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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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The proof of the weak Nullstellensatz in Qing Liu’s book

I am reading Qing Liu’s book and the whole proof on pp 30 goes like this: I do not understand the conclusion in the yellow-highlighted line. Let $f$ be the isomorphism highlighted in red. Why $f$ maps $\alpha_i$ to $\alpha_i$?
user150248
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A "fractional ideal'' over a domain is flat iff it is locally free of rank 1

Exercise 1.2.9 in Qing Liu's Algebraic Geometry and Arithmetic Curves reads as follows: Let $A$ be an integral domain and $K$ its field of fractions. Suppose $M \subset K$ is a finitely-generated $A$-submodule of $K$. Show that $M$ is flat iff it…
Sameer Kailasa
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Describe a minimal injective resolution of $R$ as an $R$-module.

Let $k$ be a field and $R=k[X,Y]_{(X,Y)}$. Describe a minimal injective resolution of $R$ as an $R$-module (in terms of the Bass numbers of $R$). Solution: Let $k$ be a field and $R=k[X,Y]_{(X,Y)}$. Then $$ k(p) = \left.R_p\middle/pR_p\right. =…
Username Unknown
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A question over Cohen–Macaulay local rings

Let $R$ be a Cohen–Macaulay local ring and $M$ be a finitely generated $R$-module. If ${\rm Hom}_R(M , R)=R$ then can we conclude that $M=R$ ?
Mohammad Bagheri
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The same ideal in different polynomial rings.

Let $J\subseteq k[x_1, \ldots, x_{n-1}]\subseteq k[x_1, \ldots, x_n]$ be an ideal such that $(x_1, \ldots, x_{n-1})$ is a minimal prime of $J$ (thinking in the polynomial ring $k[x_1, \ldots, x_n]$). Is $J$ an ideal $(x_1, \ldots, x_{n-1})$-primary…
Euler
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Minimal set of generators of an ideal

Let $R$ be a commutative local ring with maximal ideal $\mathfrak{m}$. Let $I$ be an ideal and $x\in R$ such that $x$ is not a zero divisor on $R/I$. Then a minimal set of generators for $I$ is sent to a minimal set of generators for $(I +(x))/(x)$…
user09127
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Finite ring extensions and finite field extensions

Let $R\subset S$ be two finitely generated integral domains over an algebraically closed field $k$. If $S$ is finite as $R$-module then $[L:K]<+\infty$, where $L$ and $K$ are the quotient fields of $S$ and $R$, respectively. Does the converse hold…
Vincenzo Zaccaro
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About module decomposition

I am reading p. 20 of Introduction to Commutative Algebra by Atiyah and Macdonald. There there is a module decomposition $$ A=\mathfrak{a}_1\oplus\cdots\oplus\mathfrak{a}_n $$ of a commutative ring $A$. Then the last sentence on the page says "The…
user584333
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Algebraic structure of $\{0,1,\ldots,+\infty\}$

I am reading about the dimension of a commutative ring $A$ with identity $1\ne0$, which is defined as the supremum of the heights of its prime ideals. I think it turns out that the dimension is always well-defined to be in $\{0,1,\ldots,+\infty\}$.…
user584333
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The normalization of an integral domain and its quotient

Let $k$ be a field, $A$ a finitely generated local integral domain over $k$ of Krull dimension $1$, and $A'$ be the normalization of $A$ in the fraction field of $A$. Then is $A'/A$ a torsion $A$-module, i.e., $A' /A \cong \bigoplus A/g_i $ for…
k.j.
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Does such ring extension exist?

Does there exist a commutative ring extension $A\subsetneq B$ satisfying the following conditions: $A$ is a normal local domain and $B$ is a regular domain with the same dimension; $A$ and $B$ have the same fraction field; $B$ is finitely generated…
G.-S. Zhou
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Irreducible primary ideal in Noetherian local ring (Sharp, Exercise 8.29)

Let $(R,M)$ be a Noetherian local ring, and let $Q$ be an $M$-primary ideal of $R$. Note that the $R$-module $(Q:M)/Q$ is annihilated by $M$ so it can be regarded as an $R/M$-vector space. Show that the following statement are equivalent i) $Q$ is…
Desunkid
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Let $R$ be an integral domain. If $I \cap J = IJ$ for all ideals $I,J$ of $R,$ then $R$ is a field.

Let $R$ be an integral domain. If $I \cap J = IJ$ for all ideals $I,J$ of $R,$ how do I show $R$ is a field? Hints will suffice. Thank you.
Alexy Vincenzo
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Is extending an ideal through a ring homomorphism the same as through extension of scalars?

Suppose we have commutative rings $A$ and $B$, a (maybe injective) ring homomorphism $f: A \rightarrow B$ and an ideal $I \subseteq A$. Is it true that $I^e \cong I \otimes_A B$, where $I^e$ denotes the extension of $I$ into $B$? In other words,…
beanshadow
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Does the concept of localizing at an extension of a prime ideal make sense?

If $A,B$ are commutative rings with $1$, $p$ is a prime ideal in $A$ and $f:A\rightarrow B$ makes $B$ an $A$-algebra, I want to know if it is possible to define the localization $B_p$ of $B$ at the extension of $p$. I suspect it is possible,…
dsm4
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