12
votes

Accepted

### Can an intersection of ideals in a Noetherian ring be replaced by a countable intersection?

Yes it's true.
First, in a (commutative) noetherian ring $A$, every chain of ideals is well-ordered by reverse inclusion. The supremum of ordinal types of such chains is denoted $o(A)$.
A ...

12
votes

Accepted

### Inverse of the Structure Theorem for Finitely Generated Modules over PID

There seems to be quite some literature about rings with this property, sometimes under the name "FGC domains".
From Googling, not personal knowledge:
In Theorem 14 of
Kaplansky, Irving, Modules ...

12
votes

Accepted

### A question on a Macaulay2 computation

Commutative algebra is NOT the same as algebraic geometry, especially projective algebraic geometry.
The variety in $\mathbb{P}^9$ defined by $I$ and the variety in $\mathbb{P}^9$ defined by $I_0$ are ...

11
votes

Accepted

### Reduced ring with all non-prime ideals finitely generated

Question: Let $R$ be a reduced ring with all
non-prime ideals finitely generated. Then is $R$ Noetherian?
The answer is Yes.
To lessen my typing, let me use the abbreviation
NFG to mean not-finitely-...

11
votes

Accepted

### Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't?

It need not be a sheaf. As an example, consider a space $X$ which is a disjoint union of open subspaces $X_n$, and pick $\mathcal O_X,\mathcal I,\mathcal J$ with the property that some element $c_n$ ...

10
votes

Accepted

### $I,J$ are $p$-primary ideals, but $I+J$ is not

Let $R$ be the commutative ring $k[x,y,z]$. Let $I$ be the ideal generated by the regular sequence $(x^2,y)$. Let $J$ be the ideal generated by the regular sequence $(x^2,y-xz)$. Then both $R/I$ ...

Community wiki

10
votes

Accepted

### Distribution relation in the Euler system of Heegner points

What I don't understand is why the exactly the same terms should appear in both sums.
The Galois action on CM points is described in adelic terms via the fundamental theorem of complex multiplication ...

10
votes

Accepted

### Is every 2-sided ideal in a C*-algebra hereditary?

No. Take $A = C[0,1]$ and let $I$ be the (unclosed) ideal generated by the function $f(t) = t$. This ideal is self-adjoint, but it does not contain the function $g(t) = t\sin^2(\frac{1}{t})$, so it is ...

9
votes

### Maximal subideal of an ideal

Claim:
Let $T$ be an ideal in a commutative ring. If $T$ contains
a unique maximal proper subideal, then $T$ is principal.
Proof: Indeed take any element outside the unique maximal proper subideal. ...

9
votes

### Do relatively prime polynomials $f$ and $g$ in $k[x,y]$ generate an ideal of finite codimension?

A standard undergraduate maths approach is via resultants.
I am not going to survey resultants here (but see below), I'll just say that an immediate consequence of 1) is that there exists a nonzero ...

9
votes

### Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't?

So here is a counterexample which is qcqs:
Take $X$ the affine line with double origin $a_1$ and $a_2$, then take $I_1$ and $I_2$ the ideal of functions vanishing each at one of the origins ...

9
votes

Accepted

### What is the ideal of hypersurfaces singular at a given irreducible variety?

If $X=\mathbb{V}(I)$ is given by the ideal $I$, then the $m$th symbolic power $I^{[m]}$ consists of all those functions vanishing to multiplicity $m$ at the generic point of $X$. Thus a hypersurface $\...

8
votes

Accepted

### commutative, infinite, artinian ring (with unity) in which distinct ideals has distinct index

Let $R$ be an infinite, commutative, Artinian ring with the distinct-index property for distinct ideals. The claim is that $R$ is a field.
Case 1. $R$ has a nonzero ideal $I$ such that $I^2=0$.
...

7
votes

### Transitivity of ideals of banach algebra

Here is a simple example to show that "being an ideal in" is not transitive, even for finite-dimensional commutative algebras.
Take $A$ to be the algebra generated by $1$ and an element $z$ ...

7
votes

### Do relatively prime polynomials $f$ and $g$ in $k[x,y]$ generate an ideal of finite codimension?

There's perhaps a more elementary answer than this, but here's what I have so far. Define $A := k[x,y]$ and $I := (f,g)$. I'll be using the following fact:
Fact: $A$ is a Cohen-Macaulay UFD.
To ...

7
votes

### Counterexample for Chvatal's conjecture in an infinite set

The usual definition of ideal also includes the requirement that $F$ is closed under unions.
But in this case, I claim that Chvatal's property holds for all infinite ideals.
First, if $a\in Y\in F$, ...

