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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0
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1answer
44 views

Show that $R = \mathbb{Z}[\sqrt{-17}]$ is not a Euclidean ring.

Show that $R = \mathbb{Z}[\sqrt{-17}]$ is not a Euclidean ring. To do this I tried showing that the ring is not a principal ideal domain. I wonder if this is enough and how to actually show that it is ...
0
votes
2answers
60 views

Help with understanding quotient ring structure

Let $R$ be the ring $\mathbb{Z}[x]/((x^2+x+1)(x^3+x+1))$ and $I$ be the ideal generated by $2$ in $R$. What is the cardinality of the ring $R/I$? I am having a hard time understanding what the ring ...
-1
votes
2answers
21 views

Definition of irreducible element of a ring

I found in my notes the following definition: Let $r\neq 0$, $r$ non-invertible. $r\in R$ is called irreducible iff $r=a\cdot b$ with $a,b\in R$ then either $a\in U(R)$ or $b\in U(R)$. Why does ...
1
vote
1answer
41 views

Proving $\Bbb Z/p\Bbb Z\cong R/pR$

Let $R$ be a ring of square-free order $n$. If $p \mid n$ then $\Bbb Z/p\Bbb Z\to R/pR$ is a well-defined isomorphism. I'm really unsure how to approach this problem. So we need to show that if $...
0
votes
1answer
42 views

Certain Subset of a Ring [closed]

Does there exist an infinite ring $R$ with finitely many units and an infinite $S \subseteq R$ \ $\left\{0 \right\}$ such that: There exists finite $X \subseteq S$ such that for every $y \in S$, ...
7
votes
4answers
407 views

Is my understanding of quotient rings correct?

Amidst all the rigorous constructions of quotient rings involving equivalence relations and ideals, I feel that I have finally grasped what a quotient ring is. I have applied this intuition to a few ...
0
votes
1answer
31 views

For a field $K$, show that $f(x)=x^4+x^2+1\in K[x]$ is not a unit and not irreducible.

For a field $K$, show that $f(x)=x^4+x^2+1\in K[x]$ is not a unit and not irreducible. What I tried: To show that $f$ is not a unit I did the following. Suppose that $f$ is a unit, then there exists ...
2
votes
2answers
50 views

Question on splitting field and irreducible polynomials.

Let $K$ be a field, and consider a monic irreducible polynomial $f(x) \in K[x]$. Denote $d = \deg(f)$, and let $g(x) = f(x^2)$. Furthermore, let $\alpha \in \Omega^g_K$ (the splitting field of $g$ ...
0
votes
0answers
29 views

Taylor's expansion in polynomial ring

While looking proof of Hilbert's Nullstelllensatz in M. Artin's Algebra, I faced some problems. See below: The last expression for $f(x)$ is called in the book as Taylor's expansion. ...
-4
votes
2answers
33 views

Showing the kernel of $f:\mathbb{Z} \rightarrow \mathbb{Z}_n, f(x)=[x]_n$ is the ideal $n\mathbb{Z}$? [closed]

How to display that the kernel of $f:\mathbb{Z} \rightarrow \mathbb{Z}_n, f(x)=[x]_n$ is the ideal $n\mathbb{Z}$? I cannot understand from reading "the kernel is $n\mathbb{Z}$" that it really is the ...
2
votes
0answers
30 views

Books about multivariate polynomials

I'm looking for a book on multivariate polynomials, preferably a monograph (could also be a chapter inside another book). I'm interested in what can be said about roots, factoring, irreducibility, ...
3
votes
2answers
69 views

Ring of order $n$ is isomorphic to $\Bbb Z/n\Bbb Z$, with $n$ square-free

Let $R$ be a ring of order $n$ and suppose $n$ has no square in its prime decomposition. How do I see that $R$ is isomorphic to $\Bbb Z/n\Bbb Z$? I bet that the map $\Bbb Z \to R, \, 1\mapsto 1_R$ ...
1
vote
2answers
67 views

Ring with four solutions to $x^2-1=0$

I am looking for a ring $R$ in which $2$ is invertible and there are four solutions to $x^2-1=0$. $R=\Bbb Z/8\Bbb Z$ has the four solutions $1,3,5,7$ to $x^2-1=0$, but $2$ is not invertible.
6
votes
0answers
142 views

Can we characterize all infinite Euclidean-domains having exactly one invertible element?

