Thus, $(R_I,+,\bullet )$ Indeterminacy B-Ring.

Theorem 3.7.

Consider $R_I$ as an Indeterminacy B-Ring. Then $(-\mathit{\alpha I})=\mathit{\alpha I},\forall \mathit{\alpha I}\in R_I.$

Proof:

We prove that if $R_I$ is a ring, so $\forall \mathit{\alpha I},\mathit{\beta I}\in R_I,(-\mathit{\alpha I})(-\mathit{\beta I})=\mathit{\alpha \beta I}$ and ${(-\mathit{\alpha I})}^2=\mathit{\alpha I},(-\mathit{\alpha I})(-\mathit{\alpha I})={\left(\mathit{\alpha I}\right)}^2$ . From the definition of an Indeterminacy B-Ring,

Thus, ${(-\mathit{\alpha I})}^2=(-\mathit{\alpha I})$ . But ${(-\mathit{\alpha I})}^2={\left(\mathit{\alpha I}\right)}^2$ . So, ${\left(\mathit{\alpha I}\right)}^2=(-\mathit{\alpha I})$ .

from * we get $\mathit{\alpha I}=(-\mathit{\alpha I})$ and this required.

Theorem 3.8.

Every Indeterminacy B-Ring $(R_I,+,.)$ with the characteristic $2$ has the property $2\mathit{\alpha I}=0I$ .

Proof:

Let $\mathit{\alpha I}\in R_I,$ and since $R_I$ is an Indeterminacy B-ring

Therefore,

So,

$\mathit{\alpha I}+\mathit{\alpha I}={(\mathit{\alpha I}+\mathit{\alpha I})}^2$ ( $R_I$ is a B-Ring)

Hence,

Therefore,

Then,

Thus, $h\left(R\right)=2$ .

Theorem 3.9.

Consider $R_I$ as an Indeterminacy B-Ring. Then $R_I$ is comm. under $(·)$ .

Proof:

Assume that $I,\mathit{\beta I}\in R_I$ . We need to show that

(since $\mathit{\alpha I},\mathit{\beta I}\in R_I$ and is Indeterminacy B-Ring)

But $R_I$ an Indeterminacy B-Ring we have ${\left(\mathit{\alpha I}\right)}^2=\mathit{\alpha I}$ and ${\left(\mathit{\beta I}\right)}^2=\mathit{\beta I}⟹(\mathit{\alpha I}+\mathit{\beta I})=\mathit{\alpha I}+\mathit{\text{αIβI}}+\mathit{\text{βIαI}}+\mathit{\beta I}⟹$ $0=\mathit{\text{αIβI}}+\mathit{\text{βIαI}}⟹$ $\mathit{\text{αIβI}}=-\mathit{\text{αIβI}}$ but $\mathit{\alpha I}=-\mathit{\alpha I}$ from Theorem 3.7

Hence $\mathit{\text{αIβI}}=\mathit{\text{αIβI}}$ as required.

Theorem 3.10.

If $R_I$ is an Indeterminacy ring with identity. Then, every Indeterminacy maximal ideal is Indeterminacy prime ideal.

Proof:

Assume that $R_I$ Indeterminacy ring with identity and $(P_I,+,\bullet )$ An Indeterminacy maximal ideal in $R_I$ . To verify that $(P_I,+,\bullet )$ is Indeterminacy prime ideal. Let $\mathit{\alpha I},\mathit{\beta I}\in R_I,\ni \mathit{\alpha I}\bullet \mathit{\beta I}\in P_I$ and let $\mathit{\alpha I}\notin P_I$

$(P_I,+,\bullet )$ an Indeterminacy maximal ideal in $R_I,$ and $\mathit{\alpha I}\notin P_I$ .

Then $\mathit{PI}+\left(\mathit{\alpha I}\right)=R_I$ . Hence $I\in R_I⟹I\in \mathit{PI}+\left(\mathit{\alpha I}\right)⟹I=\mathit{\delta I}+\mathit{\text{λIαI}},\mathit{\lambda I}\in R_I,\mathit{\delta I}\in P_I$ and $I^2=I$ , $I=\mathit{\delta I}+\mathit{\lambda \alpha I}\}\ast \mathit{\beta I}$ $\mathit{\beta I}\bullet I=\mathit{\delta I}\ast \mathit{\beta I}+\mathit{\text{λIαI}}\ast \mathit{\beta I}$ , $\mathit{\beta I}=\mathit{\delta \beta I}+\mathit{\text{λαβI}}$ . Therefore $\mathit{\delta I}\in P_I,\mathit{\beta I}\in R_I⟹\mathit{\delta \beta I}\in P_I.$ Hence $\mathit{\lambda I}\in R_I,\mathit{\alpha I}\bullet \mathit{\beta I}\in P_I⟹\mathit{\text{λαβI}}\in P_I$ . Then $\mathit{\delta \beta I}+\mathit{\text{λαβI}}$ $\in P_I$ . So $\mathit{\beta I}\in P_I$ . Thus $(P_I,+,\bullet )\text{is Indeterminacy prime ideal}$ .

