# Field and Algebra

What is the difference between "algebra" and "field"? In term of definition in Abstract algebra. (In probability theory, sigma-algebra is a synonym of sigma-field, does this imply algebra is the same as field?)

• The terms algebra and field indeed have different meanings in different branches of mathematics. In abstract algebra, or more precisely ring theory, a field is a commutative division ring. An algebra in this context is a vector space equipped with a bilinear product. Generally, an algebra needn't be itself a field. Commented Nov 1, 2013 at 13:58
• In the context of measures $\mathcal{A}$ is an algebra on a set $X$ if it a subset of the powerset of $X$ that is not empty and $A,B\in\mathcal{A}$ imply $A^{c}\in\mathcal{A}$ and $A\cup B\in\mathcal{A}$. Commented Nov 1, 2013 at 14:02
• This should help you understand the disparity between the two notions of algebra/field in relation to measure theory. math.stackexchange.com/questions/265893/… Commented Nov 1, 2013 at 14:20
• Thanks. The link explains. What textbook has the precise definition of "Algebra". I took a look at Mac Lane and Birkhoff but didn't find this definition. Commented Nov 2, 2013 at 4:04

An algebra over a field is like a vector space with some sort of multiplication between vectors, like 3-dimensional real space with the cross product.

A field is like a set with some notion of addition, subtraction, multiplication and division, like the field of real numbers.

Every field is an algebra because every field is a (one dimensional) vector space, but not every algebra is a field. The previous example of real 3-dimensional space with the cross product is such an algebra.

An algebra is a ring that has the added structure of a field of scalars and a coherent (see below) multiplication. Some examples of algebras:

1. M_n(F), where $F$ is any field.
2. $C(T)$, continuous real (or complex)-valued functions on a topological space $T$ (here the scalars could be either the real or the complex numbers).
3. $B(X)$, bounded operators over a Banach space $X$, with complex (or real) scalars.
4. $F[x]$, polynomials over a field $F$.

A field, on the other hand, is a commutative ring where every nonzero elements is invertible (i.e. a commutative division ring).

Note: "coherent multiplication" means that given $x,y$ in the algebra and $\alpha,\beta$ in the field, $$\alpha(x+y)=\alpha x+\alpha y,\ \ (\alpha+\beta)x=\alpha x+\beta x, \ \ (\alpha\beta)x=\alpha(\beta x),\ \ (\alpha x)y=x(\alpha y).$$