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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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1answer
61 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?
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3answers
66 views

Show that $F$ is isomorphic to $\mathbb{Z}_p$ for some prime $p$ [closed]

Here is the question: Suppose that $F$ is a field and there is a ring homomorphism from $\mathbb{Z}$ onto $F$. Show that $F$ is isomorphic to $\mathbb{Z}_p$ for some prime $p$. Please help!!!
2
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1answer
63 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
21 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 ...
1
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0answers
19 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
12 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
33 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
47 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
43 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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0answers
13 views

A question about the integral extension of a integral domain.

Suppose that $R$ is an integral domain, $S$ is integral over $R$, that is to say, any elements in $S$ satisfies a monic polynomial with coefficients in $R$. (Here $S$ is a commutative ring with ...
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2answers
35 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 $Ф(r)=r'$.
2
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1answer
27 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
34 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
13 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
25 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
40 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 ...
1
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3answers
40 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
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1answer
57 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: ...
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1answer
13 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
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1answer
28 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 ...
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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
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1answer
67 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
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1answer
16 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
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0answers
71 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. ...
0
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1answer
31 views

If D is integral domain, then $b_m$ is unit

Show that if $D$ is integral domain and $b(x) = b_0 + b_1x + ... + b_mx^m$ and $b_m \ne 0$ satisfies for all $a(x) \in D[x]$, $q(x)$ and $r(x)$ $\in D[x]$ exist such that $a(x) = b(x)q(x) + r(x)$ ...
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1answer
22 views

show the ring is an integral domain [duplicate]

Given A=$\Bbb Z$, a commutative ring with element $1$ with $\oplus$ and $\odot$, defined as: $a\oplus b=a+b-1$ and $a\odot b=a+b-ab$, show the ring is an integral domain. It is known that the ...
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1answer
24 views

Integral domain $0$ element

An integral domain is defined as a commutative ring with 1 that it's elements comply $x*y=0 \Rightarrow x=0 $ or $y=0$ Is this element $0$ the one of $\Bbb Z$ or is the neutral element of the set ...
0
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1answer
38 views

What is the volume of a domain?

In this paper, page 240, column 2, after the equation (1.8) it is written that $|\Omega|$ is the volume of $\Omega$, which is the discretized domain containing the pixels of an image where we ...
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1answer
40 views

Example for being an integral domain is not a local property.

Let $K$ be a field and $R=K\times K$ the product ring. We know this ring is not integral domain. But for all $P \in Spec(R)$, $R_P$ (the localization of $R$ at $P$) is integral domain. I know this ...
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1answer
47 views

Can we have $A \cong \mathrm{Frac}(A)[X]$?

Is there an integral domain $A$, with field of fractions $K$, such that $A$ is isomorphic to $K[X]$ ? Such a ring $A$ has to be a PID, which is not a field (i.e. PID of Krull dimension $1$), and is ...
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1answer
23 views

Why the integers form an integral domain?

This is the definition of integral domain A ring that is commutative under multiplication, has a multiplicative identity element, and has no divisors of 0. 0 is an integer. 0 can divide any ...
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1answer
34 views

Proof check and Are these conditions sufficient for p prime to imply p irreducible?

Let $D$ be an integral domain with a multiplicative norm $N:D \rightarrow \mathbb{N}$ such that $$ p|b \implies N(p) \leq N(b) \ \ \ \forall p,b \in D. $$ with equality only if $b,p$ are associates....
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2answers
33 views

Reducing the degree of a polynomial

Problem: Let R be an Integral domain. Let p $\in$ R[x]. Suppose deg(p) > 0. Show that multiplying p by other polynomials cannot bring down the degree. Attempt: This makes good intuitive sense because ...
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1answer
27 views

Are these boolean rings integral domains?

Let X be a set and let P(X) denote the Boolean ring whose elements are the subsets of X, with addition being symmetric difference and multiplication being intersection. Is P({1}) an integral domain? ...
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1answer
24 views

Ideal generated by $x_1 - a_1, \dots, x_k - a_k$ is prime in polynomial ring over integral domain

I would like to show the ideal $P_k = \langle x_1 - a_1, \dots, x_k - a_k \rangle$ of $R = F[x_1,\dots,x_n]$, where $F$ is a field, $k \leq n$ and $a_i \in F$, is a prime ideal. In the case where all ...
1
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1answer
33 views

Showing that the Krull dimension of a certain domain is 1.

