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Questions tagged [commutative-algebra]

Questions about commutative rings, their ideals, and their modules.

Commutative algebra is the area of mathematics that deals with commutative rings and their ideals, as well as modules over commutative rings.

Many results and tools of commutative algebra are cornerstones of algebraic geometry. Important tools of commutative algebra include localization and completion of rings and modules.

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Finitely many prime ideals lying over the same prime ideal

Let $A \subseteq B$ an extension of rings such that $B$ is an $A$-module finitely generated. Show that for every prime ideal $\mathfrak{p} \subseteq A$ there is only a finite number of prime ideals $\mathfrak{q} \subseteq B$ such that $\mathfrak{q}…
rgl4
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Finitely generated as a module implies f.g. as an algebra

I read some notes on commutative algebra and I got stuck on this proposition. Why finitely generated as an $R$-module implies finitely generated as an $R$-algebra by the same elements? How to deal with non-linear combinations, when we generate an…
Kirill Losev
  • 506
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Analytical Independence

I am aware of the definition of analytical independence in Noetherian rings. I am wondering if anyone knows of any generalization of the concept (or similar concept ) to non-noetherian rings.
user62198
  • 281
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topology of completion of a module

Assume $M$ is $A$ module, where $A$ is a commutative ring with unity. Assume $\{M_i,i\in I\}$ is a set of some submodules of $M$, where $I$ is a directed set (this means there is a partial order relation $\le$ on $I$, and whenever $i,j\in I$,there…
Idele
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Integral dependence and fraction fields

Consider $\mathbb{Q}[x]\subset\mathbb{Q}(x)\subset\mathbb{Q}(x)[y]=:K$, where $$y^2=x,$$ and let $O_K$ be the integral closure of $\mathbb{Q}[x]$ in $\mathbb{Q}(x)[y]$. Show that $\mathbb{Q}[x][y]=O_K$. I ask for some help/hints with this…
Heitor Fontana
  • 3,676
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Power series over Noetherian domains

Does there exist an integral Noetherian domain $R$ such that there is an injective unital ring homomorphism $R[[x]]\rightarrow R$? I thought of using Krull dimension to answer this but a ring can have lower Krull dimension than its subring (e.g.…
user691259
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Faithful flatness of equalizers

Let $R$ and $S$ be two commutative rings and $f$ and $g$ be two homomorphisms from $R$ to $S$. Also, let $E$ be their equalizer, i.e. the subring $\{x \in R \vert f(x)=g(x)\}$. If $R$ is a flat $E$-module, does this imply that it is faithfully…
Geoffrey Trang
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Associated graded is integrally closed domain implies original ring was integrally closed

Let $A$ be a local ring (I am also happy to assume it is Noetherian) with maximal ideal $\mathfrak{m}$. Claim: If the associated graded ring $$G_{\mathfrak{m}}(A)=\bigoplus_{n\geq 0} \mathfrak{m}^n/\mathfrak{m}^{n+1}$$ is an integrally closed…
654897419
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On modules which elements are annihilated by a power of the maximal

Let $(R,m,k)$ be a commutative Noetherian local ring and let $M$ be an $R$-module in which every element is annihilated by some power o $m$. It induces a structure of $\widehat{R}$-module on $M$ where $\widehat{R}$ is the $m$-adic completion of $R$:…
Rafael
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How many prime ideals in the polynomial algebra?

Let $k$ be an infinite field. Let $d$ be a positive integer. Is it true that the cardinality of the set of maximal ideals of $k[x_1, \dots, x_d]$ is equal to the cardinality of the set of prime ideals of $k[x_1, \dots, x_d]$ and is equal to the…
jon
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Does Noetherian + finite dimenasional imply essentially of finite type?

Let $R$ be a Noetherian (commutative) algebra over a field $k$. If $\dim R<\infty$ (Krull dimension), does it follow that $R$ is essentially of finite type over $k$? (Meaning: $R=S^{-1}(k[x_1,\ldots,x_n]/I)$ is the localization with respect to a…
Qfwfq
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Arbitrary intersection of Noetherian domains is not necessarily Noetherian

If we have $R_{i}$, $i\in I$, $I$ may be infinite and each $R_{i}$ is a Noetherian integral domain with the same quotient field $K$ then it seems $R = \bigcap_{i\in I} R_{i}$ is not necessarily Noetherian. Example is a non-Noetherian Krull domain…
Sharma Sharma
  • 763
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An inequality of length of module

Do you know a proof for the following inequality? Suppose that $(R,m)$ is a Noetherian local ring, $q$ is an $m$-primary ideal and $M$ is a finitely generated $R$-module. Then $$ l(q^nM/q^{n+1}M) \leq l(M/qM) \cdot \mu(q^n), $$ where $\mu(q^n)$…
mr.bigproblem
  • 860
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Checking whether certain lying-over prime is isolated

I am working through Stephen McAdam's Asymptotic Prime Divisors, and I've hit a snag on the proof of Lemma 3.1 with a question that basically amounts to: Let $A \subseteq B$ be an extension of integral domains. Let $(0)\ne \mathfrak{p} \in…
beanshadow
  • 570
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Domain such that countable intersections of nonzero ideals are nonzero.

Definition: Say that a domain $D$ with field of fractions $K$ satisfies property $(\rho)$ if countable intersections of nonzero ideals are nonzero, or equivalently if $D[[x]]_{D \setminus 0} = K[[x]]$. I'd like to get a better understanding of how…
Badam Baplan
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