The following comes from Wikipedia.    It looks very strange to me since this "ring" seems to have nothing to do with the "ring" in abstract algebra. $(\cal R, \bigcup)$ is not an Abelian group (no inverse element if empty set is chosen as the identity), and $(\cal R, \backslash)$ is not a monoid (does not satisfy associativity). Am I right?   Then why this thing is called a ring?  Thanks!
122027
"Ring" and "semiring" are concepts defined both in algebra and set theory.  In Algebra  A  is a set R equipped with two binary operations + and · called addition and multiplication, that Addition (+) is abelian, Multiplication (⋅) is associative, Multiplication distributes over addition, and Multiplicative identity (1) exists.  A ,  is an algebraic structure similar to a ring, but without the requirement that each element must have an additive inverse.   In Set theory  A  is a family of sets closed under unions and set-theoretic differences.  That is, it obeys the two properties $$A \setminus B \in \mathcal{R} $$ $$A \cup B \in \mathcal{R}$$ This implies that it is also closed under intersections, $$A \cap B \in \mathcal{R}$$  A  is a non-empty collection S of sets such that   $\emptyset \in S$ If $E \in S$ and $F \in S$ then $E \cap F \in S$. If $E \in S$ and $F \in S$ then there exists a finite number of mutually disjoint sets $C_i \in S$ for $i=1,\ldots,n$ such that $E    \setminus F = \bigcup_{i=1}^n C_i$.   I wonder, are such definitions somehow linked or equivalent? or they are just coincidence that both fields of mathematics used the same terms?
54489
In $\mathbb{Z}_n$ the elements are fully partitioned between the units and the zero-divisors. I believe this is the case, am I correct?  Now, I take it this does not hold true in general, there may be rings with elements that are neither units nor zero divisors?
21580
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?
9032
I was thinking about the following problem:     Suppose R is a ring s.t. every left ideal is also right. Is R commutative?   This actually continues the easier question:     Suppose G is a group whose all subgroups are normal. Is G commutative?   The answer to the latter is negative, since one can take the quaternion group of order 8.  Edit: One can indeed take a division ring as an counterexample. I am looking for something more interesting, in particular with some proper ideals.  Thanks
41674
In self studying abstract algebra and I've come upon the following problem which I could not solve directly. For any $d\in \mathbb{Z}$ we are asked to show that $\mathbb{Z}[\sqrt d]=\{a+b\sqrt{d} ~\colon a,b\in \mathbb{Z} \}$ is an integral domain. This reduces to showing that the ring has no zero divisors which I tried to do through algebraic manipulation.  I have seen two solutions to this problem that use higher level techniques. One defines conjugate and norm and proves the claim using simple consequences of these definitions. The other says that since the given ring is a subring of the field $\mathbb{C}$ it can have no zero divisors.  I don't like either approach because the first uses ideas that haven't been introduced in the text and isn't a natural place to start (for me). The second, while both true and simple, is not satisfying to me. That $\mathbb{C}$ is a field has not been established and assuming it just makes the problem trivial.   My question is whether or not naive tactics like algebraic manipulation are capable of demonstrating the result. Am I lacking in cleverness or is such an approach impossible?
50207
Let $R$ be a ring with unity. How can I prove that group of invertible elements of $R$ is never of order $5$?   My teacher told me and my colleagues that problem is very hard to solve. I would be glad if someone can provide me even a small hint because, at this point, I have no clue how to attack the problem.
82449
I was wondering what the notation $R[a]$ really stands for, if $a\in K$, where $K$ is a ring  and $R$ is a subring of $K$.  In my book they define $\mathbb{Z}[\sqrt2]=\{a+b\sqrt2|a,b \in \mathbb{Z}\}$.  So, my guess is that $R[a] = \{P(a)\mid P \in R[X]\}$. Since for $\mathbb{Z}[\sqrt2]$ this is the case, or is this just a coincidence?
102960
Let $p$ be a prime number. Let $R$ be the subset of all rational numbers $m / n$ such that $n \neq 0$ and $n$ is not divisible by $p$. Show that $R$ is a ring. My Proof-trying. $R1$. Associativity for under addition is clear. $R2$. $\dfrac {m} {n}+0=0+\dfrac {m} {n}$, so there exist the identitiy element for under addition. $R3$. $\dfrac {m} {n} + \dfrac {-m} {n}=\dfrac {-m} {n} +  \dfrac {m} {n}$, so there exist inverse element. $R4$. $\dfrac {m} {n} + \dfrac {m'} {n'}=\dfrac {m'} {n'}+\dfrac {m} {n}$, so commutativity. So $R$ is an additive abelian group. $R5$. $\dfrac {m} {n} ( \dfrac {r} {s} + \dfrac {t} {v})=\dfrac {mr} {ns} + \dfrac {mt} {nv}$ and $( \dfrac {r} {s} + \dfrac {t} {v})\dfrac {m} {n}=\dfrac {rm} {sn}+\dfrac {r} {s}+\dfrac {tm} {vn}$ $R6.$ $R$ is associativity for under multiplication, clear. $R7.$ $1 \dfrac {m} {n}=\dfrac {m} {n} 1 = \dfrac {m} {n}$, so there exist identity element for under multiplication. Can you check my answer? If false, can you write true answer?
21057
$R$ - commutative Ring, and $a$  non-zero-divisor  $R^x$ - multiplicative Monoid? (it's called Einheitsgruppe in German, maybe unitary group?)   I started by stating, that if $a\mid b$ and $b\mid a \rightarrow a=b$.  So for $a\mid b$, $\exists x\in R$ with $xa=b$ and for $b\mid a,\exists y\in R$ with $yb=a$  Now: $b=xa=xyb=wb$, with $w=xy$.  $\stackrel{a=b}{\rightarrow} b=wa$  Now the problem that I think I have with my proof, is that R is not a given Ring. For example $a=b$ wouldn't work in $\mathbb{Z}$. Is there any other way to do the proof?
