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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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Localization is an idempotent A-algebra

One of my homework problems asks us to prove that let $S$ be a subset of a ring $A$, the localization $A[S^{-1}]$ is an idempotent $A$-algebra. I don't remember my instructor defining an "idempotent $A$-algebra" in class. The assignment is due soon…
Coco
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Intersection of ideal in (non-noetherian) polynomial ring

Let $A$ be a (commutative, associative, unital) ring. Let $f$ be a non-zerodivisor in $A$, and let $I$ be a finitely-generated ideal in $A[\frac1f][T_1,\dotsc,T_r]$. Can it happen that $I\cap A[T_1,\dotsc,T_r]$ is no longer a finitely-generated…
Charles Denis
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Relationship between dimension, depth, codimension and $I$-depth

Let $R$ be a local ring which is not Cohen-Macaulay, and let $I\subset R$ be an ideal. If I know dim($R$), depth($R$) and codim($I$), can I know depth($I$)?
Marco Flores
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Observations on the associativity of the tensor product between two modules.

I have a little question about of the associativity of the tensor products of two modules. If $M, N, P$ are $A$-modules, we know that $(M \otimes N) \ P \cong M \otimes N \otimes P$ If i want to prove that, i have to find a bilinear-map from $(M…
Samsara814
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In the proof of extension of a homomorphism, proposition 5.23 of Atiyah-Macdonald, Introduction to commutative algebra.

I don’t understand the part of the proof of prop5.23 of Atiyah-Macdonald, Introduction to commutative algebra. The author separates the proof into two cases, the case that x is transcendental or the case that x is algebraic over A. About the former…
nat-cat
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How to use the hint of Atiyah-Macdonald, Exercise 3.12 (iv)?

Here is the question: If $M$ is any $A$-module, then $T(M)$ is the kernel of the mapping $x \mapsto 1 \otimes x$ of $M$ into $K \otimes_A M$ where $K$ is the field of fractions of $A$. Here, $A$ is an integral domain. The hint is as follows: Show…
Anthony Lee
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An easy property for extension and contraction of ideal

So given ring homomorphism $f:A\rightarrow B,$ for commutative ring with identity. My problem is about the property $\mathfrak{a}\subseteq\mathfrak{a}^{ec}.$ It's easy to make myself convinced about this property since $\mathfrak{a}$ definitely lies…
Beginner
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For $0 \neq \varphi\in \operatorname{Hom}_B(A,E)$, does there exists $a \in A$ such that $a \varphi$ is nonzero and is annihilated by $P$?

Let $(A, P)$ be a zero-dimensional local ring with maximal ideal $P$. Suppose that for some local ring $(B, P_B)$, $A$ is $B$-algebra that is finitely generated as a $B$-module and the maximal ideal of $B$ maps into that of $A$. Let $E$ is the…
Plantation
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In proof of the Proposition 21.4 in the Eisenbud's Commutative Algebra book ( Canonical module of a local zero-dimensional ring ).

I am reading the Eisenbud, Commutative Algebra, Proposition 21.4 and stuck at certain statement. In this question, we shall assume that all the rings considered are Noetherian. First I arrange assoicated definitions and theorems. Definition 1. Let…
Plantation
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Example of a fractional ideal which is locally invertible everywhere

There is a theorem (e.g. proposition 9.6 in Atiyah-MacDonald) that states that a fractional ideal $I$ of a domain $R$ is invertible if and only if it is finitely generated and it is locally invertible everywhere, that is $I_\mathfrak{p}$ is…
Anakhand
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Any prime ideal that is a contraction of an ideal of S is a contraction of a prime ideal of S

I was asked to prove the following: Let $f: R \to S$ be a morphism of rings. Then, any prime ideal $P \subseteq R$ that is a contraction of an ideal of S is a contraction of a prime ideal of S. I've got the following hint: Consider $\mathcal{F} =…
liv
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flat module and injective of ring homomorphism

Let $f: R \rightarrow S$ be an injective ring homomorphism. Suppose $S/J$ is flat over $R$ for every ideal $J$ of $S$. Show that for any ideal $I$ of $R$ for any ideal $J$ of $S$, we have $(IS) \cap J = IJ$. Here is my approach:we have injective and…
lee
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Example of ring having no superficial element.

Let $A$ be a Noetherian local ring and $I$ be an ideal of $A$. An element $x\in I$ is said to be an $A$-superficial element if there exists a natural number $c$ such that $(I^{n+1}:x)\cap I^c=I^n$ for all $n\geq c$. We know that if the residue field…
Samarendra sahoo
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Zero Ideal Having Primary Decomposition Implies Finitely Many Minimal Prime Ideals

In the second paragraph of the proof below, I don't understand why the zero ideal having primary decomposition implies that there are finitely many prime ideals in $A$ (prime ideals that happen to be minimal). I would appreciate any help.
Hoji
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Tensor product of free module of rank n not free of rank n?

Look at the abelian group $G=\mathbb{Z}1\oplus\mathbb{Z}\sqrt{2}\subset\mathbb{R}$. As a $\mathbb{Z}$-module it is isomorphic to $\mathbb{Z}^2$ so if we calculate $G\otimes_\mathbb{Z} \mathbb{R}$ it should give us $\mathbb{R}^2$ but clearly it is…
Somge
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