TITLE: What is difference between a ring and a field?
QUESTION [59 upvotes]: The ring axioms require that addition is commutative, addition and multiplication are associative, multiplication distributes over addition. 
A field can be thought of as two groups with extra distributivity law. 
A ring is more complex: with abelian group and a semigroup with extra distributivity law.
Is a ring a more basic structure than a field, or vice versa? What's the relation between them? What's the background why people study them?

REPLY [86 votes]: A ring is an ordered triple, $(R,+,\times)$, where $R$ is a set, $+\colon R\times R\to R$ and $\times\colon R\times R\to R$ are binary operations (usually written in in-fix notation) such that:

$+$ is associative.
There exists $0\in R$ such that $0+a=a+0=a$ for all $a\in R$.
For every $a\in R$ there exists $b\in R$ such that $a+b=b+a=0$.
$+$ is commutative.
$\times$ is associative.
$\times$ distributes over $+$ on the left: for all $a,b,c\in R$, $a\times(b+c) = (a\times b)+(a\times c)$.
$\times$ distributes over $+$ on the right: for all $a,b,c\in R$, $(b+c)\times a = (b\times a)+(c\times a)$.

1-4 tell us that $(R,+)$ is an abelian group. 5 tells us that $(R,\times)$ is a semigroup. 6 and 7 are the two distributive laws that you mention.
We also have the following items:
a. There exists $1\in R$ such that $1\times a = a\times 1 = a$ for all $a\in R$.
b. $1\neq 0$.
c. For every $a\in R$, $a\neq 0$, there exists $b\in R$ such that $a\times b = b\times a = 1$.
d. $\times$ is commutative.
A ring that satisfies (1)-(7)+(a) is said to be a "ring with unity." Clearly, every ring with unity is also a ring; it takes "more" to be a ring with unity than to be a ring.
A ring that satisfies (1)-(7)+(a,b,c) is said to be a division ring. Again, eveyr division ring is a ring, and it takes "more" to be a division ring than to be a ring. (5)+(a)+(b)+(c) tell us that $(R-\{0\},\times)$ is a group (note that we need to remove $0$ because (c) specifies nonzero, and we need (b) to ensure we are left with something).
A ring that satisfies (1)-(7)+(a,b,c,d) is a field.  Again, every field is a ring.
We do indeed have that $(R,+)$ is an abelian group, that $(R-\{0\},\times)$ is an abelian group, and that these structures "mesh together" via (6) and (7). In a ring, we have that $(R,+)$ is an abelian group, that $(R,\times)$ is a semigroup (or better yet, a semigroup with $0$), and that the two structures "mesh well".
We have that every field is a division ring, but there are division rings that are not fields (e.g., the quaternions); every division ring is a ring with unity, but there are rings with unity that are not division rings (e.g., the integers if you want commutativity, the $n\times n$ matrices with coefficients in, say, $\mathbb{R}$, $n\gt 1$, if you want noncommutativity); every ring with unity is a ring, but there are rings that are not rings with unity (strictly upper triangular $3\times 3$ matrices with coefficients in $\mathbb{R}$, for instance). So
$$\text{Fields}\subsetneq \text{Division rings}\subsetneq \text{Rings with unity} \subsetneq \text{Rings}$$
and
$$\text{Fields}\subsetneq \text{Commutative rings with unity}\subsetneq \text{Commutative rings}\subsetneq \text{Rings}.$$