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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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Noether Normalization Steps

I'm trying to calculate the Noether normalization of some rings (suppose $K$ is a field with $0$ characteristic). The first one is $K[x,y,z]/(x^2y+xz+z)$. I know it's possible to make an algorithm to solve this problem in general, also I'm concerned…
diff_math
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Rank of Noetherian modules

Suppose $M$ is a Noetherian left-module, why is the rank of $M$ unique? that is if $M^{r} \cong M^{s}$ as left modules then why $r=s$? Is this true if $M$ is Artinian?
user10
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Finding a subring of the reals isomorphic to $\mathbb{Z}[t]/(4t+3)$

Let $I=(4t+3)$ be an ideal in $\mathbb{Z}[t]$. Find a subring of $\mathbb{R}$ isomorphic to $\mathbb{Z}[t]/I$. If $(4t+3)$ were monic, this question would be easily answered but since it isn't I'm having problems seeing exactly how $\mathbb{Z}[t]/I$…
chris
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On the construction of the polynomial for the step of contradiction in Hilbert's basis Theorem

I don't understand a step in this proof of the Hilbert Basis Theorem. Here is the proof Planeth Math. I don't understand why $ \mathrm{deg} (f_{N+1}-g)< \mathrm{deg}(f_{N+1}) $. This can only happen if $ g $ has the same degree as $ f_{N+1} $ and…
Evolution
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About the $k$-subalgebras of $k[x]$

Still in my "commutative algebra marathon", I came across the following exercise: Any $k$-subalgebra $A$ of $k[x]$ is finitely generated as $k$-algebra; also, if $A\ne k$, then $\dim A=1$. Although I have found a related answer here, I don't get…
Renan Mezabarba
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Is $k[x][[h]]$ finitely generated as $k[[h]]$-algebra?

Is $k[x][[h]]$ finitely generated as an algebra over $k[[h]]$, where $k$ is a field, and $xh=hx$.
IORI
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Regular ring of infinite dimension?

Is it possible for a (non-local) regular ring to have an infinite dimension? As far as I know, the well known characterization of regular rings in terms of finite global dimension is only for local case. Or does it also apply to non-local regular…
ashpool
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Why do the map from elements to sum is surjective?

Let $A$ be a commutative complete ring with unit for the $I$-adic topology, where $I$ is the ideal of $A$. Let $(M_n)_{n\geq 0}$ be $A$-modules such that $I^{n+1}M_n=0$ and that there exist a surjective homomorphism $\pi_n:M_{n+1}\to M_n$ with $\ker…
student
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Are there formal power series ring in infinitely many indeterminates

There are polynomial rings in infinitely many indeterminates. Does it make sense to talk about power series rings in infinitely many indeterminates. If not, what do we get when we complete the polynomial ring $k[x_1,...]$ with respect to the maximal…
Merlin
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Application of Krull's principal ideal theorem

Let n be a positive integer, and let $P_0\subsetneq P_1\subsetneq ...\subsetneq P_n$ be a chain of prime ideals in a Noetherian ring R. Moreover, let $a\in P_n$. Prove: 1.There is a chain of prime ideals $P_0'\subsetneq P_1'\subsetneq ...\subsetneq…
somebody trivial
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$\mathbb{Q}[t]$ is integrally closed in $Quot(\mathbb{Q}[t])$

I'm having trouble trying to show that $\mathbb{Q}[t]$ is integrally closed in $Quot(\mathbb{Q}[t])$. Where $Quot(\mathbb{Q}[t])$ is the field of fractions of $\mathbb{Q}[t]$. So I'm trying to show that $\mathbb{Q}[t]=$ integral closure of…
Andrew Brick
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If $\ell_{A_\mathfrak{p}}(N)<\infty$, then is it true that $\operatorname{Hom}_A(N,E(A/\mathfrak{q}))=0$?

Let $A$ be a Noetherian ring, $\mathfrak{p},\mathfrak{q}\subset A$ distinct prime ideals of the same height, $N$ an $A_\mathfrak{p}$-module of finite length. Then is it true that $$\operatorname{Hom}_A(N,E(A/\mathfrak{q}))=0,$$ where…
ashpool
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an example of regular ring with nilpotent elements

A regular local ring is a domain. But in general, a regular ring is not domain, so you can find regular rings with nilpotent elements. I am unable to construct an example of (A, I) as A is a regular ring, I is a nilpotent ideal and A/I is…
Abbassi Arwa
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Criterion for quotient ring to be decomposable.

I read in passing that for a commutative ring $R$ and an ideal $I$, then $R/I$ is decomposable if and only if there exist proper ideals $J$ and $K$ such that $J+K=R$, and $J\cap K=I$.
Nastassja
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Prime ideals in polynomials rings

Let $A$ be a commutative ring, $\mathbb{q}\subset A$ an ideal of $A$, and $\mathbb{q}A[x]$ the ideal of $A[x]$ generated by $\mathbb{q}$ (consists of the polynomials with coefficients in $\mathbb{q}$). Show $\mathbb{q}$ is prime in $A…
chris
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