$\mathbb Z_2$ and $\mathbb Z_2[x]$ are two euclidean-domains having exactly one invertible element ; my question is ; Can we characterize all euclidean domains $D$ having exactly one invertible ...
1
vote
1answer
30 views

unit and proper divisor of zero in a ring

unit and proper divisor of zero A unit in a ring R cannot be a proper divisor of zero. Let x ∈ R be a unit. Hence there exists a y ∈ R such that x · y = y · x = 1. Suppose x · w = z for some w ∈ R.(...
0
votes
1answer
46 views

How to show the dimension of the vector space K[X]/fK[X]?

Let K be a field and f$\neq$0 $\in$ K[X] a polynom. a) Show that the Ring K[X]/fK[X] is a K-vector space with the dimension n=deg(f) b) f is called irreducible, if for g,h \in K[X] we have f=g*h $\...
2
votes
1answer
26 views

Singular ideal containing a given nilpotent ideal

Let $R$ be a ring with identity, and $Z(R_R)$ be the singular ideal. Is it true that any nilpotent ideal of $R$ lies in $Z(R_R)$? It is well known that any central nilpotent element would belong to ...
1
vote
2answers
49 views

Not fully understanding polynomial quotient rings.

This is my (informal) understanding of a quotient ring. I understand that this is very flimsy, but I hope you can get the main idea. You have some ring $R$ and you want to quotient out an ideal $I$...
0
votes
1answer
87 views

If $R$ is an integral domain and $R[x]$ is an euclidean domain, then $R$ is a field [closed]

Is this obvious? I cannot see that this is true. The converse is fairly obvious though. I tried to show $(x)$ is a maximal ideal and try the quotient but failed. I will appreciate any help.
0
votes
1answer
54 views

If $A$ is an integral domain with a finite number of primes then $Q(A)=A_a$ for some $a \in A$ [closed]

If $A$ is an integral domain with a finite number of prime ideals is it possible to get the field of fractions localizing only by a set $\{a^k\}$?
2
votes
0answers
40 views

Finding units in quadratic integer rings

I want to find the units in $\mathbb{Z}[\alpha]$, where $\alpha=\frac{1+\sqrt{-11}}{2}$. One can of course use norms to find the units in quadratic integer rings of the form $\mathbb{Z}[\sqrt{D}]$ ...
0
votes
1answer
22 views

Restricting the quotient map of rings to a subring

When $q$ maps $R$ to $R/I$ and $p$ is the restriction of $q$ to a subring $A$ of $R$, why is the image of $p$ $(A+I)/I$? $q$ maps $r$ to $r+I$, so shouldn't $p$ map $a \in A$ to $a+I$, so image of $p$...
3
votes
1answer
65 views

Bijection between compact space $K$ and maximal ideals of real-valued functions on $K$

Let $K$ be a compact topological space, and denote by $R$ the ring of continuous functions $K \to \mathbb{R}$, with addition and multiplication defined pointwise. We prove that there is a bijection ...
0
votes
2answers
46 views

Cardinal of quotient rings of gaussian integers. [duplicate]

It is known that $\mathbb{Z}[i]$ is a PID and that $\mathbb{Z}[i]/(a+bi)\mathbb{Z}[i]$ is finite for all $(a,b) \in \mathbb{Z}^2\backslash \{(0,0)\}$. My question : Is there any result on the ...
1
vote
0answers
48 views

Converse to Chinese remainder, domain, isomorphism as rings

The converse to CRT asks: does $R/(I \cap J)\simeq R/I \times R/J$ imply $I+J=R$? For me $\simeq$ is an abstract ring isomorphism, sending $1$ to $1$, not necessarily $R$ linear, i.e., one of $R$--...
0
votes
2answers
22 views

Division ring if and only if it has no proper left ideals.

