Noetherian rings as homomorphic image of finite direct product of Noetherian domains?

A theorem of Hungerford says that : Every PIR (principal ideal ring , obviously commutative ) is a homomorphic image of a finite direct product of PID s . My question is , is there a similar criteria for Noetherian rings, i.e. : Is every Noetherian ring a homomorphic image of a finite direct product … Read more

torsion free modules MM over Noetherian domain of dimension 11 for which l(M/aM)≤(dimKK⊗RM)⋅l(R/aR),∀0≠a∈Rl(M/aM) \le (\dim_K K \otimes_R M) \cdot l(R/aR), \forall 0 \ne a \in R

Let R be a Noetherian domain of Krull-dimension 1 (i.e. every non-zero prime ideal maximal). Let M be a torsion free R-module . Let K be the fraction-field of R and let r=dimKS−1M=dimKK⊗RM (where S=R∖{0} ). Suppose r is finite. Under these conditions, if M is finitely generated, then I can prove that l(M/aM)≤r⋅l(R/aR),∀0≠a∈R . … Read more

Noetherian ring with a “strange” idempotent ideal

Do you know a left-noetherian ring R with a two-sided ideal I such that: I=I.I; I is not projective as a left R-module (and better, the tensor product over R of I with itself is not a projective left R-module)? Answer Take any idempotent e in a finite dimensional quiver algebra KQ/L , then the … Read more

On the annihilator of a module

Question. Let A be a Noetherian ring and M a finitely generated A-module. Does there always exist an element s∈M such that Ann(s)=Ann(M)? Remark. The annihilator of a module is a lower bound of the annihilators of elements in the module. The question asks whether this lower bound can be reached at some element. It … Read more

Algebras such that the tensor product with any Noetherian algebra is Noetherian

Let $R$ be a Noetherian commutative unital ring. It is generally speaking not true that the tensor product of two Noetherian $R$-algebras is Noetherian (e.g. take $R$ to be a field, and consider the tensor square of some crazy field extension). What is true is that a tensor product of a finite type $R$-algebra and … Read more

an easy example of valuation ring which is not noetherian? [duplicate]

This question already has answers here: non discrete valuation ring [closed] (2 answers) Closed 7 years ago. Is there an easy example of valuation ring which is not noetherian? Answer The valuation ring of Cp is not noetherian. AttributionSource : Link , Question Author : iff , Answer Author : Martin Brandenburg

Pushouts of noetherian rings

Does the category of noetherian commutative rings have pushouts? Background: If X/S is an abelian scheme, then the relative Picard functor PicX/S is only defined on the category of locally noetherian S-schemes (as far as I know). It is a group functor and in some situations it is representable. We then get a group object … Read more

For isolated singularity algebra, is every maximal Cohen-Macaulay module locally projective?

Let $R$ be a Cohen-Macaulay noetherian local ring. Let $\Lambda$ be a noetherian $R$-algebra which is maximal Cohen-Macaulay as an $R$-module, where for every nonmaximal prime $\mathfrak{p}$, $\Lambda_{\mathfrak{p}}$ has finite global dimension. How can I prove that every $\Lambda$-module which is maximal Cohen-macaulay as an $R$-module, is locally projective on the punctured spectrum of $R$? … Read more

When does prime elements remain prime in certain integral extension

Let R be an integral domain and ˉR denote its integral closure in the fraction field (i.e. normalization). If p∈R is a prime element in R, then does p remain prime in ˉR also ? If this is not true in general, then what if we also assume R is Noetherian ? By a prime … Read more

Are local, Noetherian rings with principal maximal ideal PIR?

A question asked by a friend. I believe it’s false, but lack a decisive counterexample. This question shows that it is true for valuation rings, but I know too little about them. In the wider context, a solution to this problem would provide another proof that Artinian local rings whose maximal ideal is principal are … Read more