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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?

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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

3 Answers 3

up vote 20 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 $$\text{Fields}\subsetneq \text{Commutative rings with unity}\subsetneq \text{Commutative rings}\subsetneq \text{Rings}.$$

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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 at 15:56

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

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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

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.

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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

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