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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How to show this ideal is not principal

I have been brushing up on cubic number fields. Specifically, let $s$ be a root of the polynomial $x^3 + x^2 + 3x + 17$, and consider $K = \mathbb{Q}(s)$; we have $\mathcal{O}_K = \mathbb{Z}[s]$, and ...
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Factoring the ideal $(8)$ into a product of prime ideals in $\mathbb{Q}(\sqrt{-7})$

I am trying to factor the ideal $(8)$ into a product of prime ideals in $\mathbb{Q}(\sqrt{-7})$. I am not exactly sure how to go about doing this, and I feel I am missing some theory in the matter. ...
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Prime elements in $\mathbb{Q}[[X,Y,Z]]$ whose status as an infinite series is unchanged by arbitrary multiplication

Let's suppose $R$ is the ring $\mathbb{Q}[[X,Y,Z]]$. I'm interested in finding power series $f(x,y,z) \in R \setminus \mathbb{Q}[X,Y,Z]$ which are, first of all, prime elements in $R$, but also ...
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Augmentation ideal and the abelianization of $G$

On a qual problem recently, I came across the following fact: If $G$ is a finite group, and $\mathfrak{a}$ is the augmentation ideal of the integral group ring $\mathbb{Z}G$, then $$\mathfrak{...
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Proper ideals generated by central ideals

Let $R$ be a unital ring and denote its center by $Z(R)$. If $I$ is an ideal of $Z(R)$, then the set $RI$ (consisting of finite sums of elements of the form ra where $r\in R$ and $a\in I$) is clearly ...
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An extension of an algebraic number field which makes an integral ideal $I$, a principal ideal

I want to show that, given an ideal $I \subseteq \mathcal O_K$ (where $K/\mathbb Q$ is an algebraic number field), there is a finite extension $K'/K$ such that, $I\mathcal O_{K'}$ becomes a principal ...
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Techniques for showing an ideal in $k[x_1,\ldots,x_n]$ is prime

An affine variety $X$ over a field $k$ is irreducible if and only if its defining ideal $I(X)$ is prime (in this post we use the convention that varieties are not necessarily irreducible). Hence, it ...
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Two ideals both alike in dignity, in fair Paris where we lay our scene. (proving ideals are isomorphic)

Let $A$ be an integral domain. I have to show that two ideals $\mathfrak a$ and $\mathfrak b$ are isomorphic as $A$-modules if and only if there exist $a$ and $b$ such that $a\mathfrak b=b\mathfrak a$....
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Generators for the intersection of two ideals

Let $I=\langle a_1,\dots, a_s\rangle, J=\langle b_1,\dots, b_t\rangle$ be ideals of arbitrary commutative ring. Then we know that $I+J=\langle a_1,\dots, a_s, b_1,\dots, b_t\rangle, IJ=\langle\{...
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On those integers $n>1$ such that any commutative ring with identity having exactly $n$ ideals is a PIR

Convention : All rings are commutative with unity unless stated otherwise. By ideals we will mean to include $\{0\}$ and $R$ also. Let us call an integer $n>1$ a "principal number" if any ring $R$...
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Ideal with large Grobner basis with respect to one monomial order

What is an example of a set of at most four polynomials $f_1,\ldots,f_n$ (in any number of variables) such that $\{f_i\}$ is a Grobner basis of $I=\langle f_i\rangle$ with respect to one monomial ...
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A Gröbner Basis Computation Gone Bad

Here is the problem statement: Consider the polynomial ideal $I = \langle b-r_1-r_2, c-r_1r_2 \rangle \subset \mathbb{Q}[r_1,r_2,b,c].$ Show that $I \cap \mathbb{Q}[b,c] = \langle 0 \rangle$. ...
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The difference between the ring version and module version of Chinese Remainder Thereom.

Chinese Remainder Theorem for Commutative Rings If $R$ is a commutative ring with $1$ and $I, J$ are ideals of $R$ that are pairwise coprime or comaximal (meaning $I + J = R$), then $IJ = I \cap J$, ...
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Finitely generated idempotent ideals are principal: proof without using Nakayama's lemma

I am trying to understand Nakayama's lemma. It looks like some "fixed point theorem". Using Nakayama's lemma , I can easily solve the following question. I want another proof. Thanks. Let $A$ be a ...
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Does every prime ideal in a ring arise as kernel of a homomorphism into $\mathbb{Z}$?

Let $R$ be a commutative ring. Clearly the kernel of $h$ is a prime ideal whenever $h : R \rightarrow ...
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Example of a ring with an infinite inclusion chain of ideals [closed]

I'm trying to track down an example of a ring in which there exists an infinite chain of ideals under inclusion. (i.e. $I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq...$)
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The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not principal [closed]

The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not a principal ideal. I don't know how to consider it. Any suggestions?
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Is $(xy-1)$ a maximal ideal in $\mathbb C[x,y]$?

