boolean ring Archives - Epsilonify

Every prime ideal is a maximal ideal in a Boolean ring

The Mathematician — Fri, 12 May 2023 13:00:41 +0000

In a Boolean ring every prime ideal is a maximal ideal

At first sight, it would not be easy to prove this by using the definitions of the prime ideal and maximal ideal itself. The trick is to use propositions that are for sure introduced earlier in any introduction book of ring theory.

Proof of that every prime ideal is a maximal ideal in a Boolean ring

Let $P$ be a prime ideal of the Boolean ring $R$ . Then $R/P$ is a Boolean ring and an integral domain, which we have seen here. But we saw here that $R \cong \mathbb{Z}/2\mathbb{Z}$ . Further, we do know that $\mathbb{Z}/2\mathbb{Z}$ is a field, so is $R$ since it is isomorphic to the integer modulo two. Now by this proposition, $P$ is also a maximal ideal.

Conclusion

In a Boolean ring every prime ideal is a maximal ideal.

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The only boolean ring that is an integral domain is Z/2Z

The Mathematician — Mon, 08 May 2023 13:00:41 +0000

A Boolean ring that is an integral domain is Z/2Z

We will prove that the only boolean ring that is an integral domain is

\mathbb{Z}/2\mathbb{Z}

Proof of that the only boolean ring that is an integral domain is Z/2Z

Let

R

be a boolean ring which is an integral domain. Boolean rings are commutative, as we have seen here. Take the elements

a, 1-a \in R

which are not equal to

0

and

1

. Then we get:

\begin{align*} a(1-a) = a - a^2 = 0, \end{align*}

since

a^2 = a

. We took for the assumption that our boolean ring is an integral domain, so this implies that

a

and

1-a

are no zero divisors. This leads us that

a = 0

1 - a = 0

, which implies that

R = \{0,1\}

and this is exactly

\mathbb{Z}/2\mathbb{Z}

Conclusion

The only boolean ring that is an integral domain is Z/2Z.

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Boolean ring is commutative

The Mathematician — Wed, 28 Sep 2022 13:00:00 +0000

Show that a Boolean ring is commutative

By definition, a ring

R

is Boolean if

x^2 = x

\forall x \in R

.

Proof. We need to show that

xy = yx

for all

x,y \in R

. So first, we have:

\begin{align*} (x + y)^2 = (x + y) &\iff x^2 + xy + yx + y^2 = x + y \\ &\iff x + xy + yx + y = x + y \\ &\iff xy + yx = 0 \end{align*}

Now we have

xy = -yx

. We would like to prove that

-yx = yx

. We can check that by finding its inverse:

\begin{align*} (y + y) &= (y + y)^2 \\ &= y^2 + 2y + y^2 \\ &= y + y + y + y \\ &= 0 \end{align*}

which implies that

y + y = 0

. Now we get

-y = y

and therefore we have that

xy = -yx = yx

, which implies that Boolean ring

R

is indeed a commutative ring.

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