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Questions tagged [integral-domain]

For questions regarding integral domains, their structures, and properties. This tag should probably be accompanied by the Ring Theory tag. This tag is not for use for questions regarding integrals in analysis and calculus.

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14 views

Is this proof of a irreducibility criterion in an integral domain correct?

This is an exercise from Grillet's "Abstract Algebra" (page $145$, proposition $10.10$). Let $R$ be an integral domain, let $I$ be an ideal of $R$, and let $\pi\colon R\to R/I$ be a canonical ...
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16 views

Rotman's “Advanced Modern Algebra”, exercise 3.97: a degree function of a Euclidean domain [duplicate]

Rotman defines a degree function of an integral domain $R$ as a function $\delta\colon R\setminus\{0\}\to\mathbb{N}$ so that $(1)$ for any $a,b \in R\setminus\{0\}$ we have $\delta(a) \leq \delta(ab)$...
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0answers
50 views

Surjective Norm functions in ED [on hold]

Let $R$ be an Integral Domain with $1\neq 0$ and $S\subset$ $\mathbb{N}\cup${$0$}. Define a Norm $N:R\setminus${$0$}$\rightarrow S$ s.t. $S$ ={ $N(r)$, g.c.d.$(N(a),N(b))$ | $\forall a,b,r\in R\...
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1answer
47 views

An Ideal I of $ \mathbb Z[\sqrt 3]$ generated by an integer prime, such that $\mathbb Z[ \sqrt 3]/I$ is not an integral domain.

Find an ideal I of $ \mathbb Z[\sqrt 3]$ generated by an integer prime (i.e. a prime number in $\mathbb Z$) such that $\mathbb Z[ \sqrt 3]/I$ is not an integral domain. Thoughts: We need to find ...
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2answers
36 views

Principal ideal in an integral domain

Let $R$ be an integral domain. How to prove that these two conditions are equivalent: $(d) = (x_1, ..., x_n)$ $d | x_1, ..., x_n$ and $\exists a_1, ..., a_n \in R, d = x_1a_1+...+x_na_n$ I know that ...
2
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1answer
63 views

The localization of the ring $\mathbb{Z} \times \mathbb{Z}$ at every prime ideal is an integral domain

I want to show that the localization of the ring $\mathbb{Z} \times \mathbb{Z}$ at every prime ideal is an integral domain. $\mathbb{Z} \times \mathbb{Z}$ is not an integral domain since $(0,1)\cdot(...
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0answers
26 views

Group algebra is domain iff doesn't contain nonzero element whose square is 0

Let $G$ be a torsion-free group and let $K$ be a field. I have to prove that the group algebra $KG$ is an integral domain if and only if it doesn't contain a nonzero element whose square is equal to 0....
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2answers
67 views

Prove $(F/R) \otimes_R (F/R)=0$ for $R$ an integral domain and $F$ its field of fractions

Suppose $R$ is an integral domain and $F=\mathrm{Frac}(R)$ is considered as an $R$-module. I want to prove that $$(F/R) \otimes_R (F/R)=0.$$ My attempt: I first considered the exact sequence $$R ...
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0answers
12 views

How to find the irreducible factors in this domain?

Find the irreducible factors of $r=1-\omega^2$ with $\omega = \frac{1}{2}(-1+\sqrt{-3})$ in $\mathbb{Z}[\omega]$ I know that i can factor $1-\omega^2$ like $(1-\omega)(1+\omega)$. But, how i can ...
6
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1answer
65 views

Integral domain which contains a copy of its fraction field

Let $R$ be an integral domain with fraction field $K$. If there exists an injective ring homomorphism from $K$ to $R$ , then is it true that $R$ is a field ? Strictly speaking, I am not saying $K \...
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0answers
22 views

$R$ UFD and integral domain $\Rightarrow$ every $a\neq 0 \in R\setminus R^\times$ is a product of prime elements

Let $a$ be as in title. Then by definition of UFD we have unique factorisation $$a=\epsilon p_1\dots p_n \quad : \epsilon\in R^\times \text{ and } p_i \in R \text{ irreducible}.$$ It is known that in ...
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1answer
55 views

Lang's *Algebra*: definition of $F[\alpha]$ and why it's an integral domain?

