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Questions tagged [noetherian]

For questions on Noetherian rings, Noetherian modules and related notions.

A (commutative) ring is called Noetherian if every ascending chain of ideals becomes stationary. For non-commutative rings, the notions of left- and right-Noetherian exist, and they apply to left and right ideals respectively. A Noetherian non-commutative ring is both left- and right-Noetherian.

More generally, a module is Noetherian if each ascending chain of submodules becomes stationary.

A vector space is Noetherian if and only if it is of finite dimension, and "Noetherian" can in a vague way be considered as a generalization of "finite-dimensional."

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Grade of a maximal prime ideal in a Noetherian UFD

Here is an another problem in Commutative Rings by Kaplansky, p. 103, no. 15. Let $R$ be a Noetherian UFD. Let $(a,b) \not= R$ where $a,b \in R.$ Prove that any maximal prime of $(a,b)$ has grade of at most $2.$ Note: By a maximal prime of…
Ehsan M. Kermani
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A noetherian topological space is compact

Have to prove that every noetherian topological space $(X,\mathcal{T})$ is also compact. I don't understand the proof I found: Let $\{\mathcal{U}_\alpha\}_{\alpha\in\Lambda}$ be an open cover of $X$, and…
Smurf
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Direct product of finitely many Noetherian non-unital rings is Noetherian

Let $A_1, A_2,...,A_n$ be Noetherian rings (not necessarily unital). Is the direct product $A:=A_1×A_2×⋯×A_n$ necessarily a Noetherian ring? If $A_1, A_2,...,A_n$ are unital, then one can prove that the ideals of $A$ have the form $I_1 \times I_2…
Marco Flores
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Help with a problem about Artinian rings from Christian Peskine's book

I am stuck with problem 3, chapter 4, from the book of Peskine, An Algebraic Introduction to Complex Projective Geometry. Let $A$ be a Noetherian ring. Assume that if $a \in A$ is neither invertible nor nilpotent, then there exist $b \in A$ such…
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Is the global section ring of a Noetherian Scheme Noetherian as well?

As the title suggests, I am asked to prove that, given a Noetherian scheme $(X,\ \mathcal{O}_{X})$ and any open subset $U\subseteq X$, $\Gamma(U,\ \mathcal{O}_{X}):=\mathcal{O}_{X}(U)$ is a Noetherian ring. Up to now, I have been able to show that…
Marco Vergura
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If $k\subset R\subset k[x]$, then $R$ is Noetherian?

Is there a way to prove that any subring $R$ of the polynomial ring over a field $k$ such that $k\subset R$ is Noetherian without appealing to integral extensions, Eakin-Nagata, etc.? The reason I ask is because I found this as an exercise 15.1.9 in…
Sanaa
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Nontrivial ideal of a Noetherian domain contains a finite product of nonzero prime ideals

If $R$ is a Noetherian domain and $ 0 < U < R$ an nontrivial ideal of $R$. How to prove that there exists nonzero prime ideals $p_1,...,p_n \subset R$ such that the product $ p_1 p_2 ...p_n \subset U$? I have no idea on this one. Any help is…
user112564
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Is this ring Noetherian?

The subring of $\mathbb{C}[x,y]$ consisting of all polynomials $f(x,y)$ whose gradient vanishes at the point $x=y=0$. Is this ring Noetherian?
Dave
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Classification of commutative rings $R$ satisfying $\dim(R[T])=\dim(R)+1$

Let $R$ be a commutative ring. Then $\dim(R[T]) \geq \dim(R)+1$. Is there a classification of those commutative rings with the property $\dim(R[T])=\dim(R)+1$? Every Noetherian commutative ring has this property, but also every $0$-dimensional…
Martin Brandenburg
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If $A$ is a commutative Noetherian ring, then $D=\lbrace\mathfrak{p}\in \operatorname{Spec}(A): |A/\mathfrak{p}|\leq k\rbrace$ is a finite set.

Let $A$ be a commutative Noetherian ring (with unity) and $k\in \mathbb{N}$. Prove that $D=\lbrace\mathfrak{p}\in \operatorname{Spec}(A): |A/\mathfrak{p}|\leq k\rbrace$ is a finite set. I'm trying to prove it. We know that $A/\mathfrak{p}$ is a…
Sara
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Ideal in polynomial ring which contains no non-zero prime ideal

Let $J$ be a non-zero ideal in $\mathbb C[X,Y]$ such that $J$ contains no non-zero prime ideal. Then is it true that $J$ has height $1$ ? Possible approach: Since $\mathrm{ht}(J^n)=\mathrm{ht}(J)$ for every $n>1$ so $\mathrm{ht}(J)=1$ iff…
user521337
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Integral domain over which every non-constant irreducible polynomial has degree 1

Let $R$ be an integral domain such that any polynomial $f(X) \in R[X]$ , which is irreducible in $R[X]$, has degree $1$. Then is it true that $R$ is a field ? If this is not true in general , What if we also assume that $R$ is Noetherian (and…
user
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Global dimension of quasi Frobenius ring

Let $R$ be a quasi-Frobenius ring (so $R$ is self-injective and left and right noetherian). I want to prove that $lD(R)=0$ or $\infty$, where $lD(R)$ denotes the left global dimension. I'm unsure about how to go about proving this; the only thing…
Paul Gilmartin
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Is the function ring $C^{\infty}(M)$ noetherian?

Let $M$ be a smooth manifold and $C^{\infty}(M)$ be its function ring. Is this a noetherian ring?
user53216
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Noetherian property for exact sequence

Let $0 \to M' \xrightarrow{\alpha} M \xrightarrow{\beta} M'' \to 0$ be an exact sequence of $A$-modules. Then $M$ is Noetherian is equivalent to $M'$ and $M''$ are Noetherian. For the ''$\Leftarrow$'' case: I guess if we let $(L_n)_{n\geq 1}$ be an…
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