7
votes

Accepted

### $\mathbb{C}(u(x,y),v(x,y),f(x)+g(y))=\mathbb{C}(x,y)$ implies $\mathbb{C}(u(x,y),v(x,y))=\mathbb{C}(x,y)$?

Actually, there are still counterexamples of essentially the same nature as the one given in the previous (linked) thread: Take $u=x+xy$ and $v=x^2y$, so that $\mathbb{C}(u,v)$ is the field of ...

6
votes

Accepted

### Some questions about homogroups

The answers to these problems are the following:
(1) Yes: the ideal subgroup $I$ is unique. Indeed, if $H$ is another ideal subgroup, then $HI\subset H\cap I$, $H\cap I$ is a subgroup of $H$ and $I$, ...

6
votes

Accepted

### If $(f,g)$ and $(f,h)$ are maximal ideals, then $ag+bh=P(f)$ for some $a,b \in k, P(t) \in k[t]$?

The answer is no. Let
$$f=x, \quad g=y+xy^2, \quad h=y+1+xy^3.$$
Then
$$
(f,g)=(x,y)\quad\text{and}\quad (f,h)=(x,y+1)
$$
are maximal ideals of $k[x,y]$, but for all $a,b\in k$ not both zero, you find ...

5
votes

Accepted

### Why does the variety of ideals in this quaternion type algebra have a non-reduced structure?

It is not immediately obvious (at least to me) but it is an easy calculation. The following is too long for a comment, so I post it here:
Here is a simpler example of the same phenomenon: Let $A=\...

5
votes

Accepted

### On the set of non-zero elements in an integral domain whose generating principal ideal is of a special kind

The answer is yes if $R$ is any atomic domain, e.g., $R$ is a Noetherian domain.
Claim 1. Let $R$ be any integral domain. The set $S = S_R$ is saturated in the sense that if $ab \in S_R$ for ...

5
votes

Accepted

### Ideals of Banach algebras

Just to answer the easy part of the question quickly
1) [0,1]
2) the closed unit disc
3) SO(3)

5
votes

### A property of minimal prime ideals in commutative reduced ring

This is an interesting question. Quentel's Example provides an example.
Let $R$ be a reduced ring, $Min(R)$ its space of minimal prime ideals, and $q(R)$ its classical quotient ring.
Theorem The ...

5
votes

Accepted

### Ideal generated by two univariate, coprime, integer polynomials

For question 2, compute a Gröbner basis over $\mathbb{Z}$ for the ideal generated by $f(x)$ and $g(x)$, which gives the required generator. You can do this easily in SageMathCloud (available free to ...

5
votes

Accepted

### Ideals invariant under ring automorphisms

There are tons of examples. Put $u = x^2+xy+y^2$ and $v = xy(x+y)$. Then $u$ and $v$ are $SL_2(\mathbb{F}_2)$ invariant. If $f(s,t)$ and $g(s,t)$ are homogenous polynomials with respect to the grading ...

5
votes

Accepted

### Is the Scott topology generated by the ideals as the closed sets?

The answer is no. Consider poset consisting of infinitely many incomparable elements $a_1,a_2,\dots$ and a single element $b$ larger than them all. Then $A=\{a_1,a_2,\dots\}$ is closed in the Scott ...

5
votes

### Proof in Schertz's Complex Multiplication

Presumably, the author aimed only at showing that the order $\mathfrak{O}_t$ is a one-dimensional Noetherian domain, i.e., it is a Noetherian domain and every of its non-zero prime ideals is maximal. ...

5
votes

Accepted

### Proof in Schertz's Complex Multiplication

The statement that seems odd to you is indeed false, what (I believe) he wants to say is that $\mathfrak{D}_t$ is a Noetherian ring in which every prime ideal $\mathfrak{p}\neq 0$ is maximal.
I ...

5
votes

Accepted

### For an integral domain $R$ when does $a^2 \equiv b^2 \bmod 4R$ imply $a \equiv b \bmod 2R$?

Denote $a-b=x$, then $a^2-b^2=x(x+2b)=4z$ for $z\in R$. Assuming that $R$ is integrally closed, we see that $(x/2)^2+b(x/2)-z=0$, so $x/2$ is an algebraic integer, thus $x/2\in R$.

5
votes

Accepted

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

Examples include all non-unital algebraically simple $C^\ast$-algebras. By [Blackadar, Bruce E.; Cuntz, Joachim The structure of stable algebraically simple C∗-algebras. Amer. J. Math. 104 (1982), no. ...

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