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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.

16857 questions
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Dimension of quotients of Noetherian rings

Suppose that $A$ is a Noetherian ring containing a field $k$ and consider an ideal $I$ with $J =\sqrt{I}$ the radical of $I$. Suppose that $A/J$ is a finite-dimensional vector space over $k$. I am trying to prove that this implies that $A/I$ is…
John
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Integral basis and Integral extensions

I have two questions: a) How can I find the integral basis of the integral closure of $\Bbb Z$ in $\Bbb Q(\sqrt{3})$.b) How can I show that an integral extension is not finite, for example how to show that $\Bbb…
i.a.m
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Generators of the coordinate ring for prime ideals

One of the ways I find more useful to check if a given ideal $I$ of $K[X_1,\ldots,X_n]$ is prime, is to look at the quotient ring $K[X_1,\ldots,X_n]/I$. If I'm able to show it is isomorphic to $K[f_1,\ldots,f_n]$ where $f_i$ are polynomials in…
Gabriel Furstenheim
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A section of the quotient by the nilradical

Let $R$ be a Noetherian commutative unital ring. Does the quotient by nilradical of $R$ admit a section? I suspect that this is true for finitely generated algebras over a field.
user690882
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Quadratic extension of $\mathbb{Q}(X)$

Consider the ring $\mathbb{Q}[X]$ of polynomials in $X$ with coefficients in the field of rational numbers. Consider the quotient field $\mathbb{Q}(X)$ and let $K$ be a quadratic extension of $\mathbb{Q}(X)$ given by $K:=\mathbb{Q}(X)[d]$. Let…
Federica Maggioni
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Atiyah-MacDonald Exercise 4.17. Maximality assumption?

The exercise is copied/paraphrased below. Let $A$ be a ring with the following property. (L1) For every ideal $\mathfrak{a}\neq (1)$ in $A$ and every prime ideal $\mathfrak{p}$, there exists $x\not\in\mathfrak{p}$ such that…
654897419
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commutative diagram

Let $$\begin{array} AA & \stackrel{f}{\longrightarrow} & B \\ \downarrow{h} & & \downarrow{h'} \\ C & \stackrel{g}{\longrightarrow} & D \end{array} $$ be a commutative diagram of $\mathcal{O}$-modules ($\mathcal{O}$ principal domain) with $f$ and…
user65490
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Invertible ideals are maximal ideals?

Let $I$ be an invertible ideal in an integral domain $R$. I claim that it is a maximal ideal. Please tell my i am correct or not. Here is my attempt: If $I\subseteq J$, for some proper ideal $J$ of $R$. Then $R=II^{-1}\subseteq JI^{-1}$ and hence…
waqas mahmood
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Two definitions of exactness

Given a functor $F:A\to B$ of abelian categories we may say that $F$ is left exact if it maps exact sequences to left exact sequences, and similarily for right. For arbitrary categories, we may say that $F$ is left exact if it preserves finite…
Eivind Dahl
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Relation between associated primes and primary decomposition for non-finite modules

Theorem 6.8(ii) p.41 in Matsumura's Commutative Ring Theory, says that if $A$ is a Noetherian ring, $M$ a finite $A$-module and $N=N_1 \cap \cdots \cap N_s$ an irredundant primary decomposition of a proper submodule $N$ with $Ass(M/N_i) = \left\{P_i…
Manos
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Is there a Euclidean ring that is not a domain?

I have been reading about the equivalent definitions of PID. I learned that a domain $R$ is a PID if and only if $R$ has a Dedekind-Hasse norm. This followed from the fact that a ring $R$ is a principal ideal ring if $R$ admitted a Dedekind-Hasse…
libofmath
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Showing that nonzero prime ideals are maximal in a polynomial ring.

$F$ is a field and $F[X^2, X^3]$ is a subring of $F[X]$, the polynomial ring. I need to show that nonzero prime ideals of $F[X^2, X^3]$ are maximal. A classmate suggested taking a nonzero prime ideal $\mathfrak{p}$ of $F[X^2, X^3]$ and embedding…
Joe Wells
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What is an integral element?

I'm trying to straighten out the definition of an integral element... An integral element is not necessarily an integer itself, but is the root of a monic polynomial with integer coefficients? Does that sound right or am I off base?
user58437
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Local GCD domain and not Bezout domain

I have hard time to find the example of local GCD domain which is not a Bezout domain. Is there any simple example? Thanks
Sharma Sharma
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Intersection of a domain $R$ with an ideal $\mathfrak a$ of height $2$ in $R[X]$

Let $R$ be an integral domain, and consider the polynomial ring $R[X]$ over $R$. Suppose that ${\mathfrak a}$ is an ideal of $R[X]$ such that $\operatorname{ht}(\mathfrak a) = 2$. Q. Is the intersection $R \cap \mathfrak a \neq 0$? That is, does…
Rinmyaku
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