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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5
votes
2answers
483 views

Completion of a Noetherian ring

How can we prove that if $R$ is a commutative Noetherian ring, $\mathfrak{m} = (a_1,\ldots,a_n)$ is an ideal, then the completion of $R$ at $\mathfrak{m}$ is isomorphic to ...
7
votes
1answer
111 views

Short method to prove the irreducibility of $x^7+21x^5+35x^2+34x-8$ over $\Bbb Q$

I am given a task to prove that polynomial $f=x^7+21x^5+35x^2+34x-8$ is irreducible over $\Bbb Q$. In my algebra course we learnt reduction and Eisenstein criterion. Eisenstein doesn't seem to work ...
3
votes
3answers
30 views

Prove or disprove $\mathbb{Q}[x] /(x^5-3) \cong \mathbb{Q}[x] /(x^5-9)$

Want to prove or disprove this $\mathbb{Q}[x] /(x^5-3) \cong \mathbb{Q}[x] /(x^5-9)$ as communtative rings. I can show that $x^5-3$ and $x^5-9$ are irreducible in $\mathbb{Q}$, but I cannot go from ...
0
votes
3answers
32 views

Example of ideal generated by two elements

I have an easy example on my notes that I don't understand. My teacher said that in $\mathbb{Z}$, $(2,3)=2\mathbb{Z}+3\mathbb{Z}$ is a principal ideal, because $2\mathbb{Z}+3\mathbb{Z}=\mathbb{Z}$. ...
4
votes
2answers
70 views

What are all the integral domains that are not division rings?

A commutative division ring is an integral domain. But what are all the integral domains that are not division rings? The examples I currently know are the following: $\mathbb{Z}$, $\mathbb{Z}[i]$, ...
0
votes
0answers
11 views

Show that if $R$ and $S$ are ideal of a ring $A$ then the product $R\cdot S$ is a ideal of $A$. [duplicate]

How to prove that if $R$ and $S$ are ideal of a ring $A$ then the product $R\cdot S$ is a ideal. I can't show only that if $x, y\in R\cdot S$ then $x-y\in R\cdot S$. The other axioms of ideal I ...
1
vote
1answer
17 views

Is group of units of a polynomial ring only constant polynomial which is involved in R

Let R be a integral domain(or maybe field) edit : Let R be a field. The group of units of R[x] is $$ a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots+ a_2 x^2 + a_1 x + a_0 $$(or infinity) such ...
0
votes
0answers
25 views

Commutative ring of prime power order

Suppose $R$ is a non trivial commutative ring with identity of prime power order. What can we say about the structure of $R$? If $R$ is of prime order, then $R$ is a field?
1
vote
0answers
15 views

$x^{mn} -a$ is irreducible in F[x] iff $x^m -a$ and $x^n -a$ are irreducible.

Let F be any field, a is in F and (m,n)=1. Show that $x^{mn}-a$ is irreducible in F[x] iff $x^m -a$ and $x^n -a$ are irreducible in F[x]?
0
votes
0answers
30 views

if the sum of two units is a unit, then there is an unique maximal ideal

Let $R$ be a ring with identity element. I have to proof that if the sum of two units of $R$ is a unit, then $R$ has an unique maximal ideal. But i don't see a connection. If someone could give me a ...
1
vote
0answers
23 views

For a field $K$ is $K\subset{K[X_{1},…X_{n}]}$

Let $K$ be any field and $K[X_1,...X_n]$ the ring of polynomials in $X_1,...X_n$ with coefficients in $K$. I am wondering if $K$ is a subset of $K[X_1,...X_n]$. I believe $K\subset{K[X_1]}$ since ...
2
votes
1answer
14 views

Prove that $R$ has ACCP if and only if every non-empty collection of principal ideals of $R$ has a maximal element.

Let $R$ be a commutative ring. Prove that $R$ has ACCP if and only if every non-empty collection of principal ideals of $R$ has a maximal element. Prove further that if $R$ is an integral domain and ...
1
vote
1answer
26 views

Quotient of the ring of integers of a quadratic field by the ideal generated by a split integer prime.