I am reading Lang's Algebra, namely the chapter about Fields. The first thing which confused me is the following: how he defines $F[\alpha]$? Later he defines this as the smallest subfield of $E$ ...
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1answer
69 views

The ideal generated by an irreducible element contains a prime ideal implies that the element is prime in a non-UFD?

Let $R$ be an integral domain. Assume $R$ is a factorization domain, but the decomposition into irreducible is not necessarily unique. Prove that if the ideal $(a)$ generated by an irreducible ...
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2answers
36 views

If $\sigma^{*}\left(p\left(x\right)\right)$ is irreducible in $S\left[x\right]$, then $p\left(x\right)$ is irreducible in $R\left[x\right]$.

Let $\sigma:R\to S$ be a ring homomorphism, and $\sigma^{*}$ be the induced map $\sigma^{*}:R\left[x\right]\to S\left[x\right]$ given by $\sum a_{i}x^{i}\mapsto\sum\sigma\left(a_{i}\right)x^{i}$ (this ...
2
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1answer
43 views

Is $\mathbb{Z}[X]/(I+J)$ integral domain

Let $I=\left<X-2\right>, J=\left<X+2\right>$. Is $\mathbb{Z}[X]/(I+J)$ integral domain? We have a theorem which says: $I\triangleleft P$ is prime ideal iff $P/I$ is domain. So, in this ...
2
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0answers
34 views

On ideal quotient of intersection of fractional ideals

Let $R$ be an integral domain with fraction field $K$. For a fractional ideal $I$ (i.e. $I$ is an $R$-submodule of $K$ such that $\exists 0\ne r\in R $ with $rI \subseteq R$) of $R$, define $(R:I)=\{x\...
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1answer
50 views

ideal extension in polynomial ring over quotient field

This is probably easy, but I can't seem to figure it out. I have a $K$-algebra $B$, which is a domain and an irreducible polynomial $f \in K[x]$. Why does $B[x]/f$ embed in $Q(B)[x]/f$ where $Q(B)$ is ...
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1answer
15 views

Trouble understanding proof regarding elements of an integral domain which are not product of irreducible elements.

I'm having trouble understanding the proof of a proposition regarding elements which are not product of irreducible elements in integral domains. The proposition is the following: Let $A$ be an ...
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0answers
10 views

What does this equation means to calculate the length of domain boundary?

I have a function $\phi(x,y)$ and this function have following properties: $$ \begin{align} \phi(x,y)>0&\text{ for }(x,y)\in\Omega \\ \phi(x,y)=0&\text{ for }(x,y)\in\delta\Omega \\ \phi(x,...
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0answers
29 views

If $A$ and $B$ are integral domains and $A \subseteq B$ then $A$ has characteristic $p$ iff $B$ has characteristic $p$

Here's my attempt : The unity of $A$ and the unity of $B$ are both same, so, we'll call it $1$ unambiguously. $( \Leftarrow )$ Suppose $A$ has characteristic $p$ then $p \cdot 1 = 0$. Since $1$ is ...
2
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1answer
32 views

If A is a finite integral domain and is a cyclic group with addition alone, then order of A is prime

Here's my attempt at it: Let $A$ be a finite integral domain and $a\in A$ and suppose that $A=\langle a \rangle$. Hence, $|a|=|A|$. By definition of the order of an element, we have that $|a| \cdot a ...
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1answer
31 views

Integral Domain, PID and gcd

Let $R$ be a PID and $R'$ a Integral Domain and $R\subseteq R'$. Let $a,b,d \in R $ and $d$ is a gcd of $a$ and $b$ in $R$. Then $d$ is also a gcd of of $a$ and $b$ in $R'$. Proof: $d$ is a gcd of $...
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3answers
471 views

Why is it natural to define a prime as $p|ab$ implies $p|a$ or $p|b$?

Definition: An element $p$, not zero and not a unit, is called prime if $p|ab$ implies $p|a$ or $p|b$. I am having trouble understanding why this definition of prime follows naturally. Could someone ...
2
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0answers
52 views

Are unique prime ideal factorization domains locally noetherian?