I have to solve the following exercise. Let $R$ be an integral domain that is not a field. Suppose that for all $x\in R\setminus \{0\}$ we have the quotient $R/(x)$ is a finite set. Show that $\dim R ...
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0answers
29 views

The relation between Noetherian valuation domain, principal ideal domain, factor ring and Dedekind

I am looking for relationships between Noetherian valuation domains, principal ideal domains, factor rings and Dedekind domains. Every Noetherian valuation domain is principal ideal domain also is ...
4
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1answer
37 views

If every proper overring of $R$ is a valuation domain, then $R$ is a valuation domain?

Let $R$ be a domain and $K$ be its quotient field. Is it true that if every ring $W$ with $R \subsetneq W \subset K$ is a valuation domain, then $R$ is a valuation domain? Here, a domain $R$ is said ...
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1answer
39 views

Is this ring an integral domain? Prove or give a counterexample [duplicate]

I need to either prove this is an integral domain or give a counterexample. I have yet to find a counterexample, but can't seem to prove it on my own. Would love any advice/help! Thank you!
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2answers
30 views

Noether normalization: what is a finite map?

Let $D$ be a domain, and $R$ a finitely generated $D$ algebra. There exists a nonzero $f \in D$, and a finite injective ring map $D_f[X_1,\dots,X_n] \hookrightarrow R_f$. Here the $X_i$ are ...
0
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1answer
119 views

Show that if $R$ is an integral domain, then $R[x]$ is an integral domain

This is my own proof, I just need to know if I am doing this correctly. (By the way, I'm 100% sure this is not an original proof, I just came with it by myself for a homework question and want to ...
1
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1answer
33 views

$A = \langle p \rangle$ is a prime ideal of $\mathbb{Z}_n$ if and only if $p$ is a prime divisor of $n$

I am struggling slightly with proving the following; $A = \langle p \rangle$ (principal ideal generated by $p$) is a prime ideal of $\mathbb{Z}_n$ if and only if $p$ is a prime divisor of $n$. I am ...
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0answers
38 views

give an example of an integral domain R of characteristic p>0, such that the map R->R, r->r^p is a non-surjective ring homomorphism

I know that ring homomorphism is injective but not surjective, but I can't come up with an example with an integral domain R of characteristic p>0 such that the map is r-> r^p. I need help with this. ...
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1answer
21 views

integral domain for set of integers

In an integral domain, a product is zero only when one of the factors is zero; that is, ab=0 only when a=0 or b=0. Can we do the following steps for proving set of integers an integral domain? let a=3 ...
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0answers
17 views

Proper divisor of an element in an Integral domain

I am confused as to what the following statement means. Let R be an integral Domain. Let $\enspace a \in R, \text{then} \enspace a$ is a proper divisor if $b = aq$ where $q \in R$ and neither $a$ ...
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0answers
98 views

When integral domain is flat?

Let $A$ and $B$ be integral domain. $A$ is integrally closed and $A \rightarrow B$ is integral. -($\star$) This is sufficient to show that the going-down theorem. If $A \rightarrow B$ is flat, then ...
2
votes
2answers
141 views

Prove that $V$ is not a free module.

I am attempting to solve this: Let $R=\mathbb{Z}[\sqrt{-5}]$, and let $V$ be the R-module presented by the matrix $\begin{bmatrix} 2 \\ 1+δ \end{bmatrix}$ where $δ=\sqrt{-5}$. Prove that $V$ is not ...
1
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0answers
64 views

Green's function fundamental solution in $\mathbb{R}^2$ going to infinity?

There is porbably a rather simple answer to this, but I was wondering why it is not a problem that our Green's function fundamnetal solution in $\mathbb{R}^2$ goes to infinity as $r\rightarrow \infty$...
2
votes
2answers
93 views

An $n$ degree polynomial with more than $n$ roots?

I know that a polynomial of degree $n$ over a field has at most $n$ roots (even counted with multiplicity). My question is how this works with polynomials over integral domains. For example, $f(x) = ...
0
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1answer
31 views

The ideals of $\mathbb{Z}_{(p)}$ = {$m/n \in \mathbb{Q} : p$ does not divide $n$}.

Define $\mathbb{Z}_{(p)}$ = {$m/n \in \mathbb{Q} : p$ does not divide $n$}. If $A \neq 0$ is an ideal of $\mathbb{Z}_{(p)}$ show that $A = <p^k>$ where $k \geq 0$ is the smallest integer such ...