12679
I'm sure this is an obvious question, but I'm having trouble finding the right words to type into Google.  I know that the definition of a ring allows that the additive identity not have a multiplicative inverse, but is this a requirement?  Specifically, is something like $\mathbb{R}\!\left[\frac{1}{0}\right]$ such that $\frac{1}{0} \cdot 0 = 1$ a ring, or does some contradiction arise from allowing the additive identity to have a multiplicative inverse?
115283
Let $R$ be an associative ring with $1$. Prove that $R$ is a division ring if and only if $R$ has no proper left ideals.    Clearly, if $R$ is a division ring and $I\neq\{0\}$ is a left ideal, then every nonzero $z\in I$ satisfies $1=z^{-1}z\in I$, hence $I=R$.  Now if $z\in R$ is nonzero and $R$ has no proper left ideals, then $I:=\{rz:r\in R\}$ is equal to $R$ and there exists $r\in R$ such that $rz=1$. However, we still need to prove $zr=1$. How can we do this?  Thank you.
45049
a+b and ab are the usual integer addition and multiplication. You can assume that this new operation forms a ring, say R is the set of integers with these operations. Then does R have zero-divisors? And Are R and Z (the integers) isomorphic?  Any help will be appreciated!
11036
Given some Abelian group $(G, +)$, does there always exist a binary operation $*$ such that $(G, +, *)$ is a ring? That is, $*$ is associative and distributive:  \begin{align*} &amp;a * (b * c) = (a*b) * c \\ &amp;a * (b + c) = a * b + a * c \\ &amp;(a + b) * c = a * c + b * c \\ \end{align*}  We also might have multiplicative identity $1 \in G$, with $a * 1 = 1 * a = a$ for any $a \in G$. Multiplication may or may not be commutative.  Depending on the definition, the answer could be no in the case of the group with one element: then $1 = 0$. But the trivial ring is not a very interesting case. For cyclic groups the statement is certainly true, since $(\mathbb{Z}_n, +, \cdot)$ and $(\mathbb{Z}, +, \cdot)$ are both rings. What about in general? Is there some procedure to give arbitrary abelian groups ring structure?
1899
How to prove the following?  $R$ is an associative ring with identity. $R$ contains element $r$. The element is not invertible on the right and is not a left divisor of zero. Then the ring $R$ cannot be Artinian on the right.
106906
I am studying ring theory and I am having a hard time understanding a part of a proof of a certain theorem. Note that in my lecture notes, a ring is by definition commutative with $1$. Let $R$ be a ring, then the set of units of $R$ is denoted by $R^*$. Consider the following theorem:  Let $R$ be a principal ideal domain (PID) and $\pi\in R$ irreducible. For any $a,b\in R$ with $\pi|ab$ we have $\pi|a$ or $\pi|b$.  The following is the first part of the proof from my lecture notes:  Consider the ideal $(a,\pi)$. Since $R$ is a PID this ideal is generated by a single element, which we call g. Note that $g|\pi$. Hence either $g$ is a unit, or $g=\epsilon\pi$ with $\epsilon\in R^*$. In the second case, since $g|a$ we also get $\pi|a$ and our assertion is proven.  The second part of the proof covers the case in which $g$ is a unit; I understand what's going on there (under the assumption that $g$ is indeed a unit) so I've omitted it. Please note that this is the proof word by word, i.e. I have quoted it from my lecture notes. First of all, is it true that, since $(g)=(a,\pi)$ we have $\{rg\ |\ r\in R\}=\{r_1a+r_2\pi\ |\ r_1,r_2\in R\}$ and by choosing $r_1=0$ and $r_2=1$ we have $rg=\pi$ for some $r\in R$, thus $g|\pi$? If this is true, then by similar reasoning one can proof that $g|a$ (which is stated in the last sentence of the proof). Next, the proof says that $g$ is either a unit or $g=\epsilon\pi$ for some unit $\epsilon$. By definition, we see from the factorisation above that either $g$ or $r$ is a unit. The first case corresponds to the statement in the proof that $g$ can be a unit. But if $g$ is not a unit, why is $g=\pi\epsilon$ for some unit $\epsilon$? Lastly, how does $\pi|a$ follow from all that has been said in the proof? EDIT: As requested by a user in the comments, here is the second part of the proof in my lecture notes, along with the first one for easy reading:  Consider the ideal $(a,\pi)$. Since $R$ is a PID this ideal is generated by a single element, which we call g. Note that $g|\pi$. Hence either $g$ is a unit, or $g=\epsilon\pi$ with $\epsilon\in R^*$. In the second case, since $g|a$ we also get $\pi|a$ and our assertion is proven. Now suppose  that $g$ is a unit and hence $(a,\pi)=R$. In this case there exist $x,y\in R$ such that $ax+\pi y=1$. Multiply by $b$ to get $abx+\pi by=b$. Both terms on the left are divisible by $\pi$. Hence the sum $b$ is also divisible by $\pi$ which proves our theorem.
121228
My question if it possible for us to have, for example, a monoid(any algebraic structure really) with a class as the collection of the elements, instead of a set.  An example would be the class of sets and set union as addition. Clearly, it is closed under the operation, is associative and it has an identity element, so it is a monoid. Is it possible to say so?
99979
I can't think of a ring without unity that has a subring with unity. There must be some element in the parent ring that doesn't work with the subring's identity, but I'm struggling to see how that would be possible. Any suggestions?