In this question I asked: "Are unique prime ideal factorization domains noetherian?". In this answer Badam Baplan pointed out that locally noetherian domains are unique prime ideal factorization ...
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1answer
100 views

Are unique prime ideal factorization domains noetherian?

Let $A$ be a domain satisfying the following condition: If $\mathfrak p_1,\dots,\mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $\mathbb N^k$, ...
3
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1answer
376 views

What will be the smallest ring containing two rings?

Let $A$ be a commutative ring with $1$. Suppose $R$ and $S$ are two subrings of $A$ containing the same multiplicative unity. Then what is the description of the smallest ring containing $R$ and $S$ ...
5
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1answer
78 views

Unique prime ideal factorization in domains?

This is a follow-up to this question. Let $A$ be a domain; let $\mathfrak p_1,\dots,\mathfrak p_k$ be distinct prime ideals of $A$ such that $\mathfrak p_i^{j+1}\ne\mathfrak p_i^j$ for all $1\le i\...
6
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1answer
142 views

Unique prime ideal factorization in noetherian domains?

[I changed the title and the body of the question. Below I explain why I did so, and paste the previous version.] Let (UPIF) (for "Unique Prime Ideal Factorization") be the following condition on a ...
2
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1answer
84 views

Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous?

Can anyone make me understand in simple language why the second condition for being an Euclidean domain is superfluous ? Why $v(a) \leq v(ab)$ is not needed? How we can deduce from the first one?
2
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1answer
79 views

Show that $f(x,y,z)=x^2-y^2z$ is irreducible in $\mathbb{C}[x,y,z]$.

Let $p\in\mathbb{C}[x,y,z]$ be defined by $p(x,y,z)=x^2-y^2z$. Goal: Prove that $p$ is irreducible. Let $I\subset\mathbb{C}[x,y,z]$ be the ideal defined by $$I:=(p).$$ My approach is to show that ...
0
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1answer
24 views

Is it possible for an integral domain to have an ideal that cannot be generated by a countable set?

This question came up when I was working on the following problem Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain: i) any two ...
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0answers
41 views

Isomorphism between quotient fields of polynomial rings [closed]

Let $R$ be an integral domain and $F$ be its field of fractions. If $X$ is a nonempty set, prove that there is a ring monomorphism from $R[X]$ to $F[X]$ that extends to an isomorphism of their ...
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0answers
13 views

How to prove an integral domain with this property is a field [duplicate]

I would like to prove the following: Let $A$ be an integral domain such that, given a descending set of ideals $I_1 \supseteq I_2 \supseteq \ldots \supseteq I_n \supseteq \ldots$, there is an $m \in \...
2
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1answer
52 views

Associate elements in non-integral domains. [duplicate]

Elements $a$ and $b$ of an integral domain are associates if $a\mathrel{\vdots}b$ and $b\mathrel{\vdots}a$ I have proved this fact. Then I tried to find out $a$ and $b$ which divide each other and ...
2
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1answer
60 views

$A\subset B\subset K$, $K$ field of fractions of $A$, $B$ flat over $A$ $\implies B=S^{-1}A\ ?$

Let $A\subset B\subset K$ be domains such that $K$ is the field of fractions of $A$, and $B$ is flat over $A$. Is there a multiplicative subset $S$ of $A$ such that $B=S^{-1}A\ ?$
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1answer
44 views

Integral Domains And Fields

How Can I Prove (a+b√2,+,*) is an Integral Domain where a and b belong to rational numbers? All I know is R is an integral Domain if it is a commutative ring and has no zero divisors.
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2answers
45 views

How to prove that $Ф(1) = 1'$ if $R'$ is an integral domain?

