Questions tagged [ring-theory]

This tag is for questions about rings, which are a type of algebraic structure studied in abstract algebra and algebraic number theory.

A ring $R$ is a triple $(R,+,\cdot)$ where $R$ is a nonempty set such that $(R,+)$ forms an abelian group, $(R,\cdot)$ forms a semigroup, and the two operations are related by the distributive laws: $a\cdot(b+c)=a\cdot b+a\cdot c$ and $(b+c)\cdot a=b\cdot a+c\cdot a$.

Important examples of rings include domains (such as the integers), fields (such as the real numbers), square matrix rings, polynomial rings, and rings of functions. Rings are studied in their own right in abstract algebra, but they are also prominently used in number theory, geometry, algebraic geometry, and logic.

Many authors require the semigroup $(R,\cdot)$ to have an identity, often denoted $1_R$ or $1$. Many other authors do not make that requirement. This difference is something that students and posters should be aware of. Scholars of the former school call the structures not necessarily having a unit element . Scholars of the latter school call $R$ a ring with identity, when $1_R$ exists. This difference of opinions has an impact on the definition of a ring homomorphism. The scholars who include the presence of $1_R$ as an axiom assume that it is preserved under ring homomorphisms. The scholars who don't insist on the existence of $1_R$ obviously cannot make this requirement.

The operation $\cdot$ does not have to be commutative, but when it is, $R$ is called a commutative ring.

There are numerous types of rings studied in different ways. An ideal in a ring is the ring-theoretic analogue of a normal subgroup of a group. The study of ideals is an important component of ring theory.

This tag often goes along with the and/or tags.

19658 questions
37
votes
1 answer

A ring isomorphic to its finite polynomial rings but not to its infinite one.

I was messing with the ring $k[x_1,\dots,x_n,\dots]$ of polynomials in numerable many variables in order to solve an exercise of Atiyah, and the following question came to me and made me curious: Is there a commutative unitary ring $A$ isomorphic…
Jimmy Page
  • 576
  • 3
  • 7
37
votes
6 answers

Proving that surjective endomorphisms of Noetherian modules are isomorphisms and a semi-simple and noetherian module is artinian.

I am revising for my Rings and Modules exam and am stuck on the following two questions: $1.$ Let $M$ be a noetherian module and $ \ f : M \rightarrow M \ $ a surjective homomorphism. Show that $f : M \rightarrow M $ is an isomorphism. $2$. Show…
Anon
  • 1,271
  • 3
  • 12
  • 24
36
votes
5 answers

Why is the commutator defined differently for groups and rings?

The commutator of two elements in a group is defined as $[g, h] = g^{−1}h^{−1}gh.$ In a ring, the commutator of two elements is $[a, b] = ab - ba.$ I'm asking because a ring is a (abelian) group under addition, so I would have expected it to be $[g,…
man_in_green_shirt
  • 2,368
  • 18
  • 36
36
votes
4 answers

Structure of ideals in the product of two rings

$R$ and $S$ are two rings. Let $J$ be an ideal in $R\times S$. Then there are $I_{1}$, ideal of $R$, and $I_{2}$, ideal of $S$ such that $J=I_{1}\times I_{2}$. For me it's obvious why $\left\{ r\in R\mid \left(r,s\right)\in J\text{ for some } s\in…
IIJ
  • 361
  • 1
  • 3
  • 3
35
votes
6 answers

Examples of prime ideals that are not maximal

I would like to know of some examples of a prime ideal that is not maximal in some commutative ring with unity.
35
votes
3 answers

What does the topology on $\operatorname{Spec}(R)$ tells us about $R$?

Let $R$ be a commutative ring with a unit. $\newcommand{\spec}{\operatorname{Spec}}\spec(R)$ denotes the set of all prime ideals in $R$, and it can be topologized using the Zariski topology. Last year I took a course in commutative algebra, and we…
Asaf Karagila
  • 370,314
  • 41
  • 552
  • 949
35
votes
7 answers

How can one talk about gcd in the context of complex numbers where order doesn't exist?

I was reading this: GCD of gaussian integers but the thing that we are calculating, the GCD, doesn't make any sense in the complex number world since there is no ordering of numbers there. So, how can we define and calculate the GCD there?
34
votes
2 answers

Why are groups "abelian" but rings "commutative"?

I have never seen, in any text, a ring whose multiplication is commutative being called an "abelian ring", even though this would make perfect sense, because this term would necessarily refer to multiplication (addition is commutative by definition,…
34
votes
7 answers

Why is a finite integral domain always field?

This is how I'm approaching it: let $R$ be a finite integral domain and I'm trying to show every element in $R$ has an inverse: let $R-\{0\}=\{x_1,x_2,\ldots,x_k\}$, then as $R$ is closed under multiplication $\prod_{n=1}^k\ x_i=x_j$, therefore…
34
votes
4 answers

$\mathbb C[X]/(X^2)$ is isomorphic to $\mathbb R[Y]/((Y^2+1)^2)$

This question led me to the following: Prove that $\mathbb C[X]/(X^2)$ is isomorphic to $\mathbb R[Y]/((Y^2+1)^2)$.
user26857
33
votes
6 answers

Is $\mathbb{Z}[x]$ a principal ideal domain?

Is $ \mathbb{Z}[x] $ a principal ideal domain? Since the standard definition of principal ideal domain is quite difficult to use. Could you give me some equivalent conditions on whether a ring is a principal ideal domain?
33
votes
1 answer

When is a group ring an integral domain

If $R$ is an integral domain (I am having $\mathbb{Z}$ or a field in mind) and $G$ a (not necessarily finite) group then we can form the group ring $R(G)$. Note that if $g^{n+1} = e$ then $(e-g)(e+g\ldots + g^n) = e - g^{n+1} = 0$. This means if…
mna
  • 615
  • 4
  • 10
33
votes
3 answers

A Laskerian non-Noetherian ring

A Laskerian ring is a ring in which every ideal has a primary decomposition. The Lasker-Noether theorem states that every commutative Noetherian ring is a Laskerian ring (as an easy consequence of the ascending chain condition). And I've found the…
Pieter
  • 453
  • 3
  • 8
33
votes
3 answers

Why is the localization of a commutative Noetherian ring still Noetherian?

This is an unproven proposition I've come across in multiple places. Suppose $A$ is a commutative Noetherian ring, and $S$ a multiplicative subset of $A$. Then $S^{-1}A$ is Noetherian. Why is this? I thought about taking some chain of…
Buble
  • 1,509
  • 12
  • 16
32
votes
2 answers

Motivation for the ring product rule $(a_1, a_2, a_3) \cdot (b_1, b_2, b_3) = (a_1 \cdot b_1, a_2 \cdot b_2, a_1 \cdot b_3 + a_3 \cdot b_2)$

In a lecture, our professor gave an example for a ring. He took it out of another source and mentioned that he does not know the motivation for the chosen operation. Of course, it's likely that somebody just invented an arbitrary operation…
Qi Zhu
  • 6,822
  • 1
  • 14
  • 34