I am wondering about primes $p$ in $\mathbb Z$ that are split in $\mathcal O_{K}$, $K=\mathbb Q(\sqrt d)$. Let $\omega=\sqrt d$ if $d \equiv 2,3 \mod 4$ and $\omega=\frac{1+\sqrt d}{2}$ if $d \equiv 1 ...
1
vote
1answer
17 views

The norm of Gaussian integers and the irreducible element $ 1 + i $.

Note: Let $ \text{N}(a + bi) \stackrel{\text{df}}{=} a^{2} + b^{2} $. Observe that $ \text{N}(1 + i) = 2 $. Is it always true that if $ 1 + i $ divides a Gaussian integer, then the norm of $ 1 + i $ ...
0
votes
1answer
99 views
+500

Fermat's last theorem and $\mathbb{Z}[\xi]$

I heard that one can prove special cases of FLT by using unique factorization in $\mathbb{Z}[\xi]$ (whenever this is possible), where $\xi$ is a primitive $n$-th root of unity. How can one do this in ...
1
vote
1answer
19 views

How to prove subfield generated $K(u_1,u_2,..u_{n-1},u_n)=K(u_1,u_2,..u_{n-1})(u_n) $

This is problem in Hungerford chapter 5: Fields and Galois Theory. Prove $K(u_1,u_2,..u_{n-1},u_n)=K(u_1,u_2,..u_{n-1})(u_n)$ and $K[u_1,u_2,..u_{n-1},u_n]=K[u_1,u_2,..u_{n-1}][u_n] $ My ...
18
votes
7answers
4k views

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 ...
1
vote
2answers
22 views

Let $q$ be a prime congruent to 3 mod 4, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements

Let $q$ be a prime congruent to 3 mod 4, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements The field portion I understand. $\mathbb{Z}[i]$ is a PID and because $q$ is ...
4
votes
2answers
426 views

If coprime elements generate coprime ideals, does it imply for any $a,b\in R$ that $\langle a\rangle+\langle b\rangle=\langle \gcd (a,b)\rangle$?

Some users, including me, were thinking in chat about the following conditions for a commutative ring $R$ with $1$: $$(\forall a,b\in R)\;\;\langle a\rangle+\langle b\rangle=\langle\gcd(a,b)\rangle ...
8
votes
3answers
241 views

sufficient conditions for being a PID

Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, ...
1
vote
1answer
22 views

Show that $m = \pm 2$ or $m = \pm 3.$

Let $$R = \left\{\frac{a + b\sqrt{-19}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\} = \mathbb{Z} \left[\dfrac{1+\sqrt{-19}}{2} \right] = \mathbb{Z}[\alpha].$$ and define $d:R \setminus \{0\} ...
1
vote
1answer
36 views

Prove that if $z$ is good then so is $z + r$ for every $r \in R$.

Let $$R = \left\{\frac{a + b\sqrt{-19}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\} = \mathbb{Z} \left[\dfrac{1+\sqrt{-19}}{2} \right] = \mathbb{Z}[\alpha].$$ Note that $R$ is an integral ...
1
vote
0answers
37 views

Let $\mathbb{Z}[i]$ denote the Gaussian integers. The set of units of $\mathbb{Z}[i]$ is $\{\pm 1, \pm i\}.$

Let $\mathbb{Z}[i]$ denote the Gaussian integers. The set of units of $\mathbb{Z}[i]$ is $\{\pm 1, \pm i\}.$ A proof from: https://proofwiki.org/wiki/Units_of_Gaussian_Integers (Proof 2) Proof. ...
2
votes
2answers
86 views

Verify that R is a ring

Let $\alpha = \frac{1}{2}(1+\sqrt{-19}) \in \mathbb{C}$ and $R = \{a+b\alpha\mid a,b \in \mathbb{Z}\} \subseteq \mathbb{C}$. Let, $x = (a+b\alpha), y = (c+d\alpha)$ (I am good with showing $x - y \in ...
1
vote
5answers
381 views

Let $R$ be a commutative ring with 1 then why does $a\in N(R) \Rightarrow 1+a\in U(R)$?