Since $R'$ is an integral domain , $Ф(b) = Ф(b.1) = Ф(b).Ф(1) = Ф(1).Ф(b)$ .But i can prove this only for those $r'∈R'$ for which there $∃r∈R$ such that $Ф(r)=r'$.
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1answer
36 views

Integrally Closed (Normal) Domains

Being integrally closed is an important condition in commutative algebra. I'm trying to develop an understanding of this. So, I would like to know more about its motivation and intuitive significance. ...
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0answers
45 views

Example of a Noetherian domain which is not equidimensional

A finite dimensional commutative ring $R$ with unity is called equidimensional if all its minimal prime ideals have same dimension (dimension of a prime ideal $\mathfrak p$ is defined to be Krull ...
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1answer
27 views

Non-zero Ideal in an Integral Domain is indecomposable

Need to prove that : any non-zero ideal in an integral domain is indecomposable. Now if $I=A\bigoplus B$, then $A$ and $B$ are subgroups of $I$, if I am not wrong. Does it say anything about having ...
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0answers
28 views

Rings that are UFD except for the fact that they are not integral domains.

In the definition of an UFD $R$, it is said that $R$ should in particular be integral. For me, this hints that people are not so interested in rings that are UFD except for the fact that they are not ...
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1answer
47 views

When $\mathbb{Q}[\sqrt{n}i]$ and $\mathbb{Z}[\sqrt{n}i]$ normal?

Denote $\mathbb{Q}[\sqrt{n}i]=\{a+b\sqrt{n}i | a,b \in \mathbb{Q}\} \subset \mathbb{C}$ and $\mathbb{Z}[\sqrt{n}i]=\{a+b\sqrt{n} | a,b \in \mathbb{Z}\} \subset \mathbb{C}$. The above two rings are ...
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3answers
90 views

Let $R$ be an integral domain. If $x \in R$ is prime, then $x$ is irreducible.

I am trying to understand the proof for the following theorem: Let $R$ be an integral domain. If $x \in R$ is prime, then $x$ is irreducible. Here is the proof: I typed this a while ago and I don'...
1
vote
1answer
59 views

Prove for every $i∈\Bbb Z∪{0}$ that $N(b^i)<N(b^{i+1})$ . [closed]

Let $D$ be a Euclidean domain with euclidean valuation $N$ . Suppose $b∈D$ is neither zero nor a unit. Prove for every $i∈\Bbb Z∪{0}$ that $N(b^i)<N(b^{i+1})$ . Euclidean valuation is a function: ...
1
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1answer
17 views

Irreducible Polynomial definition

I was reading Gallian's Contemporary Abstract Algebra where I came across the following : Let $D$ be an integral domain. A polynomial $f(x)\in D[x]$ that is neither the zero polynomial nor a unit ...
2
votes
1answer
29 views

If $b$ is not a prime in $R_1$ then is $R_2$ an integral domain or a field?

We have the rings $R_1, R_2$, the elements $a,b\in R_1$, the ideal $I=(a)$ of $R_1$ and a ring homomorphism $f:R_1\rightarrow R_2$. It is given that $\ker f=I$. It holds in $R_1$ that $b\mid a$ but ...
1
vote
0answers
23 views

if $p = charD \ \ $ then $ px = 0 \ \ \ \forall x \in D $.

The characteristic of an integral domain D is defined as: $charD := min\{ n \in \mathbb{N} : na = 0 , \text{for some} \ 0 \neq a \in D \} $ I want to prove that if $p = charD \ \ $ then $ px =...
4
votes
1answer
74 views

Are there any non-orientable integral domains?

Let $R$ be an integral domain. Let $R_0=R-\{0\}$ and $R^*$ be the unit group of $R$. An orientation of $R$ (my terminology) is a submonoid $N\subseteq R_0$ which intersects each associate equivalence ...
1
vote
1answer
19 views

Extending a ring hom from $R \to L$ to $K \to L$, where $K$ is fraction field of domain $R$.

Will this work as a proof? Let $R$ be a domain and $L$ a field. Let $f : R \to L$ be a ring hom. Let $K$ be the field of fractions of $R$. Then to extend $f$ to $K$ means there is a map $f^* : K ...
7
votes
0answers
111 views

$(x)$ prime ideal in $R[x]$ iff $R$ integral domain by contrapositive

I've done this proof a few ways and I like this one but since it wasn't the "official" one, I wanted to ask if anyone sees a reason it's invalid. It just makes more sense to me on a concrete level. ...