I learnd that the maximal ideals in $\mathbb C[x,y]$ have the form $(x-z_1, y-z_2)$ by the Nullstellensatz. But if we set $I=(xy-1)$ then $\mathbb C[x,y]/I$ is isomorphic to $\mathbb C[x,1/x]$ which ...
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$\mathbb Z\times\mathbb Z$ is principal but is not a PID

I need to find an example of a ring that is not a PID but every ideal is principal. I know that $\mathbb Z\times\mathbb Z$ is not an integral domain, so certainly is not a PID, but here every ideal is ...
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Show that the ideal of all polynomials of degree at least 5 in $\mathbb Q[x]$ is not prime

Let $I$ be the subset of $\mathbb{Q}[x]$ that consists of all the polynomials whose first five terms are 0. I've proven that $I$ is an ideal (any polynomial multiplied by a polynomial in $I$ must ...
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An ideal that is maximal among non-finitely generated ideals is prime.

I've been doing some old exam problems and I've come across a problem that I've answered, but my gut is telling me that there's something I'm glossing over. Let $R$ be a commutative ring with ...
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Ideal of the twisted cubic

The twisted cubic is the image of the morphism $\phi : \mathbb{P}^1 \to \mathbb{P}^3 , (x:y) \mapsto (x^3:x^2 y:x y^2:y^3)$, it is given by $X = V(ad-bc,b^2-ac,c^2-bd)$. Now I would like to compute $I(...
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Prime ideals in $C[0,1]$

Are there any prime ideals in the ring $C[0,1]$ of continuous functions $[0,1]\rightarrow \mathbb{R}$, which are not maximal? Perhaps, I duplicate smb's question, but this is an interesting problem! ...
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Are distinct prime ideals in a ring always coprime? If not, then when are they?

Essentially as the title suggests - in some commutative ring $K$ (with 0,1), if we have 2 distinct proper prime ideals $\mathfrak{p}_1 \neq \mathfrak{p}_2$, is it necessarily the case (or if not, when ...
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$I$ is maximal $\implies I$ is prime

Been asked to show this is true with hints $R/I$ field $\Longleftrightarrow$ $I$ is maximal and $R/I$ integral domain $\Longleftrightarrow$ $I$ prime. Can you check this please, I have had a ten ...
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Show that $\mathbb{Z}[x]=\lbrace \sum_{i=0}^{n}{a_ix^i}:a_i \in \mathbb{Z}, n \geq 0 \rbrace$ is not a principal ideal ring.

Show that $\mathbb{Z}[x]=\lbrace \sum_{i=0}^{n}{a_ix^i}:a_i \in \mathbb{Z}, n \geq 0 \rbrace$ is not a principal ideal ring. I know the definition of principal ideal ring is that every ideal is ...
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If $I\subseteq J\subseteq A$ have same image in localization by all maximal ideals, then $I=J$

I will state my question first: Suppose $I\subseteq J\subseteq A$ are two ideals in a commutative ring $A$. Furthermore, assume that for every maximal ideal $\mathfrak{m}$ of $A$, the image of $...
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Having trouble with just one line in a proof on why nonzero prime ideals are maximal in a Dedekind domain

http://planetmath.org/?op=getobj&from=objects&name=ProofThatADomainIsDedekindIfItsIdealsAreInvertible In the PlanetMath article above, in the second paragraph of the proof of the first lemma, ...
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Why is $(2, 1+\sqrt{-5})$ not principal?

Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$? Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb Z[\sqrt{-5}]...
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The maximal ideal in a local ring is finitely generated

Assume $m<R$ is the maximal ideal of a commutative local ring with identity, such that $m=m^2$. Is $m$ finitely generated? Is the condition $m=m^2$ redundant? I am trying to apply Nakayama's lemma ...
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Is an ideal which is maximal with respect to the property that it consists of zero divisors necessarily prime?

This is in follow-up to this question. Let $R$ be a commutative ring with identity and consider the set $Z \subset R$ of zero divisors. If the ideal $I\subset Z$ is maximal with respect to the ...
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Norm and square of the ideal $(2,1+\sqrt{-5})$ in the ring of integers

Let $I=(2,1+\sqrt{-5})$ be an ideal of the ring of integers of $\mathbb Q(\sqrt{-5})$. What is its norm $N(I)$? And is $I^2$ principal? My notes say: $1$, $\sqrt{-5}$ is a $\mathbb Z$-basis for $...
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Is every proper nontrivial ideal in a Noetherian ring not flat?