Let $R$ be a commutative ring with 1, we define $$N(R):=\{ a\in R \mid \exists k\in \mathbb{N}:a^k=0\}$$ and $$U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.$$ Could anyone help me prove that if ...
6
votes
1answer
36 views

Is $f(x)$ reducible if $f(a)=0$

I am confused about this seemingly trivial question: If $f(a) = 0$ for some $a\in D$, then when is $f(x)$ reducible in $D[x]$? ($D$ is an integral domain). My answer: Always. Let $f(a)=0$. ...
3
votes
1answer
35 views

Show that if $r$ is nilpotent in a ring with identity, then $1-r$ is a unit in $R$ [duplicate]

Let $R$ be a ring. An element $r \in R$ is called nilpotent if $r^n=0$ for some integer $n \ge 1$. Show that if $r$ is nilpotent in a ring with identity, then $1-r$ is a unit in $R$. Proof. Recall ...
1
vote
2answers
36 views

What is $Q(x)$?

I do not really understand what $\mathbb{Q}(\pi)$ is here: Ofcourse we see that $\mathbb{Q}(\pi)$ is a field. But I have to "guesses" of what they mean, is one of them correct? 1. ...
0
votes
0answers
38 views

Find elements in the center of $n × n$ matrix ring $M_n (R) $ for any $n ≥ 2$. [duplicate]

Let $R$ be a ring. The center of $R$ is the set $C(R) = \{c ∈ R : cr = rc, ∀r ∈ R\}$. Determine elements in the center of the $n × n$ matrix ring $M_n (R) $ for any $n ≥ 2$. So, we have that ...
3
votes
0answers
30 views

Let $R$ be a ring with identity. Prove that if $1-ab$ is invertible for some $a,b \in R$, then $1-ba$ is also invertible. [duplicate]

Let $R$ be a ring with identity. Prove that if $1-ab$ is invertible for some $a,b \in R$, then $1-ba$ is also invertible. Ok, si if $R$ is a ring with unity, then we have $R$ with $1 \ne 0$ We have ...
8
votes
2answers
499 views

subideals of an ideal

If $R$ is a ring, J is an ideal in $R$, and $I$ is an ideal of $J$ (with $J$ considered as a ring), does it follow that $I$ is an ideal of $R$? That is, is $I$ necessarily closed under multiplication ...
0
votes
0answers
12 views

Direct product of Rings isomorphism

I was reading chapter five in my Abstrat algebra book about finite Abelian groups. In Proposition 6 part (1). It states that $Z_{m} \times Z_{n} \cong Z_{mn}$ if and only if $gcd(m,n)=1$. This ...
0
votes
1answer
30 views

What are some examples of principal, proper ideals that have height at least $2$?

Krull's principal ideal theorem states that in a Noetherian ring $R$, any principal proper ideal $I$ has height at most $1$. Presumably the Noetherian hypothesis is required, so what are some ...
0
votes
0answers
19 views

Semiprime group rings

$R$ is commutative semiprime ring, $(R,+)$ abelian group without torsion. Then $RG$ is semiprime. Proof by contradiction. If $x$ is nilpotent element in $RG$, then $x=r_1g_1+r_2g_2+...r_ng_n$. ...
5
votes
2answers
44 views

Uniqueness of prime ideals of $\mathbb F_p[x]/(x^2)$

What are the prime ideals of $\mathbb F_p[x]/(x^2)$? I have been told that the only one is $(x)$, but I would like a proof of this. I want to say that a prime ideal of $\mathbb F_p[x]/(x^2)$ ...
2
votes
1answer
25 views

Example of a domain where all irreducibles are primes and that is not a GCD domain

One has the following relations for a domain $R$: $R$ GCD domain $\Rightarrow$ All irreducible elements are prime $R$ PID $\Rightarrow$ $(R$ GCD domain $\land$ $R$ statisfies ACCP$)$ $R$ UFD ...
0
votes
1answer
36 views

