# Are all rings groups? [closed]

Also are all groups rings?

The book I'm reading is saying that it's not necessary a ring if you have a group. However, that is strange because rings are used to define groups. A ring is group that is a monoid. Yet how can't it be a ring?

Any decent books on ring theory? the book I'm using Langs Algebra is confusing as hell. Liek don't even know what a group is from that book. Like what is a field? It mentions it without giving an example of a field, what the hell is a monoid?. Is Z a field? like it looks like a field.

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A group is a set G combined with binary operator *, and a ring is a set G combined with two binary operators +,$\times$, where the G,+ is a commutative group. So rings are defined using groups. – Dimitri Surinx Sep 22 '11 at 17:09
What is your definition of a ring? Usually a ring is a set $R$ equip with two operations $+$ and $\cdot$. $(R, +)$ is an abelian group, and $(R, + \cdot)$ has other properties such as associativity of the multiplication and distributivity. Also I think every group is a monoid. – William Sep 22 '11 at 17:10

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Also: If you take 2 minutes and browse the site a bit, you will immediately see that no one says «what the hell is X?» here. Please be civil. In any case, it is rather clear that you do not know the basics of abstract algebra, so this line of questioning is more or less useless (among other things, Lang was not writing the book with you in his intended audience). You should simply pick some other book, which starts at a more basic level. – Mariano Suárez-Alvarez♦ Sep 22 '11 at 17:16

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-1 for the way the question is asked. Moreover, you could easily find clear definitions of group, ring, field, etc., by looking at, e.g., Wikipedia. – Keenan Kidwell Sep 22 '11 at 17:17

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I feel like complexity is a troll... – Dimitri Surinx Sep 22 '11 at 17:19
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## closed as not a real question by Qiaochu YuanSep 22 '11 at 19:00

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

## 2 Answers

A ring is a set $R$ with two binary operations, usually dentoed by $+$ and $\times$, such that $(R,+)$ is an abelian group, $(R,\times)$ is a semigroup (if you like your rings to have identities, then you require it to be a monoid), and such that the two operations are connected via the distributive laws: $$a\times(b+c) = (a\times b)+(a\times c)\quad\text{and}\quad (x+y)\times z = (x\times z) + (y\times z)\quad\text{for all }a,b,c,x,y,z\in R.$$

So given a ring $(R,+,\times)$, you get a group by "forgetting" about the operation $\times$; and you get a semigroup/monoid by "forgetting" about the operation $+$. In that sense, every ring is also a group (under addition).

It is wrong to say "A ring is a group that is a monoid", because you are not precise enough in refering to the operations, and to the connection between the two operations.

I am also confused by your claim that "rings are used to define groups". How so? A group is defined to be a set $G$ together with an operation $\cdot$ that is associative, has a two-sided identity element, and has two-sided inverses for each element. No notion of ring is harmed in the production of this definition. It is true that many structures that are in fact rings are used to provide examples of groups, but you don't define groups in terms of rings.

Lang's Algebra defines groups on page 7 (Revised 3rd Edition, Springer-Verlag GTM 211) as "a monoid such that for every element $x\in G$ there exists an element $y\in G$ such that $xy=yx=e$. It defines monoid in page 3 as a set $G$ with a "law of composition" (a binary operation) that is associative and has a unit element.

Lang defines "field" on page 84: a field is a commutative ring (the operation $\times$ is associative) in which there is an identity for $\times$, denoted $1$, such that $1\neq 0$ ($0$ is the identity for $+$), and in which every nonzero element has a multiplicative inverse: for every $a$, if $a\neq 0$ then there exists $b$ such that $a\times b = b\times a = 1$. $\mathbb{Z}$ is not a field, because $2$ does not have a multiplicative inverse. It is a ring (in fact, an integral domain).

Common examples of fields are: $\mathbb{Q}$, with the usual addition and multiplication; $\mathbb{R}$ with the usual addition and multiplication; $\mathbb{C}$, with the usual addition and multiplication; $\mathbb{Z}/p\mathbb{Z}$, the integers modulo $p$ with $p$ a prime, using modular addition and modular multiplication.

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What is Q and R? Are they just numbers? – complexity Sep 22 '11 at 17:34
Q and R are the set of all rational numbers and the set of all real numbers, respectively.  C refers to complex numbers.  Z refers to integers. – Vivek Viswanathan Sep 22 '11 at 18:13
$\mathbb{N}$ine $\mathbb{Z}$ulu $\mathbb{Q}$ueens $\mathbb{R}$uled $\mathbb{C}$hina: the natural numbers are contained in the integers are contained in the rationals are contained in the reals are contained in complex numbers. – anon Sep 22 '11 at 18:39
@anon: Heh, I wanted to remember the quaternions as well back in the day, so I came up with "$\mathbb N$o $\mathbb Z$ulu $\mathbb Q$ueens $\mathbb R$ate $\mathbb C$ountless $\mathbb H$ours"... – J. M. Sep 22 '11 at 19:22

I hope this doesn't confuse you further, but there is also such a thing as a "group ring". Given any group $G$ and a ring $R$, you define $R[G]$ as the set of "linear combinations of elements of $G$ with coefficients in $R$", i.e. elements of $R[G]$ are sums of terms $r g$ with $r \in R$ and $g \in G$. Multiplication is defined by the distributive laws and $(r g) \cdot (s h) = (r s)(g h)$ for $r, s \in R$ and $g, h \in G$.

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