Theorem 3.11.

Let $R_I$ be an Indeterminacy B-Ring and $S_I$ be an Indeterminacy ideal in that Indeterminacy ring. Then $P$ is Indeterminacy prime ideal iff it is Indeterminacy maximal ideal.

Proof:

$⟹$ Let $S_I$ Indeterminacy prime ideal. We need to prove that $S_I$ is Indeterminacy maximal ideal. Take an $P_I$ Indeterminacy ideal in $R_I$ . s.t $(S\subset P\subseteq$ R) $S_I\subset P_I\subseteq R_I,$ To show that $P_I=R_I$ .

Hence $S_I\subset P_I⟹\exists \mathit{\alpha I}\in P_I,\mathit{\alpha I}\notin S_I.$ Then $\mathit{\alpha I}\in P_I⟹\mathit{\alpha I}\in R_I$ since $P_I\subseteq R_I$ .

But $R_I$ is Indeterminacy B-Ring, then ${\left(\mathit{\alpha I}\right)}^2=\mathit{\alpha I}$

$\mathit{\alpha I}(I-\mathit{\alpha I})=0\in S_I$ . So $\mathit{\alpha I}\notin S_I,S_I$ Indeterminacy prime ideal.

Therefore, $(I-\mathit{\alpha I})\in S_I$ , Hence $S_I\subset P_I,$ And so, $(I-\mathit{\alpha I})\in P_I,$ . Also, $(I-\mathit{\alpha I})+\mathit{\alpha I}\in P_I.$

Then, $I\in P_I,$ So $P_I=R_I$ . Thus, $S_I$ is Indeterminacy maximal ideal.

Let $S_I$ be an Indeterminacy maximal ideal. To prove $S_I$ is an Indeterminacy prime ideal. Hence $R_I$ is an Indeterminacy B-Ring. Also, $R_I$ abelian Indeterminacy ring with identity and by Theorem 3.11 (if $R_I$ Indeterminacy ring with identity. Then, every Indeterminacy maximal ideal is Indeterminacy prime ideal). Thus $S_I$ Indeterminacy prime ideal.

Remark 3.12.

(1) Let $(R_I,+,\bullet )$ be a comm. Indeterminacy ring with $\left(1I\right),$ The $S_I\subseteq R_I$ such that

Then,

$(S_I,+,\bullet )$ is an Indeterminacy B-Ring.

(2) Multiplication: If $\mathit{\alpha I},\mathit{\beta I}\in S_I$ , then ${\left(\mathit{\alpha \beta I}\right)}^{²}={\left(\mathit{\alpha I}\right)}^{²}{\left(\mathit{\beta I}\right)}^{²}=\mathit{\alpha \beta I}$ , so $\mathit{\alpha \beta I}\in S_I$ .

(3) Addition: If $\mathit{\alpha I},\mathit{\beta I}\in S_I,$ then $\mathit{\alpha I}+\mathit{\beta I}=\mathit{\alpha I}+\mathit{\beta I}-2\mathit{\alpha \beta I}.$

Also,

(4) Comm.:

Associativity: holds by expansion.

Identity of $(S_I,+)$ is $0$ s.t $I+0=0+\mathit{\alpha I}=\mathit{\alpha I}$ .

Inverses in $(S_I,+)$ , $\mathit{\alpha I}+{\alpha }^{-1}I=0$ , so every element is its own inverse.

Thus $(S_I,+)$ is an abelian group.

(5) Distributive: For $\mathit{\alpha I},\mathit{\beta I},\mathit{\gamma I}\in S_I$ :

$\mathit{\alpha I}(\mathit{\beta I}+\mathit{\gamma I})=\mathit{\alpha \beta I}+\mathit{\alpha \gamma I}-2\mathit{\text{αβγI}}=\left(\mathit{\alpha \beta I}\right)+\left(\mathit{\alpha \gamma I}\right).$ So distributivity holds.

(6) Boolean property: For every $\mathit{\alpha I}\in S_I$ : each element is idempotent under multiplication, and $(S_I,+,\bullet )$ satisfies all ring axioms and every element is idempotent under multiplication. Hence $(S_I,+,\bullet )$ is a B-Ring.

Corollary 3.13.

If $(R_I,+,.)$ is an Indeterminacy ring with identity $I$ . Then, $(R_I/P_I,+,.)$ is also Indeterminacy ring with identity $I+\mathit{\delta I}$ .

