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

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Comaximal ideals in a commutative ring

Let $R$ be a commutative ring and $I_1, \cdots, I_n$ pairwise comaximal ideals in $R$, e.g. $I_i + I_j = R$ for $i \neq j$. Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r ...
7
votes
2answers
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Units and Nilpotents

If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit. I've been working on this problem for an hour that I tried to construct an element $x \in R$ such that $x(u+a) ...
9
votes
3answers
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Characterizing units in polynomial rings

I am trying to prove a result, for which I have got one part, but I am not able to get the converse part. Theorem. Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots ...
31
votes
6answers
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Quotient ring of Gaussian integers

A very basic ring theory question, which I am not able to solve. How does one show that $\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}$. Extending the result: $\mathbb{Z}[i]/(a-ib) \cong ...
9
votes
4answers
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A ring is a field iff the only ideals are $(0)$ and $(1)$

Let $R$ be a commutative ring with identity. Show that $R$ is a field if and only if the only ideals of $R$ are $R$ itself and the zero ideal $(0)$. I can't figure out where to start other that I ...
16
votes
1answer
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Why is $\mathbb{Z}[\sqrt{-n}]$ not a UFD?

I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not an UFD. I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able ...
21
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5answers
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Why can't the Polynomial Ring be a Field?

I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This ...
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7answers
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How to show that every Boolean ring is commutative?

A ring $R$ is a Boolean ring provided that $a^2=a$ for every $a \in R$. How can we show that every Boolean ring is commutative? Thanks in advance.
15
votes
2answers
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A subring of the field of fractions of a PID is a PID as well.

Let $A$ be a PID and $R$ a ring such that $A\subset R \subset \operatorname{Frac}(A)$, where $\operatorname{Frac}(A)$ denotes the field of fractions of $A$. How to show $R$ is also a PID? Any ...
6
votes
3answers
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Zero divisor in $R[x]$

Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+\cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b \ne 0$ in $R$ such that ...
11
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2answers
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$A^m\hookrightarrow A^n$ implies $m\leq n$ for a ring $A\neq 0$

I'm trying to prove that if $A\neq 0$ is a commutative ring and there is an injective $A$-module homomorphism $A^m\hookrightarrow A^n$ then $m\leq n$ must necessarily hold. This is exercise 2.11 ...
31
votes
1answer
2k views

An example of a division ring $D$ that is **not** isomorphic to its opposite ring

I recall reading in an abstract algebra text two years ago (when I had the pleasure to learn this beautiful subject) that there exists a division ring $D$ that is not isomorphic to its opposite ring. ...
0
votes
1answer
127 views

Prove $R \times R$ is NOT an integral domain

I have a question, and this is it in entirety: Let R be a non-zero ring(that is, R contains at least one element other than the zero element). Prove that $R \times R$ is NOT an integral domain. ...
11
votes
3answers
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A ring that is not an Euclidean domain

Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that: Lemma: If $(D,g)$ is an ...
30
votes
3answers
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Does every Abelian group admit a ring structure?

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*} &a * (b * c) = ...
32
votes
2answers
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Why are the only division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?

Why are the only (associative) division algebras over the real numbers the real numbers, the complex numbers, and the quaternions? Here a division algebra is an associative algebra where every ...
11
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5answers
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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 ...
10
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2answers
770 views

Find all subrings Of $\mathbb{Z}^2$

This may be a simple question: Find all subrings of $\mathbb{Z}^2$.
6
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2answers
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What are the left and right ideals of matrix ring? How about the two sided ideals?

What are the left and right ideals of matrix ring? How about the two sided ideals?
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2answers
732 views

In a finite ring extension there are only finitely many prime ideals lying over a given prime ideal

I'm trying to solve the exercise 6.7 of Miles Reid's Undergraduate Commutative Algebra (pag 93). How can I prove that if $B$ is a finite ring extension of $A$, there are only finitely many prime ...
3
votes
1answer
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Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so ...
12
votes
4answers
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Every nonzero element in a finite ring is either a unit or a zero divisor

Let $R$ be a finite ring with unity. Prove that every nonzero element of $R$ is either a unit or a zero-divisor.
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3answers
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Why doesn't 0 being a prime ideal in Z imply that 0 is a prime number?