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 ...
1
vote
1answer
47 views

Any subring of $A$ is an ideal. If $A$ is an integral domain then $A$ is commutative

Any subring of $A$ is an ideal. If $A$ is an integral domain then $A$ is commutative. Is my proof correct? So let $a$ and $b$ nonzero elements of $A$. $C(a)=\{ x\in A \mid ax=xa\}$ Is a subring ...
0
votes
3answers
63 views

Show that $R_P$ has a unique maximal ideal

Problem is: Let $R$ be a commutative ring and let $P$ be a prime ideal. (a) Prove that the set of non-units in $R_{P}$ is the ideal $P_{P}$. (b) Deduce that $R_{P}$ has a unique ...
3
votes
0answers
50 views

How quickly can we find a value that has large multiplicative order modulo $n$?

If we're trying to find an element modulo $n$ that has multiplicative order at least $\sqrt{n}$, how quickly can we do this? We don't know if $n$ is prime or composite, only that $n$ definitely has a ...
-1
votes
2answers
65 views

If $I$ and $J$ are ideals in a ring $R$ with $1$ such that $I+J = R$, show that $I^m$ and $J^n$ are co-maximal for all $m,n \in \mathbb{N}$ [duplicate]

If $I$ and $J$ are ideals in a ring $R$ with 1 which are co-maximal, i.e $I+J = R$, show that $I^m$ and $J^n$ are co-maximal for all $m,n$ in $\mathbb{N}$ Work done: Should I proceed using Zorn'...
0
votes
0answers
25 views

Inverse elements in a certain monoid

Let $R$ be a ring with unity, and $Z(R_R)$ be its right singular ideal, i.e. the set of elements of $R$ whose right annihilators are essential in the right module $R_R$. My question: If $x\in Z(...
2
votes
4answers
75 views

The principal ideal $(1+i)$ in $\mathbb{Z[i]}$ is maximal.

I'm trying to prove that the principal ideal $(1+i)$ in $\mathbb{Z[i]}$ is maximal. My approach: $m+in \equiv (m-n)$ mod $(1+i), \forall m,n \in \mathbb{Z}$ and $2 \equiv 0$ mod $(1+i)$ then what?
8
votes
3answers
278 views

What is an easy example of non-Noetherian domain?

Keep in mind, I'm strictly an amateur, though a very old one. I learned about imaginary numbers barely two years ago and ideals a year ago, and I'm still decidedly a novice in both topics. In the ...
0
votes
2answers
86 views

Proving $\mathbb R[x]/\langle 1+x^2\rangle$ $\cong$ $\mathbb C$ without using 1st isomorphism theorem

I've seen many the proofs of this by making use of First isomorphism theorem, by considering the map,$$\phi:\mathbb R[x]\rightarrow\mathbb C$$ defined by $\phi(a+bx)=a+bi$. My questions are ...
0
votes
0answers
26 views
0
votes
1answer
30 views

If $A$ a commutative ring has identity element then $ ann(A)=\{0\}$

If $A$ a commutative ring has identity element then $ ann(A)=\{0\}$ Why is the identity element necessary? Since $a\cdot 0=0$ for all $a\in A$. And how can I prove that $ann(\emptyset)=A$? Every ...
2
votes
1answer
61 views

$A$ local Noetherian ring with principal maximal ideal implies PIR?

Suppose that $A$ is a local Noetherian ring with principal maximal ideal. Can we prove that every ideal of $A$ is principal? I tried to exploit the Noetherian property on the set of non-principal ...
1
vote
2answers
54 views

Let $R$ be a ring, $S$ a subring and $I$ an ideal. If $R$ is Noetherian, are then $S$ and $R/I$ also Noetherian?

