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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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Formal Differentiation and Evaluation of Polynomials

Let $R$ be a commutative ring and consider the ring of polynomials $R[x]$. Let $D:R[x]\to R[x]$ be the formal derivative. For any element $a\in R$ we also have an evaluation homomorphism $\varphi_a:R[x]\to R$ defined by $\varphi_a f=$…
Drew Armstrong
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Does a 1-dimensional noetherian domain obey cancellation law?

Someone calls this "order" which puzzles me, because I can't understand it's name. I was wondering whether these rings obey cancellation law, i.e. if $\mathfrak a\mathfrak b=\mathfrak a\mathfrak c$ then $\mathfrak b=\mathfrak c$ for $\mathfrak…
Yoshinobu Osawa
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Characterizing Dedekind domains by ideal-theoretic identities

$\newcommand{\fa}{\mathfrak{a}}\newcommand{fb}{\mathfrak{b}}\newcommand{\fp}{\mathfrak{p}}$ Motivation: In elementary number theory, one proves that for $a,b\in\mathbb{N}$, $a b=\gcd(a,b)\operatorname{lcm}(a,b)$. This has an obvious…
Daniel Miller
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Failures of Nakayama's lemma or Krull theorem

In this post, if $I$ is a finitely generated ideal of a commutative ring $A$ with 1 such that $I^2=I$, then we know $I$ is principal. To answer this question without using Nakayama lemma, I have a complicated method, first reduce to Noetherian…
wxu
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transitivity of finitely generated condition

Let $A \subseteq B \subseteq C$ be rings. I know that if $B$ is a finitely generated $A$-module and $C$ is a finitely generated $B$-module, then $C$ is a finitely generated $A$-module. (Proof is in Atiyah or here) But I have some silly questions,…
Gobi
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Going up theorem (basic question)

If $S \subset R$ are commutative rings with $1$ and $R$ is an integral extension of $S$ then they have the same dimension. Basically the proof uses the going up theorem. But I have a question about a part of the proof: Let $P_{0} \subset P_{1}…
user6495
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Noetherian ring and the same prime divisors

Let $A$ be a Noetherian ring and let $x \in A$ be an element which is neither a unit nor a zerodivisor. Why the ideals $xA$ and $x^{n}A$ (where $n \in \mathbb{N}$) have the same prime divisors? i.e. $$\operatorname{Ass}(A/xA) =…
user6495
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Hilbert's Nullstellensatz: intersection over maximal ideals?

In Reid's commutative algebra, the author gives some exposition about the Nullstellensatz, $I(V(J))=\operatorname{rad}(J)$. But I can't understand it: The Nullstellensatz says that we can take the intersection just over maximal ideals of…
Gobi
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$S/I$ is an integral extension of $R/R\cap I$

I have some probelems with the following statement: Let $S, R$ be rings. Suppose $S$ is an integral extension of $R$ and $I$ is an ideal in $S$ then $S/I$ is an integral extension of $R/(R\cap I)$. These are my thoughts so far. Let $i$ be…
harajm
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Is there an example of a non-noetherian one-dimensional UFD?

Or in the contrapositive form Is every one-dimensional UFD noetherian? I know how to construct a non-noetherian UFD (polynomials in infinite number of variables over a field) and I know that it is even possible to construct a finite dimensional…
KotelKanim
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Show that $M=\bigcap_{\mathfrak{p}\in\operatorname{Spec}(R)}M_\mathfrak{p}=\bigcap_{\mathfrak{m}\in\text{Max}(R)}M_\mathfrak{m}$ for certain $M$.

$\newcommand{\Spec}{\operatorname{Spec}}$ $\newcommand{\mSpec}{\operatorname{Max}}$ This is a homework from my algebra course. I am in a situation where I think I have found a solution, though somehow there's a condition in the question that I don't…
Louis
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Is it possible to say $\hat{I^n}=\hat{I}^n$ in general?

Consider a commutative unital ring $A$ and its ideal $I$. Let $\hat{A}$ be a completion of $A$ with $I$-adic topology. I want to know whether $\hat{I^n}=\hat{I}^n$ holds or not in general. (Note that $I$-adic topology of $I^n$ concides with its…
nessy
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Do Groebner bases give the smallest generating set for Ideals?

Given a Reduced Groebner Basis $(f_1,\ldots,f_n)$ for an ideal $I$, can there be another basis $(g_1,\ldots,g_m)$ for $I$ where $m
Steven-Owen
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Counterexample for tensor product of modules and completion

I'm reading Qing Liu's Algebraic geometry book, and here's a problem that I can't find an answer: Find a ring $A$ and a $A$-module $M$ such that $M\otimes_{A} \hat{A} \to \hat{M}$ is not surjective, where $\hat{A} = \lim A/I^{n}$ and $\hat{M} =\lim…
Seewoo Lee
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Artinian affine $K$-algebra

Let $K$ be a field and $A$ an affine $K$-algebra. Show that $A$ has (Krull) dimension zero (is artinian) if and only if it is finite dimensional over $K$.
user66598
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