Questions about commutative rings, their ideals, and their modules.

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Computing Krull dimension of $\mathbb{Z}[X_1,\ldots,X_n]/I$ [closed]

Let $I$ be an ideal of $\mathbb{Z}[X_1,\ldots,X_n]$. How does one compute the Krull dimension of $\mathbb{Z}[X_1,\ldots,X_n]/I$? Are there any general methods? Or methods which work in special cases?
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2answers
89 views

Let $(R,M)$ be a local ring. Suppose that $R$ is noetherian and let $I,J \unlhd R$ such that $J \subseteq I$. Prove that the following are equivalent.

Let $R$ be a local ring with maximal ideal $M$. Suppose that $R$ is noetherian and let $I,J$ be ideals of $R$ such that $J \subseteq I$. Consider the following statements: 1) Every minimal set of ...
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1answer
100 views

Is $\mathbb{R}[x,y,z]/(x^2+y^2+z^2)$ a UFD?

As the title says, I am curious as to whether $A =\mathbb{R}[x,y,z]/(x^2+y^2+z^2)$ is a UFD. I believe the answer is yes. A thought I had was to apply Nagata's criterion, say by localizing ...
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66 views

Commutative algebra text that solely contains 200+ exercises

I am looking for a textbook that I came across awhile ago that I have been unable to find for the last week or so of periodic searching. The textbook had nothing other than, from memory 247 exercises ...
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1answer
62 views

Given $I=\langle xy, xz+z(y^2-z^2)\rangle$, prove that $I=\langle x, z(y^2-z^2)\rangle \cap \langle y, xz-z^3)\rangle $.

This is Exercise 3c. from Chapter 9, Section 7 of Ideals, Varieties, and Algorithms by Cox et al. Given $I=\langle xy, xz+z(y^2-z^2)\rangle$, prove that $I=\langle x, z(y^2-z^2)\rangle \cap ...
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1answer
57 views

Does $S = R \cap K$ of a field extension $K \subseteq L = Q(R)$ satisfy $Q(S) = K$?

If $K$ is finite field, then one can easily show that there is no proper subring $R$ with $Q(R) = K$, where $Q(R)$ is the field of fractions of $R$. As a consequence, algebraic extensions $K$ of ...
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118 views

If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$? [migrated]

I was describing to a friend the result that a compact Hausdorff space is determined up to homeomorphism up to by its ring of continuous functions, and he asked how one could see the genus of a ...
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0answers
21 views

Locally presentable sheaves and the associated module functor

Let $R$ be a commutative ring. Any $R$-module has a presentation $R^{(J)}\rightarrow R^{(I)}\rightarrow M\rightarrow 0$. The associated module functor $M\mapsto \tilde M$ is exact and so preserves ...
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15 views

Counter example that Artinian k-algebra is not finite k-vector space [duplicate]

Let k be a field, A be a k-algebra. If A is not a finitely generated k-algebra,then the following two conditions are NOT equivalent: (i) A is Artinian; (ii) A is a finitely k-algebra, i.e. A is ...
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If $f: A\to B$ is faithfully flat and $B$ is an Artinian ring then $A$ is also Artinian.

Let $f : A → B$ be a map of rings. The map $f$ is called faithfully flat if $B$ is flat $A$-module ($B$ is $A$-module w.r.t. multiplication defined by $ab := f(a)b$) and if for any $A$-module $M, M ...
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0answers
52 views

Krull dimension of quotient ring

What is the Krull dimension of $B = A[x,y,z]/\langle x^2y + x + 1, y^3 + 2z + 1 \rangle $ given $A$ is a Noetherian, commutative ring? (Assuming that all coefficients are non zero in $A$)
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1answer
48 views

Maximal ideal in $\mathfrak{R}[x]$

Let $\mathfrak{R}$ be a commutative ring with identity. Show that if there exists a monic polynomial $p(x)\in \mathfrak{R}[x]$ of degree at least one such that the ideal $(p(x)) ...
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2answers
50 views

