Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ring more basic or field more basic, what's the relation between them? what's the background why people study them?

share|cite|improve this question
A field is a ring where the multiplication is commutative and every nonzero element has a multiplicative inverse. There are rings that are not fields. For example, the ring of integers $\mathbb{Z}$ is not a field since for example $2$ has no multiplicative inverse in $\mathbb{Z}$. – Henry T. Horton May 5 '12 at 4:54
Technically, the multiplicative structure of a field is not a group, since $0$ does not have an inverse. – Arturo Magidin May 5 '12 at 4:56
Note that every group is also a semigroup, so saying "two groups" is *more complex" than saying "a group and a semigroup"; it's 'easier' to have a group and a semigroup than two groups, because whenever you have two groups you also have "a group and a semigroup", but you can have a group and a semigroup and not also have two groups. – Arturo Magidin May 5 '12 at 4:58
The Wikipedia articles on Fields and Rings appear to answer all your questions. Did you not consult them already? – Bill Dubuque May 5 '12 at 5:01
@BillDubuque yes, you are right. I am reading them through, I should have read them through and asked the question. – zinking May 5 '12 at 5:12
up vote 35 down vote accepted

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:

  1. $+$ is associative.
  2. There exists $0\in R$ such that $0+a=a+0=a$ for all $a\in R$.
  3. For every $a\in R$ there exists $b\in R$ such that $a+b=b+a=0$.
  4. $+$ is commutative.
  5. $\times$ is associative.
  6. $\times$ distributes over $+$ on the left: for all $a,b,c\in R$, $a\times(b+c) = (a\times b)+(a\times c)$.
  7. $\times$ 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 $$\text{Fields}\subsetneq \text{Commutative rings with unity}\subsetneq \text{Commutative rings}\subsetneq \text{Rings}.$$

share|cite|improve this answer
Hope to ask a little bit more: "domain". I guess the answer is (1)-(7) + (a). No guarantee (d). Correct? Thanks! – sleeve chen Sep 20 '14 at 14:11
Great presentation to succintly lay out the distinctions starting from the core properties of the ring's (1-7), adding the "extra credits" a-d to distinguish the terms of field, abelian/division groups. – javadba Feb 2 '15 at 15:56
the - in the (1)-(7)+(a,b,c) is a little ambiguous. – alvin Dec 10 '15 at 13:32

There's a whole range of algebraic structures. Perhaps the 5 best known are semigroups, monoids, groups, rings, and fields.

  • A semigroup is a set with a closed, associative, binary operation.
  • A monoid is a semigroup with an identity element.
  • A group is a monoid with inverse elements.
  • An abelian group is a group where the binary operation is commutative.
  • A ring is an abelian group (under addition, say) that happens to have a second closed, associative, binary operation as well. And these two operations satisfy a distribution law. (You may or may not require rings to have an identity with the second operation)
  • A field is a ring where both operations commute, where every element has both an additive (i.e. the first operation) and a multiplicative (i.e. the second operation) inverse (and thus there is a multiplicative identity), and the extra requirement that if $xy = 0$ for some $x \not = 0$, then we must have $y = 0$ (we call this having no zero-divisors).

People study these, and maps between them, because it is stunning how often things can be given a group or ring-like structure. So knowing how these things behave carries a lot of information about many things.

share|cite|improve this answer
The existence of a multiplicative inverse for every nonzero element automatically implies that there are no zero divisors in a field. – Henry T. Horton May 5 '12 at 5:06
Yes, this is true. – mixedmath May 5 '12 at 5:09
Good video at – SIslam Apr 30 at 10:54

A field has multiplicative inverses, rings don't need to have that- Just additive ones. Rings are the more basic object. ${Fields}\subset {Rings}$

share|cite|improve this answer
Note that a ring such that every nonzero element has a multiplicative inverse is just a skew field or division ring. Fields are defined to be commutative under multiplication. – Henry T. Horton May 5 '12 at 4:58
Yeah, you're right. I thought he said in his definition that multiplication was communitive – Chris Dugale May 5 '12 at 5:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.