I guess my general question is exactly what's in the title, but let me explain why I'm asking and how I came to it. Consider the ideal $I=\langle x,y \rangle \subset k[x,y]$ for a field $k$. Just to ...
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Show that an integral domain with finitely many ideals is a field

I know that an integral domain with finite number of elements is a field, but, how do relate this with the finitude of the number of ideals?
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Prove $M$ is a Maximal Ideal in $\Bbb Z\times \Bbb Z$

A problem from introduction to abstract algebra by Hungerford. It asks: If $p$ is a prime integer, prove that $M$ is a maximal ideal in $\mathbb Z \times \mathbb Z$, where $M =\{(pa,b)\mid a,b\in \...
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Ideal of a ring

I'm trying to describe an ideal of the ring $R=\left\{ \begin{pmatrix}a & b\\ 0 & c \end{pmatrix}:a,b,c \in \mathbb{R}\right\} $ It's easy to prove that $I=\left\{ \begin{pmatrix}0 & a\\ 0 ...
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Norm of ideals in quadratic number fields

I do not really understand how to factor ideals in a quadratic field $K = \mathbb{Q}(\sqrt{d})$, mainly because I have some trouble computing the norm of ideals. I think I understand what is going on ...
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Finding the ideals in a ring of fractions

I am dealing with the ring $$R=\left\{\frac{a}{b} \mid a,b\in\mathbb{Z}\mbox{, $b$ is not divisible by 3}\right\}$$ with addition and multiplication as defined in $\mathbb{Q}$ and I'm trying to find ...
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In $K[X,Y]$, is the power of any prime also primary?

I've recently been reading about primary decomposition, and was browsing the questions here. From this, I know that it is not true that every primary ideal is the power of a prime ideal. I'm curious ...
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What phenomenon is this? $(2\Bbb{Z} + 1)\cup 3\Bbb{Z} = 2\Bbb{Z} \cup 3\Bbb{Z} + 3$

$(2\Bbb{Z} + 1)\cup 3\Bbb{Z} = 2\Bbb{Z} \cup 3\Bbb{Z} + 3$ Proof: $$ \begin{align*} 2\Bbb{Z} &= \bullet \circ \bullet \circ \bullet \circ \bullet \circ \dots \\ 3\Bbb{Z} &= \bullet \circ \...
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How to show that $\mathbb{C}[x_1,x_2,x_3, x_4]/(x_1x_2 - x_3x_4, x_1x_3 - x_2x_4, x_1x_4 - x_2x_3)$ is an integral domain?

I am looking for a way to show that the ring $\mathbb{C}[x_1,x_2,x_3, x_4]/I$ where $$I = (x_1x_2 - x_3x_4, x_1x_3 - x_2x_4, x_1x_4 - x_2x_3)$$ is an integral domain. In other words I want to show $I$...
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In an extension of finitely generated $k$-algebras the contraction of a maximal ideal is also maximal

Let $k$ be a field and let $A \subset B$ be two finitely generated $k$-algebras. Prove that the contraction of any maximal ideal of $B$ is a maximal ideal of $A$. thank you very much again!
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Is there an example of commutative ring with exactly three prime ideals for which this property holds?

Is there an example of commutative ring with exactly three non zero prime ideals $P_i$ which satisfies the following statement: $P_1P_2=0$ and for an ideal $I\neq 0$ such that $I\neq P_i$ we have $...
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Definition of primary ideal question

A primary ideal (in a commutative ring with unity) is an ideal $J$ for which if $ab\in J$, then either $a\in J$ or $b^n\in J$ for some integer $n\geq 1$. So it also implies (due to commutativity) that ...
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Points and maximal ideals in polynomial rings

Let $k$ be a field, then I want to prove the following statement: for every $P=(b_1,\ldots,b_n)\in K^n$, the ideal $\mathfrak{m}_P=(x_1-b_1,\ldots,x_n-b_n)$ is maximal in the polynomial ring $k[x_1,\...
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The ideal $(x,y)$ is not a free $K[x,y]$-module

Given a field $K$ we have the polynomial ring $K[x,y]$ in $2$ variables, which is also a left module (over itself). How can we prove that the ideal $(x,y)$ is not a free module?
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Artin's Algebra exercise special case of some theorem/problem?

The following exercise is from Artin's Algebra Text: Show that there is a one to one correspondence between maximal ideals of $ \bf R$$[x]$ and complex upper half plane. Solution: Follows from ...
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For which countable successor ordinals $\alpha$ is the reverse order isomorphic to the ideals of a PID ordered by inclusion?

Let $\alpha$ be a countable successor ordinal and $\alpha^{\mathrm{op}}$ the reverse order. For which $\alpha$ is there a commutative principal ideal ring $R$ such that the ideals of $R$ form a chain ...
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An ideal that is radical but not prime.

I'm preparing for an exam and, as part of this preparation, I'm looking for an ideal $I$ in an integral domain $R$ that is radical but not prime. Here is an example I'm fooling around with: ...
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When is k[x,y]/I complete for the (x,y)-adic topology?

Let $k$ be a field. If necessary, add assumptions on $k$ or just take $k=\mathbb{C}$. It is easy to classify the ideals $I \subseteq k[x]$ such that $k[x]/I \to k[[x]]/(I)$ is an isomorphism, namely $...