Localising Dedekind domains

I'm wondering if the following is true: Let $A\subset B$ be two Dedekind domains with $B$ integral over $A$. Let $Q$ be a non-zero prime ideal in $B$ and $P=Q\cap A$. Then the localisation of $B$ ...
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100 views

Description of the equalizer of $\prod _j R/I_j \rightrightarrows \prod _{i,j}R/(I_i+I_j)$

This will hopefully be my last question about the chinese remainder theorem. I have asked several questions in an attmept to get a general version without conditions on the ideals which will trivially ...
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1answer
58 views

prove the inverse image of a maximal ideal is also a maximal ideal

The following is the problem: Let $K$ be a field, and $A$ is a commutative ring containing $K$. $\phi:A \rightarrow K[X]$ is a ring homomorphism which is the identity on $K$. If $M$ is a maximal ...
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0answers
36 views

$X \times \mathbf{A}^1$ is regular in codimension 1

This is Proposition II.6.6 in Hartshorne. We assume that $X$ is a Noetherian, integral, and separated scheme which is regular in codimension 1, i.e. every local ring of dimension one is regular. For ...
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40 views

Spec $\mathbb Q[x]$ [duplicate]

I am trying to find Spec $\mathbb Q[x]$. Since $\mathbb Q$ is a field, the prime ideal is just generated by irreducible polynomial with coefficients in $\mathbb Q$. I know the case of $\mathbb R$, ...
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1answer
84 views

Dimension of variety $V(I)$ where $I=\langle xy,xz \rangle\in k[x,y,z]$ and $k$ is an infinite field.

I have some doubts about this question although I know the answer is $2$. The dimension of an affine variety is defined in many different ways. One of the definitions defines it as the maximum ...
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1answer
183 views

If $R[x]$ and $R[[x]]$ are isomorphic, then are they isomorphic to $R$ as well? [duplicate]

There are examples of commutative rings $R \neq 0$ such that $R[x]$ is isomorphic to $R[[x]]$ (see this question; an example would be $R=S[x_1, x_2, \ldots][[y_1, y_2, \ldots]]$, with $S \neq 0$ any ...
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Does the resultant of two coprime polynomials measure something?

Let $R[X]$ be the polynomial ring over a commutative ring $R$, $f$ and $g \in R[X]$ two coprime polynomials, and $r(f,g)\in R$ their resultant. Because $f$ and $g$ are coprime, $r(f,g)\neq 0$. But ...
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1answer
31 views

Proving that there only finitely many minimal prime ideals of any ideal in Noetherian commutative ring

Currently, I'm trying to solve a problem from a textbook: Let $R$ be a commutative Noetherian ring with identity, and let $I \subset R$ be a proper ideal of $R$. Then we know that set of prime Ideals ...
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2answers
52 views

Is the Zariski topology the same as the cofinite topology?

Let $R$ be a commutative ring, $spec(R)$ be the set of all prime ideals on $R$. For any ideal $I$ on $R$, we define the $V_I$ to be the set of all prime ideals containing $I$. We define the Zariski ...
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2answers
66 views

Ring of Invariants of symmetric group

The symmetric group $S_n$ acts on $\mathbb C^n$ by permuting the coordinates. In this case the ring of invariants is generated by elementary symmetric polynomials in n-variables. Now consider the ...
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2answers
43 views

How do we prove the “lying over” property for integral extensions?