I know that 1 is not a prime number because $1\cdot\mathbb Z=\mathbb Z$ is, by convention, not a prime ideal in the ring $\mathbb Z$. However, since $\mathbb Z$ is a domain, $0\cdot\mathbb Z=\{0\}$ ...
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3answers
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The subring test

This is how the wikipedia article on subring defines the subring test The subring test states that for any ring $R$, a nonempty subset of $R$ is a subring if it is closed under addition and ...
8
votes
3answers
209 views

Ring Inside an Algebraic Field Extension

Let $E|F$ be an algebraic field extension and a ring $K$ such that $F\subseteq K\subseteq E$. It is true that $K$ is a field?
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2answers
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necessary and sufficient condition for trivial kernel of a matrix over a commutative ring

In answering Do these matrix rings have non-zero elements that are neither units nor zero divisors? I was surprised how hard it was to find anything on the Web about the generalization of the ...
17
votes
7answers
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Show that $\langle 2,x \rangle$ is not a principal ideal in $\mathbb Z [x]$

Hi I don't know how to show that $\langle 2,x \rangle$ is not principal and the definition of a principal ideal is unclear to me. I need help on this, please. The ring that I am talking about is ...
14
votes
3answers
1k views

Complement of maximal multiplicative set is a prime ideal

Let $R$ be a commutative ring with identity. I've been trying to prove the following: If $S \subset R$ is a maximal multiplicative set, then $R \setminus S$ is a prime ideal of $R$. I have spent ...
5
votes
3answers
924 views

Ring of trigonometric functions with real coefficients

Let $R$ be the ring of functions that are polynomials in $\cos t$ and $\sin t$ with real coefficients. Prove that $R$ is isomorphic to $\mathbb R[x,y]/(x^2+y^2-1)$. Prove that $R$ is not a unique ...
13
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2answers
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Is $A \times B$ the same as $A \oplus B$?

When $A, B$ are $K$-modules, then $A \times B$ is the same as $A \oplus B$. Let $A, B$ be two $K$-algebras, where $K$ is a field. Is $A \times B$ the same as $A \oplus B$? Thank you very much. ...
15
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6answers
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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 ...
4
votes
3answers
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Left inverse implies right inverse in a finite ring

Let $R$ be a finite ring, and assume $\exists x,y\in R$ such that $ xy=1$. How can I show it implies $yx=1$?
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6answers
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$ p^{\frac1n} $ is irrational if $p $ is prime and $n>1$ [closed]

How can we prove that $ p^{\frac1n} $ is irrational if $p $ is prime and $n>1$?
52
votes
5answers
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Why “characteristic zero” and not “infinite characteristic”?

The characteristic of a ring (with unity, say) is the smallest positive number $n$ such that $$\underbrace{1 + 1 + \cdots + 1}_{n \text{ times}} = 0,$$ provided such an $n$ exists. Otherwise, we ...
17
votes
3answers
891 views

Motivation for Eisenstein Criterion

I have been thinking about this for quite sometime. Eisentein Criterion for Irreducibility: Let $f$ be a primitive polynomial over a commutative unique factorization domain $R$, say $$f(x)=a_0 + ...
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votes
2answers
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Methods to check if an ideal of a polynomial ring is prime or at least radical

I am looking for methods to check whether a given ideal in $K[x_0,\dots,x_n]$ is prime. I mean something you can effectively use in some concrete non-trivial example. To be more explicit, I am working ...
25
votes
3answers
979 views

A finite ring is a field if its units $\cup\ \{0\}$ comprise a field of characteristic $\ne 2$

Suppose $R$ is a finite ring (commutative ring with $1$) of characteristic $3$ and suppose that for every unit $u \in R\:,\ 1+u\ $ is also a unit or $0$. We need to show that $R$ is a field. Is this ...
8
votes
1answer
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Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$?

