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Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorTobyBartels
• CommentTimeJun 10th 2018

Note that indeed any idempotent magma in $Ab$ is commutative.

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeJun 10th 2018

There’s something just a bit odd though about the notion of idempotent monoid in $Ab$. When one says that a ring is a monoid in $Ab$, one means a monoid in the monoidal category $(Ab, \otimes)$ with the standard tensor product. However, this tensor product isn’t cartesian, and so expressing the notion of an idempotent monoid (where the axiom $x x = x$ involves duplication of a variable) doesn’t go down so smoothly.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeJun 10th 2018

Gave a meaning to “idempotent monoid” in concrete monoidal categories.

• CommentRowNumber4.
• CommentAuthorJohn Baez
• CommentTimeMay 21st 2023

The free Boolean ring on a set $X$ can be identified with $\mathcal{P}_f \mathcal{P}_f X$, where $\mathcal{P}_f \colon Set \to Set$ assigns to any set the set of all its finite subsets. In fact $\mathcal{P}_f \colon Set \to Set$ can be made into a monad in two different ways: the monad for semilattices and the monad for vector spaces over the field with 2 elements. These two monads are related by a distributive law, making $\mathcal{P}_f \mathcal{P}_f$ into the monad for Boolean rings.

Also mentioned:

The category of Boolean algebras is discussed further in BoolAlg, but some of the results about this category are proved there by working with the equivalent category of Boolean rings.

• CommentRowNumber5.
• CommentAuthorJohn Baez
• CommentTimeMay 22nd 2023

• CommentRowNumber6.
• CommentAuthorJ-B Vienney
• CommentTimeMay 22nd 2023

Added interpretation of the connectors in the idea section.

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeMay 22nd 2023
• (edited May 22nd 2023)

J-B, the additive inverse in a Boolean ring does not interpret negation. (In fact, additive inversion is the identity!) Negation is given by $x \mapsto 1 - x$, or equivalently by $x \mapsto 1 + x$.

• CommentRowNumber8.
• CommentAuthorJ-B Vienney
• CommentTimeMay 22nd 2023
Oh, thanks, I've learned something! So it is really ring because x (exclusive or) x = "false".
• CommentRowNumber9.
• CommentAuthorJ-B Vienney
• CommentTimeMay 22nd 2023

Added in the idea section “The fact that the exclusive disjunction of $x$ and $x$ is the truth value “false” makes the commutative additive monoid an abelian group where $-x = x$.”

• CommentRowNumber10.
• CommentAuthorJ-B Vienney
• CommentTimeMay 22nd 2023
• (edited May 22nd 2023)

Corrected the assertion that a boolean ring is of characteristic 2 by the correct one that its characteristic divides 2, so it is 2 except when the boolean ring is trivial, in which case it is 1.

• CommentRowNumber11.
• CommentAuthorGuest
• CommentTimeMay 22nd 2023

• CommentRowNumber12.
• CommentAuthorTodd_Trimble
• CommentTimeMay 22nd 2023

While not exactly wrong, I think it’s quite rare to refer to “characteristic 1”. Since “characteristic” usually refers to fields or perhaps algebras over fields, I’ve made an adjustment and hope no one minds.

• CommentRowNumber13.
• CommentAuthorGuest
• CommentTimeMay 22nd 2023

added statement that one needs to use the set of all decidable subsets $2^S$ instead of the set of all subsets $\mathcal{P}(S)$ in constructive mathematics to get a Boolean ring

• CommentRowNumber14.
• CommentAuthorJohn Baez
• CommentTimeMay 22nd 2023

Corrected description of the monad for Boolean rings.

• CommentRowNumber15.
• CommentAuthorGuest
• CommentTimeMay 22nd 2023

Let $\mathrm{boolean}$ denote the type of booleans or decidable truth values. Since Booleans rings are $\mathbb{F}_2$-algebras, and $\mathbb{F}_2$ is equivalent to $\mathrm{boolean}$, one could say that Boolean rings are $\mathrm{boolean}$-algebras.

• CommentRowNumber16.
• CommentAuthorTodd_Trimble
• CommentTimeMay 23rd 2023

Ba-dum pah. Or: bool-yah!

• CommentRowNumber17.
• CommentAuthorJ-B Vienney
• CommentTimeMay 23rd 2023
• (edited May 23rd 2023)

It would be nice to describe explicitly the equivalence of categories between BoolAlg and BoolRing.

It is described in this entry how the two functors act on objects but not how they act on morphisms. Even less what are the natural isomorphisms $F \circ G \cong Id$ and $G \circ F \cong Id$.

• CommentRowNumber18.
• CommentAuthorTodd_Trimble
• CommentTimeMay 23rd 2023

I’ll put something in.

• CommentRowNumber19.
• CommentAuthorTodd_Trimble
• CommentTimeMay 23rd 2023

Done.

• CommentRowNumber20.
• CommentAuthorGuest
• CommentTimeMay 23rd 2023

Quick question. Are there any non-associative unital $\mathbb{F}_2$-algebras? Or are all unital $\mathbb{F}_2$-algebras associative?

• CommentRowNumber21.
• CommentAuthorTodd_Trimble
• CommentTimeMay 23rd 2023
• (edited May 23rd 2023)

By $\mathbb{F}_2$-algebra, I mean an associative one. But there are many examples of what you want, of unital bilinear magma structures that are nonassociative.

All you need to do is take any nonassociative unital magma $M$. (For example, take $M = \{b, c, d, e\}$. and then create a multiplication table so that the first row and first column, each indexed by $e$, makes $e$ the identity element, and then fill in the rest of the multiplication table by randomly choosing entries. Most of those tables will be nonassociative.) Then, take the free vector space $\mathbb{F}_2[M]$ on $M$. The unique bilinear extension $\mathbb{F}_2[M] \times \mathbb{F}_2[M] \to \mathbb{F}_2[M]$ of the multiplication table then gives a unital bilinear magma structure which is nonassociative because $M$ is.