Let $R \subset S$ be an integral extension of commutative rings. Then if $P \subset R$ is prime, there exists a prime ideal $Q \subset S$ such that $Q \cap R = P$. My D&F book says look at ...
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1answer
40 views

Codimension and height of prime ideals

Definition: Let $Z$ be an irreducible closed subset of $X$. Then the codimension $\textrm{codim} (Z,X)$ is the supremum of integers $n$ such that there exists a chain $$ Z = Z_0 < Z_1 < ...
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1answer
48 views

Exercise from Kaplansky - Commutative Rings (1.2.3)

Exercise 3 in section 1-2: Let $R$ be an integral domain, $P$ a finitely generated non-zero prime ideal in $R$, and $I$ an ideal in $R$ properly containing $P$. Let $x$ be an element in the ...
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1answer
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Short exact sequences of quasi-coherent sheaves and closed subschemes

I am confused by this exercise in Ravi's notes: 16.3F (paraphrasing) Suppose $i : p \to A_k^1$ is the inclusion of the origin. Consider the associated short exact sequence of quasi-coherent modules: ...
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46 views

Idempotent direct summands of rings

I know that if an ideal $I$ is a direct summand of a ring $R$ then it is an idempotent ideal, i.e. $I^2=I$. My question concerns the rings all of whose idempotent ideals are direct summands. ...
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Example of a module which is free at an isolated point

I'm looking for the most simple example of a quasicoherent sheaf $\mathcal{F}$ over a scheme $X$ (preferably affine for simplicity) which has a free stalk $\mathcal{F}_x$ at a point $x \in X$ and yet ...
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1answer
40 views

In what sense are complete local rings finitely generated modules?

In the first paragraph of section 18.4 of Eisenbud's Commutative Algebra, there is the following comment. Most interesting Noetherian rings can be written as finitely generated modules over ...
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2answers
53 views

Norm of powers of a maximal ideal (in residually finite rings)

Let $A$ be a residually finite integral domain and $M$ a maximal ideal in $A$. Is this true that $$|A/M^k|=|A/M|^k \quad (k\in\textbf{N}) \quad ?$$ In Hirano's article On Residually Finite Rings we ...
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2answers
31 views

Norm of powers of a maximal ideal

Let $A$ be a integral domain and $M$ a maximal ideal in $A$ such that the quotient $A/M$ is a finite ring (and thus a finite field). Is it true, in general, that $$|A/M^k|=|A/M|^k \quad ...
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1answer
23 views

Help to check a proof about local prime ideal being principal?

Let $K$ be a number field, call it's ring of integers $\mathcal O_K$ and take a - possibly nonprincipal - prime ideal $\mathfrak q$. I have shown that $\mathcal O_K$ is Noetherian integral domain and ...
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0answers
24 views

The real points of $\operatorname{mSpec}( C(\mathbb{R}))$?

Known fact: If $X$ is a compact, Hausdorff space, then $X$ is homeomorphic to the max spectrum of $C(X)$ with the Zariski topology. This fails for non-compact spaces, as for instance there may be too ...
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1answer
126 views

Chinese remainder theorem as sheaf condition?

The chinese remainder theorem in its usual version says that for a finite set of pairwise comaximal ideals $R/\bigcap _jI_j\cong \prod _j R/I_j$. In the binary case, the following general statement ...
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17 views

Torsion-free submodule of maximal rank

Let $A$ be a free $\mathbb{Q}[t_1^{\pm 1},\dots,t_\mu^{\pm 1}]$-module of rank $n$. If $M$ is a submodule of $A$, I know that it must be torsion free. Assuming the rank of $M$ is $n$ can I conclude ...
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1answer
25 views

Idempotent subideals of $J(R)$

If $R$ is a unital ring, it is well-known that its Jacobson radical $J(R)$ contains no non-zero idempotent element of $R$. My question: Is there a ring $R$ such that $J(R)$ contains a non-zero ...
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31 views

Relation between stalks of twisted sheaf and structure sheaf

Let $A$ be a ring, $B = A[T_0,\dots, T_d]$, and $X = \textrm{Proj } B$. Then at every point $x \in X$, $$\mathcal{O}_X (n)_x \cong \mathcal{O}_{X,x}$$ Let $x$ correspond to a homogeneous prime ...
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3answers
70 views

Is $A / \mathfrak{m}$ flat if $A$ is a local ring?