If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a commutative ring, does it follow that $m \geq n$? This is obviously true for vector spaces over a field, but how would one show ...
4
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1answer
390 views

Is the number of prime ideals of a zero-dimensional ring stable under base change?

Let $A$ be a zero-dimensional ring of finite type over a field $k$ and let $X= \textrm{Spec} \ A$ be its spectrum. Note that $X$ is a finite set. Suppose that $k\subset K$ is a finite field extension ...
6
votes
2answers
776 views

$\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain

How can I prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain? Also, I need to prove that its field of fractions is isomorphic to the field of rational functions ...
6
votes
3answers
671 views

Ring of order $p^2$ is commutative.

I would like to show that ring of order $p^2$ is commutative. Taking $G=(R, +)$ as group, we have two possible isomorphism classes $\mathbb Z /p^2\mathbb Z$ and $\mathbb Z/ p\mathbb Z \times \mathbb ...
4
votes
2answers
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A ring element with a left inverse but no right inverse?

Can I have a hint on how to construct a ring $A$ such that there are $a, b \in A$ for which $ab = 1$ but $ba \neq 1$, please? It seems that square matrices over a field are out of question because of ...
1
vote
1answer
178 views

$1+a$ and $1-a$ in a ring are invertible if $a$ is nilpotent [duplicate]

Let $(A, +, \cdot)$ be a ring with $1$. An element $a\in A$ is nilpotent if there exists $n\in \mathbb{N}$ so that $a^n=0$. Show that if $a$ is nilpotent then $1+a$ and $1-a$ are invertible.
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2answers
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What do prime ideals in $k[x,y]$ look like?

Suppose that $k$ is an algebraically closed field. Then what do the prime ideals in the polynomial ring $k[x,y]$ look like? As far as I know, the maximal ideals of $k[x,y]$ are of the form ...
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votes
4answers
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Why are maximal ideals prime?

Could anyone explain to me why maximal ideals are prime? I'm approaching it like this, let $R$ be a commutative ring with $1$ and $A$ be a maximal ideal. Let $a,b\in R:ab\in A$ I'm trying to ...
5
votes
1answer
515 views

Irreducible Components of the Prime Spectrum of a Quotient Ring and Primary Decomposition

Recently I encountered a problem (the first exercise from chapter four of Atiyah & McDonald's Introduction to Commutative Algebra) stating that if $\mathfrak{a}$ is a decomposable ideal of $A$ (a ...
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2answers
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$x$ not nilpotent implies that there is a prime ideal not containing $x$.

Let $\mathscr{N}(R)$ denote the set of all nilpotent elements in a ring $R$. I have actually done an exercise which states that if $x \in \mathscr{N}(R)$, then $x$ is contained in every prime ideal ...
5
votes
4answers
487 views

Maximal ideals in $K[X_1,\dots,X_n]$

Let $K$ be a field, and $a_1,\dots,a_n \in K$. Prove that the ideal $$(X_1-a_1,\dots,X_n-a_n)$$ is maximal in $K[X_1,\dots,X_n]$. I tried proving that the only elements outside the ideal are the ...
5
votes
1answer
524 views

Constructing Idempotent Generator of Idempotent Ideal

Exercise 2.1 in Matsumura's Commutative Ring Theory reads as follows: "Let $A$ be a commutative ring and $I$ an ideal that is finitely generated and $I=I^2$. Then $I$ is generated by an idempotent." ...
4
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1answer
171 views

Commutative integral domain with d.c.c. is a field [duplicate]

If $R$ is a commutative integral domain and it also satisfies descending chain condition on its ideals then how will we show that such ring $R$ will be a field?