# Questions tagged [ideals]

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, ideals in Lie algebras.

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### Why are maximal ideals prime?

Could anyone explain to me why maximal ideals are prime? I'm approaching it like this, let $R$ be a commutative ring with $1$ and let $A$ be a maximal ideal. Let $a,b\in R:ab\in A$. I'm trying to ...
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### Classification of prime ideals of $\mathbb{Z}[X]$

Let $\mathbb{Z}[X]$ be the ring of polynomials in one variable over $\Bbb Z$. My question: Is every prime ideal of $\mathbb{Z}[X]$ one of following types? If yes, how would you prove this? ...
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### Is quotient of a ring by a power of a maximal ideal local?

Say I have a commutative ring $R$ with a maximal ideal $m$. Then $m/m^k$ is a maximal ideal in $R/m^k$ for any $k$. Is it the only maximal ideal, i.e. is $R/m^k$ a local ring? This is a well known ...
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### Finitely many prime ideals lying over $\mathfrak{p}$

Let $A$ be a commutative ring with identity and $B$ a finitely generated $A$-algebra that is integral over $A$. If $\mathfrak{p}$ is a prime ideal of $A$, there are only finitely many prime ideals $P$...
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### Verifying that the ideal $(x^3-y^2)$ is prime

How to prove that the ideal $I=(x^3-y^2)$ in $k[x,y]$ is prime? I have constructed a map from $k[x,y]$ to $k[t]$, which maps $x$ to $t^2$, and $y$ to $t^3$. Then, I want to show that the kernel is ...
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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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### Maximal ideals in $K[X_1,\dots,X_n]$

Let $K$ be a field, and $a_1,\dots,a_n \in K$. Prove that the ideal $$(X_1-a_1,\dots,X_n-a_n)$$ is maximal in $K[X_1,\dots,X_n]$. I tried proving that the only elements outside the ideal are the ...
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### What do prime ideals in $k[x,y]$ look like?

Suppose that $k$ is an algebraically closed field. Then what do the prime ideals in the polynomial ring $k[x,y]$ look like? As far as I know, the maximal ideals of $k[x,y]$ are of the form $(x-a,y-b)$...
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### An integral domain whose every prime ideal is principal is a PID

Does anyone has a simple proof of the following fact: An integral domain whose every prime ideal is principal is a principal ideal domain (PID).
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### Maximal ideals in the ring of real functions on $[0,1]$ [closed]

Let $S$ be the ring of all continuous functions from $[0,1]$ to $\mathbb R$. How to prove that all maximal ideals of $S$ have the form $M_{x_0}=\{f\in S \mid f(x_0)=0\}$? Thanks in advance.
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### Why are ideals more important than subrings?

I have read that subgroups, subrings, submodules, etc. are substructures. But if you look at the definition of the Noetherian rings and Noetherian modules, Noetherian rings are defined with ideals ...
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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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### 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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### An integral domain $A$ is exactly the intersection of the localisations of $A$ at each maximal ideal

This result appears to be ubiquitous as an algebra exercise. How do you prove this result? Let $A$ be an integral domain with field of fractions $K$, and let $A_{\mathfrak{m}}$ denote the ...
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### Can an ideal in a commutative integral domain be its own square?

If $I$ is a non-zero proper ideal of a commutative integral domain, is it possible for $I$ to be its own square?
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### The set of all nilpotent elements is an ideal

Given that R is commutative ring with unity, I want show that set of all nilpotent elements is an ideal of R. I know how to show ideal if set is given but here set is not given to me. Can anyone ...
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### Ideals in direct product of rings [duplicate]

I am trying to solve this problem: Let $R_1,...,R_n$ be rings with identity. Every ideal of $R=\prod_{i=1}^n R_i$ is of the form $\prod_{i=1}^n I_i$ where $I_i$ is an ideal of $R_i$. The ...
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### Primary decomposition of a monomial ideal

Can anyone give me an idea about the primary decomposition of the ideal $I=(x^3y,xy^4)$ of the ring $R=k[x,y]$? I am trying to connect the primary decomposition with the set Ass(R/I) which i have ...
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### Ideals of $\mathbb{Z}[X]$

Is it possible to classify all ideals of $\mathbb{Z}[X]$? By this I mean a preferably short enumerable list which contains every ideal exactly once, preferably specified by generators. The prime ...
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### Explaining the product of two ideals

My textbook says that the product of two ideals $I$ and $J$ is the set of all finite sums of elements of the form $ab$ with $a \in I$ and $b \in J$. What does this mean exactly? Can you give examples?
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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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### if R is a commutative ring in which all the prime ideals are finitely generated then R is Noetherian [duplicate]

Prove that if $R$ is a commutative ring in which all the prime ideals are finitely generated, then $R$ is Noetherian. Here is what I been told to do: Suppose that $R$ is not Noetherian, and use Zorn’...
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### Can the square of a proper ideal be equal to the ideal?

Let $R$ be a ring, commutative with $1$, let $\mathfrak{i}$ be an ideal, not the whole ring. In general $\mathfrak{i}^2\subseteq\mathfrak{i}$. Can this inclusion be an equality, or it is always a ...
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### Cardinality of quotient ring $\mathbb{Z_6}[X]/(2X+4)$

Let $R$ be the ring obtained by taking the quotient of $\mathbb{Z_6}[X]$ by principal ideal $(2X+4)$. Then 1) $R$ has infinite elements 2) $R$ is field 3) $5$ is unit in $R$ 4) $4$ is unit in $R$. ...
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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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### Converse to Chinese Remainder Theorem

So as seen on this question Converse of the Chinese Remainder Theorem, we know that if $(n,m) \neq 1$, then $\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, ...
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### Unique factorization domain and principal ideals

If R was a unique factorization domain, can we deduce that for a nonzero element d in R, d has a finite number of divisors? I need this in solving this question "If R is a unique factorization ...
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### In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal. What could be wrong in this approach?

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal Attempt: Let $R$ be the principal ideal domain. A principal ideal domain $R$ is an integral domain in which ...
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### Number of elements in the ring $\mathbb Z [i]/\langle 2+2i\rangle$

The question is : Show that $I=\langle 2+2 i\rangle$ is not a prime ideal of $\mathbb Z[i]$. Also find the number of elements in $\mathbb Z[i]/I$ and its characteristic. My try: I started with ...
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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 ...
If $K$ is a field, any non-zero ideal in the ring of formal power series $A=K[[X]]$ is of the form $AX^n$ with $n\geq 0$, so $A=K[[X]]$ is a principal ideal ring. I can't see why. Usually when we ...