Modules over Itself

Let $F$ be a field and let $R$ be the ring of polynomials over $F$ in infinitely many variables $x_1, x_2, \dots, x_n, \dots$. (Of course, each element of $R$, being a polynomial, will involve only ...
0
votes
1answer
65 views

Nilpotent elements in group algebra

Suppose $FG$ -- is group algebra and $F$ is field with characteristic $p>0$. $G$ - is finite $p$-group. Thus, it's clear that $(e-g)$ is nilpotent. But how to show that $(e-g)g_1$ is nilpotent for ...
1
vote
1answer
24 views

Factoring the Ring of Integers into Ideals

Let $K$ be a number field. Let $\frak p$ be a prime ideal in $\mathcal O_K$. Let $u\in \mathcal O_K$ and $m\in \mathbb N$. I've been told that $|u|_{\frak p} = |m|_{\frak p} = 1$ where $|\cdot|_{\frak ...
0
votes
1answer
20 views

Showing $\hat{A} \otimes_{A} M \cong \hat{M}$ when $M$ is a finitely generated free $A$-module.

I had a reading question on Proposition 10.13 from Atiyah-MacDonald. The proposition is the following PROPOSITION. For any ring $A$, if $M$ is finitely-generated, $\hat{A} \otimes_{A} M \rightarrow ...
0
votes
0answers
40 views

Quotient of Ideals in matrix rings

I'd like to know where could I find some info about the quotient $I:J=\{a\in R\mid aJ\subseteq I\}$ ($R$ a ring) in matrix rings? Or for example, in a matrix ring over $\mathbb{Z}$. I would like to ...
1
vote
2answers
166 views

Prove that $\mathcal{O}_3$ and $\mathcal{O}_7$ are euclidean domains

For a non-square integer $d$ such that $d \equiv 1 \mod 4,$ we define the set $$\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}.$$ Prove that ...
3
votes
1answer
33 views

Set of units in ring a group?

I am supposed to prove that given a commutative ring $R$, the set of units $R^{\times}$ is a group. I checked the axioms of a group and it all came down to noting that if $a,b\in R^{\times}$, then ...
1
vote
0answers
91 views

How to prove that an ideal can not be generated by 2 elements

In Kunz's "Introduction to commutative algebra and algebraic geometry", page 137-139, particular monomial affine curves are described. Here is the link. In case the curve is not an ideal ...
1
vote
1answer
39 views

Problems with understanding the proof of noetherian ring

If $M$ is an $R$-module, the the following are equivalent: 1. M is finitely generated 2. M satisfies the ascending chain condition 3. Every non-empty set of submodules of M contains at least one ...
0
votes
1answer
32 views

how to show that an ideal is convex [on hold]

I need to show that the ideal $J=(i)$ in $C(\mathbb R)$ where $i$ is the identity function, $C(\mathbb R)$ is the ring of all continuous functions on the real numbers, is a convex ideal.
1
vote
1answer
611 views

Smith Normal Form

Would the Smith Normal Form of the following matrix over $\mathbb Q[x]$ $$\begin{pmatrix}   (x+a)(x+b) & 0 & 0 &0 \\  0 & (x+c)(x+d) & 0 & 0 \\   0 ...
0
votes
0answers
9 views

$U_1(\mathbb{Z}G)$ is a finitely generated FC-group.

If each member in support of an element in $\mathbb{Z}G$ is centralized by a subgroup of finite index in $G$, then why does it imply that $U_1(\mathbb{Z}G)$ is a finitely generated FC-group., where ...
-1
votes
0answers
32 views

If $A$ is a submodule of $B$ and $B$ is a submodule of $C$, is $A$ a submodule of $C$?

Let $C$ be a commutative ring (with 1, if this matters). If $A$ is a submodule of $B$ and $B$ is a submodule of $C$, is $A$ a submodule of $C$? I can't really prove that it is true because it is ...
4
votes
2answers
174 views

Converse to Chinese Remainder Theorem

So as seen on this question Converse of the Chinese Remainder Theorem, we know that if $(n,m) \neq 1$, then $\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, ...