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.
781 questions
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Does any infinite set of elements in a GCD Domain have a GCD?
I know this holds in a UFD, since we can choose an element $a$ for any subset of it, the common divisors are also divisors of $a$ and then there are at most a finite number of them, so there is a ...
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Which integral domains $R$ are flat over every subring?
It is well-known that if every integral domain containing a given integral domain $R$ is flat over $R$, then $R$ is a Prüfer domain. So I would like to ask the question about the other direction: ...
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uniqueness of embedding of an integral domain in its field of fractions
Is it possible to prove that the only embedding of an integral domain to its field of fractions is the natural one?
$ a \mapsto [(a,1)]$
From what I was able to come up with, when considering an ...
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Infinite integral domain with a finite number of invertible
Let $ A $ be an infinite integral domain with a finite number of invertible elements. Show that every $ a \in \operatorname{Jac}(A) $ has finitely many multiples, and deduce that $ \operatorname{Jac}(...
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Understanding Transcendence Degree in the Context of Noether's Normalization Lemma
I’m currently studying algebraic geometry and working through David Mumford's The Red Book of Varieties and Schemes. The book assumes familiarity with the concept of transcendence degree early on, ...
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Uncountable, commutative ring with unity is an integral domain if the factor ring is countable for any non-zero ideal.
The problem is as follows:
Suppose $R$ is a commutative, uncountable ring with unity, with the property that for any non-zero ideal $I$ of $R$, the factor ring $R/I$ is countable. Show that $R$ is an ...
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Every ideal is prime implies that $R$ is a field [duplicate]
Let $R$ be a commutative ring with $1$ such that every proper ideal is a prime ideal. Prove that $R$ is a field. [Hint: for $a \neq 0$, consider the ideals $(a)$ and $(a^2)$].
Attempt.
Let $a \in R$ ...
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If $a/1 | r/1$ in the localization $R_I$ for every prime ideal $I$ in an integral domain $R$, then $a | r$ in R. [duplicate]
Translated to just $R$, this comes down to proving that if for every prime ideal $I$ in $R$, there exists some $d \in R \setminus I$ with $a | rd$, then $a|r$.
I found a proof in case R is a UFD:
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Find all integral domains $R$ such that $x^{4} = x$ for all $x \in R$.
At the moment, I am not even studying ring theory but I was thinking of the following question.
Question: Find all integral domains $R$ such that $x^{4} = x$ for all $x \in R$.
My initial thought was ...
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Where does this integral closure proof go wrong?
Let $A = \mathbb{Z}[x,y]/(2y-x)$, $R = \mathbb{Z}[x]$, $S = \{1,x,x^2,x^3,...\}$.
Is $S^{-1}A$ integral over $S^{-1}R$? It is not, and I was made to see it was not since it would imply via some ...
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Is $\mathbb C[[x, y]]/(x^3+y^4)$ an integral domain?
The polynomial $x^3+y^4$ is irreducible in $\mathbb C[x, y]$, thus $\mathbb C[x, y]/(x^3+y^4)$ is an integral domain. My question is: Is $\mathbb C[[x, y]]/(x^3+y^4)$ also an integral domain?
Further ...
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Does there exist an HFD which is not a Mori domain?
A half factorial domain (HFD) is an integral domain such that every nonzero nonunit $x$ admits a factorisation into irreducibles, and any two factorisations of $x$ have the same number of factors.
A ...
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Let $A$ be an integral domain, is it true $A=\cap_{p \text{ prime }}A_p$? [duplicate]
Let $A$ be an integral domain, is it true that $$A=\cap_{ht(p)=1}A_p=\cap_{p \text{ maximal}}A_p=\cap_{p \text{ prime }}A_p$$
My idea:
Claim1: $$A=\cap_{p \text{ prime}}A_p$$
'$\subset$': Since $A$ ...
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Why isn't $\mathbb{Q}$ the integral closure of $\mathbb{Z}$ in $\mathbb{Q}$ if every element satisfies this equation?
I can't understand this quote from Wikipedia:
Integers are the only elements of $\mathbb{Q}$ that are integral over $\mathbb{Z}$. In other words, $\mathbb{Z}$ is the integral closure of $\mathbb{Z}$ ...
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If $\text{Hom}_R (M, R/I ) = {0}$ for all non-zero ideals, then $\text{Hom}_R (M, R ) = {0}$
This question comes from Rotman's Introduction to Homological Algebra (page 68, second edition).
2.22 Let $R$ be a domain and suppose that $M$ is an $R$-module with $\text{Hom}_R (M, R/I ) = \{0\}$ ...
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