Proof:

Since $R_I$ is an Indeterminacy ring with identity $I,$ Then $+\mathit{\delta I}$ $\in R_I/P_I$ . Hence $I+\delta \mathrm{I}$ is an identity element of $R_I/P_I$ with respect multiplication, Since $\forall \mathit{\alpha I}+\mathit{\delta I}\in R_I/P_I$ . Hence $(\mathit{\alpha I}+\delta \mathrm{I}).(I+\delta \mathrm{I})=(\mathit{\alpha I}.I)+\mathrm{pI}=\mathit{\alpha I}+\delta \mathrm{I}$ and $(I+\delta \mathrm{I}).(\mathit{\alpha I}+\delta \mathrm{I})$ $=(I.\mathit{\alpha I})+\delta \mathrm{I}$ = $(\mathit{\alpha I}+\delta \mathrm{I})$ . Therefore, $R_I/P_I$ is Indeterminacy ring with identity element $+\mathit{\delta I}$ .

Theorem 3.14.

For $(R_I,+,.)$ , an Indeterminacy B-Ring. Then, $(\raisebox{1ex}{$R_I$}\!\left/ \!\raisebox{-1ex}{$P_I$}\right.,+,.)$ is also Indeterminacy B-Ring.

Theorem 3.15.

Consider $(R_I,+,·)$ as an Indeterminacy B-Ring. For any $\mathit{\alpha I}\ne 0,\mathit{\beta I}\ne 0,\mathit{\gamma I}\ne 0\in R_I$ , Then $(\mathit{\alpha I}+\mathit{\beta I})(\mathit{\beta I}+\mathit{\gamma I})(\mathit{\gamma I}+\mathit{\alpha I})=0$ .

Proof:

In an Indeterminacy B-Ring, every element is idempotent, that is ${\left(\mathit{\alpha I}\right)}^{²}=\mathit{\alpha I}$ for all $\mathit{\alpha I}\in R_I.$ ⁷ From idempotency, one derives that the ring has characteristic $2$ . Indeed, ${(\mathit{\alpha I}+\mathit{\alpha I})}^{²}=\mathit{\alpha I}+\mathit{\alpha I}$ implies $2\mathit{\alpha I}=0$ , hence $\mathit{\alpha I}+\mathit{\alpha I}=0$ for all $\mathit{\alpha I}\in R_I.$ ⁹ Since ${\left(\mathit{\alpha I}\right)}^{²}=\mathit{\alpha I}$ and ${\left(\mathit{\beta I}\right)}^{²}=\mathit{\beta I}$ , we have $\mathit{\alpha \beta I}+\mathit{\beta \alpha I}=0$ . In characteristic $2$ , this simplifies to $\mathit{\alpha \beta I}=\mathit{\beta \alpha I}$ . an Indeterminacy B-Ring is comm. So, ${(\mathit{\alpha I}+\mathit{\beta I})}^{²}={\left(\mathit{\alpha I}\right)}^{²}+\mathit{\alpha \beta I}+\mathit{\beta \alpha I}+{\left(\mathit{\beta I}\right)}^{²}=\mathit{\alpha I}+\mathit{\beta I}.$ ⁵ Multiplying this

Now

Now

Then $(\mathit{\alpha I}+\mathit{\beta I})(\mathit{\beta I}+\mathit{\gamma I})(\mathit{\gamma I}+\mathit{\alpha I})=0$ .⁶

Corollary 3.16.

Let $(P_I,+,\bullet )$ be a proper ideal in the Indeterminacy B-Ring $(R_I,+,·)$ . Then $P_I$ is maximal iff $(R_I/P_I,+,·)\cong (Z_{2I},+2,·2)$ .

Proof:

Since $(R_I,+,·)$ is an Indeterminacy B-Ring, the quotient Indeterminacy ring $(R_I/P_I,+,·)$ is also Indeterminacy Boolean. Moreover, as $R_I$ is a comm. Indeterminacy ring with identity, $R_I/P_I$ inherits these properties and remains a comm. Indeterminacy ring with identity.

For any element $\mathit{\alpha I}+P_I\in R_I/P_I$ , we have: ${(\mathit{\alpha I}+P_I)}^{²}=(\mathit{\alpha I}+P_I)(\mathit{\alpha I}+P_I)={\mathit{\alpha I}}^{²}+P_I=\mathit{\alpha I}+P_I$ .

Hence, $R_I/P_I$ is an Indeterminacy B-Ring. It is well known that an ideal $P_I$ is maximal in $R_I$ iff $R_I/P_I$ is an Indeterminacy field.

Corollary 3.17.