I'd like to prove the following: if $A$ is a local ring and $\mathfrak{m} \subset A$ its maximal ideal, then $A / \mathfrak{m}$ is a flat $A$-module. How can I do this? I've tried to find a suitable ...
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2answers
52 views

Proper ideal $I \implies \exists $ prime ideals $P_i$ such that $P_1 \cdots P_n \subset I$.

Let the below ideals be in a commutative Noetherian ring $R$. Corollary 22. (3) There are prime ideals $P_1, \dots, P_n$ (not necc. distinct) $\supset I$ such that $P_1\cdots P_n \subset I$. (Out ...
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2answers
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If $R$ is a noetherian ring, then minimal primes of $R[x]$ are exactly the ideals $P[x]$ of $R[x]$ where $P$ is a minimal prime of $R$

Show that if $R$ is a noetherian ring, then minimal primes of $R[x]$ are exactly the ideals $P[x]$ of $R[x]$ where $P$ is a minimal prime of $R$. Definition of a minimal prime ideal of a ring ...
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1answer
37 views

$\mathbb{F}_p$ algebra with many $p$th roots of unity

An old qual problem reads Let $\mathbb{F}_p$ denote the finite field of $p$ elements. Consider the covariant functor $F$ from the category of commutative $\mathbb{F}_p$ algebras with ...
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2answers
66 views

Germs and local ring.

I'm having trouble understanding the following argument (which I believe to be somewhat incomplete or flawed). Let $A=C(X)$ be the set of continuous functions from the topological space $X$ to the ...
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2answers
93 views

Applying the Yoneda-Lemma to prove the existence of Tensor-products

In class the professor said when he came to prove the existence of the tensor-product for $A$-modules ($A$ any ring) that the existence and properties of the tensor-product would be one-liners having ...
4
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1answer
69 views

Nullstellensatz for non-algebraically closed fields

I'm trying to prove that the Nullstellensatz holds for non algebraically closed fields, when the variety is taken over the algebraic closure. Let $R=K[x_1,...,x_n]$ and $\overline{K}$ the algebraic ...
3
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1answer
67 views

Hom / tensor adjunction for $O_X$ modules?

Does the hom-tensor adjunction hold for $O_X$ modules also? With sheaf hom and sheaf tensor product, the statement would consist of a natural transformation $Hom_O (M \otimes_O N, K)\cong_{nat} ...
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1answer
38 views

Do the strongly vanishing elements of $R[[x]]$ form an ideal?

I've always been a bit annoyed by expressions like $$\sum_{n:\mathbb{N}} a_n$$ when the relevant limit doesn't converge, for the following reason: if you're not going to tell the reader what ring this ...
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0answers
30 views

(Atiyah) If $A \subseteq B$ and $B \setminus A$ is multiplicatively closed then $A$ is integrally closed in $B$ [duplicate]

I've tried proof by contradiction, with $y \in B\setminus A$ and considering an integral expression $y^n = a_{n-1} y^{n-1} +\dots + a_0$ of least degree (hence $a_0 \neq 0).$ Then $a_{n-1} y^{n-1} ...
3
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1answer
61 views

Exercise from Kaplansky - Commutative Rings (1.1.3)

Exercise 3 in section 1-1: Let $P$ be a finitely generated prime ideal with annihilator 0. Prove that the annihilator of the module $P/P^2$ is $P$. (Hint: If $p_1,\cdots,p_n$ generate $P$ and $x$ ...
2
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1answer
33 views

Does $\operatorname{Hom}_A (A / P, M) \not = 0$ imply that $P$ is an associated prime of $M$?

$A$ is Noetherian, $M$ is finitely generated. Does $\operatorname{Hom}_A (A / P, M) \not = 0$ imply that $P$ is an associated prime of $M$? I am trying to prove that associated primes ...