I am aware of the difference between field and ring, and I have also read other posts on this site posing similar questions on Boolean Algebra but have been unsatisfied with the focus of the questions and the responses so I am posing a basic question of my own.
Here is my question: given a Boolean Algebra B on two elements where B = ({0,1}, +, x) and the multiplication matrix 1x0=0,0x1=0,0x0=0,1x1=1, is this a multiplication matrix for a Boolean Ring, Field, or both?
Many authors on the subject, such as Halmos or Sikorski, say it is a ring since it satisfies the axioms of a ring and the idempotent property p x p = p. But I have heard some profs say it is not a ring since in the case of 1x1=1 the non-null identity element '1' is its own inverse. And since the Boolean multiplication on the two-element set has an inverse for the non-null element, it cannot be a ring but is a field. This argument is extended by pointing out that only at the next level does the Boolean algebra as field become a ring since the four-element set {<0,0>, <0,1>, <1,0>, <1,1>} produces a multiplication matrix where there are non-null elements that have no inverse.
Who is right? Is the two-element Boolean Algebra B=({1,0}, +, x}) a field, ring, or both?
Fields are types of rings. They are the commutative rings for which every non-zero element has a multiplicative inverse.
An intermediary type of ring is called an integral domain. These are commutative rings for which elements may not have multiplicative inverses, but there's still no way to multiply two non-zero elements $a,b$ and have $ab=0$. More succinctly, $ab =0 \implies a=0$ or $b=0$.
Now it turns out that if you have an integral domain with only finitely many elements, such a ring is automatically a field. To prove this, note that if $a$ is any element, $a$ defines a set function from the ring to itself by sending $a(x) = ax$. This map is injective for non-zero $a$ because $a(x)=a(y) \implies ax = ay \implies a(x-y)=0$, and since this is an integral domain and we assumed $a$ is not zero, we must have $x=y$ as desired. On the other hand, because this is an injective map from a finite set to itself, basic set theory says it must also be surjective. In particular, $a(x) = 1$ for some $x$. This $x$ is the multiplicative inverse for $a$. Since $a$ is an arbitrary element other than $0$, every such element has a multiplicative inverse, and the ring is actually a field.