Every Indeterminacy B-Ring $(R_I,+,·)$ is semisimple, that is, $\mathit{r\alpha d}\left(R_I\right)=\left\{0\right\}.$

Proof:

Assume that $(R_I,+,·)$ be an Indeterminacy B-Ring. We aim to prove that $R_I$ is semisimple, i.e., $\mathit{r\alpha d}\left(R_I\right)=\left\{0\right\}$ . Let, for contradiction, that $\mathit{r\alpha d}\left(R_I\right)\ne \left\{0\right\}$ . Then there exists a nonzero element $\mathit{\alpha I}\in \mathit{r\alpha d}\left(R_I\right).$

From the auxiliary lemma, there exists an Indeterminacy ring homomorphism $\psi :R_I\to Z_{2I}$ such that $\psi \left(\mathit{\alpha I}\right)=I$ . Consequently, $\mathit{ker}\left(\psi \right)$ is a proper ideal of $R_I$ , and hence there exists a maximal ideal $P_I$ in $R_I$ such that $\mathit{ker}\left(\psi \right)\subseteq P_I$ .

Since $1I-\mathit{\alpha I}\in \mathit{ker}\left(\psi \right)\subseteq P_I$ and $\mathit{\alpha I}\in P_I$ (because $\mathit{\alpha I}\in \mathit{r\alpha d}\left(R_I\right)=P_I$ -maximal), we get $1I=\mathit{\alpha I}+(1I-\mathit{\alpha I})\in P_I$ , which implies that $P_I=R_I$ , a contradiction. Therefore, $\mathit{r\alpha d}\left(R_I\right)=\left\{0\right\}$ , and $R_I$ is semisimple.

Theorem 3.18.

Every Indeterminacy B-Ring is isomorphic to a direct product of copies of $Z_{2I}$ . Formally, $R_I\cong \prod _{\{i\in I\}}{Z}_{2I}$ .

Proof:

Every Indeterminacy B-Ring can be viewed as an Indeterminacy ring of functions from some index set $P_I$ to $Z_{2I}$ . This representation arises because each element of $R_I$ corresponds to unique Indeterminacy boolean combination of projections onto $Z_{2I}$ .

Proposition 3.19.

Every ideal in an Indeterminacy B-Ring is Indeterminacy radical ideal.

Proof:

Assume that $P_I$ an ideal of $R_I$ , and suppose $\mathit{\alpha I}\in \sqrt{P_I}$ , meaning $\mathit{\alpha ⁿ}\in P_I$ for some $n\ge 1$ . But in an Indeterminacy B-Ring, $\left(\mathit{\alpha I}\right)ⁿ=\mathit{\alpha I}$ , so $\mathit{\alpha I}\in P_I$ . Hence, $\sqrt{P_I}=P_I$ .

Theorem 3.20.

Every Indeterminacy B-Ring is reduced (contains no nonzero nilpotent elements) and therefore semi-simple.

Proof:

If $\mathit{\alpha I}\in \mathit{r\alpha d}\left(R_I\right)$ , then a is nilpotent. But in an Indeterminacy B-Ring, ${\left(\mathit{\alpha I}\right)}^2=\mathit{\alpha I}$ , implying $\mathit{\alpha I}=0$ or $1I$ . Since $1I\notin \mathit{r\alpha d}\left(R_I\right)$ , it follows that $\mathit{\alpha I}=0$ . Thus, $\mathit{r\alpha d}\left(R_I\right)=0$ , and $R_I$ is semisimple.

Proposition 3.21.

For each element $\mathit{\alpha I}$ in an Indeterminacy B-Ring $R_I$ , there exists an onto Indeterminacy ring homomorphism $\psi :R_I\to Z_{2I}$ such that $\psi \left(\mathit{\alpha I}\right)=1I$ .

This shows that Indeterminacy B-Ring possess many surjective homomorphisms to $Z_{2I}$ , allowing $R_I$ to decompose as a direct product of copies of $Z_{2I}$ .

Corollary 3.22.

Up to isomorphism, there exists only one Boolean field, namely $Z_{2I}$ .

Proposition 3.23.

An Indeterminacy B-Ring $(R_I,+,·)$ is an Indeterminacy field iff $(R_I/P_I,+,·)\cong (Z_{2I},+2,·2).$

Proof:

Assume that $(R_I/P_I,+,·)$ is an Indeterminacy B-Field. For any $\mathit{\alpha I}\in R_I$ , the following holds:

Thus, $R_I=\{0,1I\}$ , and consequently $R_I$ $\cong$ $Z_{2I}$ . Therefore, $P_I$ is maximal in $R_I$ iff $R_I/P_I$ is an Indeterminacy field, which occurs precisely when $R_I/P_I$ ≅ $Z_{2I}$ .