## 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

## How can we find a monic polynomial with the smallest degree in left ideal of $\mathrm{Mat}(F[x])$?

Let $F$ be a finite field, $R=F[x]$ be a polynomial ring and $K = \mathrm{Mat}_n(R)$ be a full matrix ring over $R$. We identify the ring $K$ with the ring $\mathrm{Mat}_n(F)[x]$, for example  \left( \begin{smallmatrix} x&0\\ x^2+1&1 \end{smallmatrix} \right)= \left( \begin{smallmatrix} 0&0\\ 1&0 \end{smallmatrix} \right)x^2+ \left( \begin{smallmatrix} 1&0\\ 0&0 \end{smallmatrix} \right)x+ \left( \begin{smallmatrix} 0&0\\ … 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

## AA is a commutative connective dg-algebra satisfying H0(A)=kH^0(A)=k. Is it true that a dg ideal generated by elements of negative degree acyclic?

Let k be a field of characteristic zero and A be a graded commutative dg-algebra over k with differential of degree +1 satisfying H0(A)=k,Hi(A)=0 for i<0. Denote by J a dg ideal of generated by Ak,k<0 (Ak denotes elements of degree k). Is it true that J is acyclic? In other words, is the natural … Read more

## Can a minimal generating set for an ideal always be made into a Groebner basis?

Let I⊆k[x0,…,xn] be an ideal, generated by some polynomials F1,…,Fr, all homogeneous and of the same degree. Suppose r is the smallest number of generators that will suffice to generate the ideal. Can one choose a monomial order or change coordinates to ensure that this generating set is also a Groebner basis for the ideal? … Read more

## Finite distributive lattices as lattice of ideals of a finite ring

Is there a finite distributive lattice that is not isomorphic to the lattice of ideals of a finite ring? Answer The answer is yes, there is a finite distributive lattice which is not isomorphic to the lattice of right ideals in a non-commutative ring with identity, of ideals in a commutative ring with identity. Let … Read more

## Condition for a monomial to belong to a particular ideal

Consider the polynomial ring R[x1,x2,…,xn], where R be an algebraically closed field (preferably C) and the ideal J=⟨m1,m2,…,mn⟩ generated by monomials . The monomials are homogeneous and each variable has maximum degree 1. Let |mi| denotes the number of variables in a monomial mi Is there a way to determine whether the monomial power ∏ni=1xd−1i … 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

## Is there a C*-algebra whose Pedersen ideal is not proper?

In [1, 5.6.3] Pedersen states without proof or reference that there are non-unital C*-algebras whose Pedersen ideal is the whole algebra. Does anyone know where can I find such an example? Is it possible to characterize algebras with this property?  Pedersen, Gert K., C*-algebras and their automorphism groups, London Mathematical Society Monographs. 14. London … Read more

Every semigroup containing an ideal subgroup is called a homogroup. Let $(S,\cdot)$ be homomgroup, hence it contains an ideal $I$ that is also a subgroup. It is easy to see that $I$ is the least ideal, a maximal subgroup of $S$, and its identity (denoted by $e_I$) is a central idempotent of $S$. Now, (1) … Read more