Let $R$ be a commutative ring. Then its ring of idempotents $I(R)$ consists of the idempotent elements of $R$, with the same multiplication as in $R$, but with the new addition $x \oplus y := ...
I remember reading in a textbook that there can exist a homomorphism between a ring which is a boolean algebra and one which is not. Can anyone give me some example of this.
While going through a book (Lectures on Boolean algebra, Halmos) I got struck at the following question : Prove that every Boolean ring without a unit can be embedded in a Boolean ring with a unit. ...
I'm trying to solve following question: If $a^2=a$ for all $a \in R$ where $R$ is a commutative ring, then $a+a=0$. I have tried to solve this problem for a while now and I'm more or less stuck. I ...
I've proved that a Boolean ring is a Boolean algebra but I am having trouble with the converse. The operation for + is defined as the symmetric difference for elements $a$ and $b$ from the Boolean ...
Could someone explain what an atom in Boolean algebra means? I am acquainted with ring theory and group theory but not Boolean algebra. As far as I can tell from browsing around, it is something like ...
If $B$ is a finite boolean alebgra and $a_1,\ldots,a_k$ are the atoms of $B$: $\forall i$ $a_ix=a_i x$, why is $x=a_1+\ldots +a_k$
Let $B$ be a finite boolean algebra. Define for $a,b\in B$ $a\leq b$ if $ab=a$ If $x\in B$ and $a_1,\dots,a_k$ are the atoms of B (e.g. $a\neq 0$ and if $b\in B$ such that $0\leq b \leq a$ then ...
While reading a probability paper titled Gröbner bases and factorisation in discrete probability and Bayes (can't find free version, sorry), I came across the explanation: The set of indicator ...