# Tagged Questions

An ideal is a subset of ring such that it is possible to make a quotient ring with respect to this subset. This is the most frequent use of the name ideal, but it is used in other areas of mathematics too: ideals in set theory and order theory (which are closely related), ideals in semigroups, ...

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### Multiplicity and regular sequences

We define multiplicity of a module $M$ of dimension $d>0$ as $$e(M) := \operatorname{lc} (P_M) (d-1)!,$$ where $P_M$ denotes the Hilbert polynomial of $M$ and $\operatorname{lc}(P_M)$ its leading ...
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### On maximal ideal spaces of a banach algebra

I am reading this article on maximal ideal spaces and there is this part that I don't quite understand very well, hope you guys can help me out. "Let $M(A)$ denote the maximal ideal space of a ...
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### How are varieties related polynomials?

My teacher says that varieties and ideals are related to each other while I tend to mix polynomials and varieties in my terminology. Could some explain how varieties are related to polynomials? And ...
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### $\mathbb{Z}[\sqrt{10}]$ is noetherian

How can we prove that $\mathbb{Z}[\sqrt{10}]$ is noetherian except by using Hilbert basis theorem? How can we find a sequence of ideals that satisfy the ACC?
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### $R=\{ m+nr\sqrt{2} \mid m,n \in \Bbb Z \}$ and $I_{a,b}=\{ ma+n(b+r\sqrt{2}) \mid m,n \in \Bbb Z \}$

Let $r$ be a natural number and $R=\{ m+nr\sqrt{2} \mid m,n \in \Bbb Z \}$. We can show that $R$ is a subring of the ring $\Bbb Q [\sqrt{2}]$. My questions are as follows: $(1)$ Suppose that a ...
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### Intersection $I \cap \mathbb{Z}$ where $I$ is an ideal of $\mathbb{Z}[X]$

is there a reasonable algorithm that allows, given finitely many generators of an ideal $I$ of $\mathbb{Z}[X]$, to find the intersection $I \cap \mathbb{Z}$? Thank you.
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### Please help in proving $ab=0 \space \forall a,b$ in a ring $R$ where the only right ideals of $R$ are the trivial ones and $R$ is not a division ring.

I have the following question I am trying to solve: Let $R$ be a ring such that the only right ideals of $R$ are $(0)$ and $R$. Prove that either $R$ is a division ring or that $R$ is a ring with ...
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### Unique prime ideal containing $(2)$ in $\mathbb{Z}[\sqrt{-3}]$

I'm having trouble with an algebraic number theory problem. Let $R = \mathbb{Z}[\sqrt{-3}]$. The problem is to show that $(2, 1 + \sqrt{-3})$ is the unique prime ideal containing the ideal $(2)$, and ...
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### A Problem for Nil-Ideals

Consider a ring $R$ and $I$ be a finitely generated nil-ideal of $R$. Is $I$ a nilpotent ideal? I have proved this for commutative rings. But for non-commutative rings I think this may not be true. ...
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### Module endomorphisms with the same kernel

Let $R$ be a finite commutative principal ideal ring. Let $n$ be a positive integer. For $i=1, \ldots, n-1$ we let \begin{align*} w_i := w_i(x_{i+1}, \ldots, x_n) = \sum_{j=i+1}^n t_{ij}x_j \in R[x_{...
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Let $R$ be a 2-dimensional complete regular local ring $R$ over an algebraically closed field $k$, that is $R\cong k[[x,y]]$. Now look at the the following subring $A$ of $M_2(R)$: $A=\begin{pmatrix} ... 0answers 65 views ### Order of prime ideals over split primes in the class group. Let$P$be a prime ideal of$\mathcal O_K$($K$a quadratic field) and let$P$have norm$p$where$p$is a split prime. Is it possible for the ideal class$[P]$to have order less than three? I feel ... 0answers 78 views ### Is there a classification of ideals of$\mathcal O_K$($K$quadratic) over ramified and split primes depending on$d \pmod 4$? I am unsure if the following argument is correct. I have not seen something like this in my course, so I'm a bit skeptical, since this seems like a very simple way of computing norms of ideals. If ... 0answers 44 views ### Show that the ideal generated by$4$in$\mathbb Z_{12}$is not a prime ideal. Show that the ideal generated by$4$in$\mathbb Z_{12}$is not a prime ideal. Hint: Give a counter-example This is my rough proof to this question. I was wondering if anybody can look over it and ... 0answers 147 views ### Can a local ring have more than one prime ideal? A local ring is defined as a ring which has a unique maximal ideal. This unique maximal ideal consists of only non-units and contains all the non-units of the ring$R$. So examples of local rings ... 0answers 88 views ### Graded Betti Numbers of a Graded Ideal with Linear Quotients Exercise 8.8(a) in Monomial Ideals by Herzog and Hibi: Let$I\subset S=K[x_{1},...,x_{n}]$be a graded ideal which has linear quotients with respect to a homogeneous system of generators$f_{1},......
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I want to find the value $|\mathbb{Z}({\sqrt{2})/(3+\sqrt{2})}|,|\mathbb{Z}({\sqrt{13})/(5+\sqrt{13})}|$ and also the number of ideals of $\mathbb{Z}({\sqrt{13})/(5+\sqrt{13})}$. But still not ...
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### A question about the consequence of Prime Avoidance.

I have found the following statement: Let $R$ be a Noetherian ring and $x$ is a non-zero divisor of $R$. Let $P$ be a prime ideal associated to $xR$. Then by Prime Avoidance there exists a non-zero ...
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Ok so I have that $k$ is algebraically closed and $F$ is an element of $k^n$, and it is a reduced polynomial. We have that $V = V(F)$. In the book it says prove that $F$ generates $I(V)$ but in my ...
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### Proof of going-up theorem

Can you tell me if my proof is correct? Thanks. (I'm using propositions 5.6 and 5.10 from Atiyah-Macdonald which I proved separately.) Theorem: Let $R$ be integral over $S$. Let $p_1 \subset p_2$ be ...
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Let $E/K$ be a separable field extension of degree $n$, let $A$ be a Dedekind Domain which quotient field is $K$, and let $B$ be the integral closure of $A$ in $E$. Then we have that the ideal ...
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### Existence of minimal prime ideal contained in given prime ideal and containing a given subset

Let $R$ be a unital commutative ring, $P$ $\subseteq$ $R$ a prime ideal, $X\subseteq P$ a subset. Show there exists a minimal (inclusion minimal) prime ideal contained in $P$ which contains $X$. My ...
### $J$ maximal ideal of $A$ $\iff$ $A/J$ is field
I'm wondering where did this complicated proof that $J$ is a maximal ideal $\iff$ $A/J$ is a field. Is there an easy to look case where we can clearly see that when we take the quotient of the ring ...