# What is difference between a ring and a field?

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?

• 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 $\mathbb{Z}$. May 5, 2012 at 4:54
• Technically, the multiplicative structure of a field is not a group, since $0$ does not have an inverse. May 5, 2012 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. May 5, 2012 at 4:58
• The Wikipedia articles on Fields and Rings appear to answer all your questions. Did you not consult them already? May 5, 2012 at 5:01
• @BillDubuque yes, you are right. I am reading them through, I should have read them through and asked the question. May 5, 2012 at 5:12

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

• Hope to ask a little bit more: "domain". I guess the answer is (1)-(7) + (a). No guarantee (d). Correct? Thanks! Sep 20, 2014 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. Feb 2, 2015 at 15:56
• the - in the (1)-(7)+(a,b,c) is a little ambiguous. Dec 10, 2015 at 13:32
• @ScottStaniewicz: $+$ is the wrong symbol. The domain of the addition function is not "$R+R$" (which has no set theoretic meaning), it's $R\times R$, because the addition function takes as input an ordered pair of reals and outputs their sum. Jul 19, 2021 at 16:43
• @ArturoMagidin ah sorry about that! my mistake, thanks for changing it back Jul 19, 2021 at 21:16

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.

• The existence of a multiplicative inverse for every nonzero element automatically implies that there are no zero divisors in a field. May 5, 2012 at 5:06
• Yes, this is true. May 5, 2012 at 5:09
• Good video at youtube.com/watch?v=WwndchnEDS4 Apr 30, 2016 at 10:54

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

• 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. May 5, 2012 at 4:58
• Yeah, you're right. I thought he said in his definition that multiplication was communitive May 5, 2012 at 5:58