Let $R$ be a ring, $S$ a subring and $I$ an ideal. If $R$ is Noetherian are then $S$ and $R/I$ also Noetherian? I have done the following: $R$ is Noetherian iff each increasing sequence of ideal $...
0
votes
2answers
76 views

Is the only homomorphism $\phi:\mathbb{R} \rightarrow \mathbb{R}$ onto the ring of reals the identity? [duplicate]

Let $\phi:\mathbb{R} \rightarrow \mathbb{R}$ be a ring homomorphism onto. Prove or disprove: $\phi(r)=r$ for all $r\in \mathbb{R}.$ Attempt: I believe it is true! Since $\phi(1)=\phi(1\cdot 1)=\phi(...
3
votes
1answer
44 views

Collecting examples of all rings R such that $x^3=x \forall x \in R$

$\mathbb{Z_{3}}$ or its Cartesian products.. The Boolean ring R which is defined as a ring in which every element is idempotent, that is $ \forall x \in R$ we have $x^2=x$ and hence $x^3=x^2x=xx=x$ in ...
1
vote
1answer
18 views

Let R be a Boolean ring with unity. Show that $0$ is the only nilpotent element and $1$ is the only unit.

A ring is said to be Boolean if $x^2=x$ for all $x \in R$ . I tried it this way: let $x \neq 0$ , $x^n=0$ and $x^2=x$ so $x^{2n}=x^{n}=(-x)^n$ How do I show that $x=0$? For unit part: Let x be a ...
1
vote
1answer
28 views

Proposition about rings of fractions

This is taken from Atiyah-Macdonald's Commutative algebra book page 41. Someone please explain to me what is the meaning of "$a$ meets $S$". This is the first time I'm seeing this in the book.
3
votes
0answers
49 views

How a complex root $\eta$ of $x^2 + x + A$ affects the ring $\mathbb{Z}[\eta]$

While reading a statement in P. Pollack's Not Always Buried Deep: A Second Course in Elementary Number Theory I came across a statement that seemed obvious and I am wondering if I am oversimplifying ...
2
votes
2answers
54 views

I need to prove that $\langle n \mathbb{Z},m\rangle =\langle m\mathbb{Z},n\rangle =d\mathbb{Z}$

Let $n, m\in \mathbb{Z}$ and $d=gcd(n,m)$, prove that: $$\langle n \mathbb{Z},m\rangle =\langle m\mathbb{Z},n\rangle =d\mathbb{Z}$$ What is $\langle n\mathbb{Z},m\rangle$ here? The intersection of ...
2
votes
1answer
72 views

$A^\times/k^\times$ is a free $\mathbb{Z}$-module of rank of at most $r - 1$

Consider an algebraically closed field $k$, a finite extension $K$ of $k(T)$, the integral closure $A$ of $k[T]$ in $K$, the integral closure $A'$ of $k[1/T]$ in $K$, and the integral closure $A''$ of ...
2
votes
2answers
56 views

Basic question about ideals in a polynomial ring.

I would be very grateful if someone would verify or refute the following solution. Many thanks! Q) Find infinitely many distinct ideals of $\mathbb{C}[X,Y]$ which contain the principal ideal $(X^3-...
0
votes
1answer
33 views

Proof that primary submodules of $R$ are primary ideals of $R$

I want to prove this: Let $R$ be a commutative ring with identity. If $Q$ is a primary submodule of $R$ (as an $R$-module), then $Q$ is a primary ideal. $Q$ is a primary submodule of $R$ if $r \in R$...
4
votes
2answers
92 views

Element of a ring without unity which divides every other element

Question. Is there an example of a ring $R$ (commutative or not) without unity and an element $x \in R$ such that for every $y \in R$ there exists a $z \in R$ such that $y = x z$? In other words, is ...
-1
votes
1answer
36 views

Let $f\in \mathbb{Z}_n [[X]] $ be a non-zero power series all of whose coefficients are nilpotent. Show that $f$ is nilpotent. [closed]

Let $f\in \mathbb{Z}_n [[X]] $ be a non-zero power series all of whose coefficients are nilpotent. Show that $f$ is nilpotent. I don't know how to proceed.. Thanks for any help!
1
vote
0answers
24 views

On a stronger property than being an Armendariz ring

A ring $R$ is said to be Armendariz if $f(x), g(x) \in R[x]$ are such that $f(x)g(x) = 0$, where $f(x) = a_nx^n + \dots a_0, g(x) = b_mx^m + \dots + b_0$, then $a_ib_j=0$ for all $i,j$. In other ...