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

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25
votes
4answers
1k views

Why isn't $\mathbb{C}[x,y,z]/(xz-y)$ a flat $\mathbb{C}[x,y]$-module

Why isn't $M = \mathbb{C}[x,y,z]/(xz-y)$ a flat $R = \mathbb{C}[x,y]$-module? The reason given on the book is "the surface defined by $y-xz$ doesn't lie flat on the $(x,y)$-plane". But I don't ...
4
votes
1answer
114 views

$C$ is irreducible iff $C=\mathscr{Z}(\mathfrak{p})$ for some prime ideal $\mathfrak{p}$?

Let $A$ be a commutative Noetherian ring, and $C$ a closed subset of $\operatorname{Spec}(A)$. In some reading, it is an unproven proposition that $C$ is irreducible iff ...
5
votes
1answer
270 views

Finite injective dimension of the residue field implies that the ring is regular

Let $(R,\mathfrak m,k)$ be a noetherian local ring. If $\operatorname{inj dim}_R k$ is finite, then $R$ is regular. This is exercise 3.1.26 from Bruns and Herzog, Cohen-Macaulay Rings. I don't ...
9
votes
1answer
647 views

Construct ideals in $\mathbb Z[x]$ with a given least number of generators

How do you construct, for each $n\geq 1$, an ideal in $\mathbb Z[x]$ of the form $(a_1,a_2,\dots,a_n)$ with $a_i\in \mathbb Z[x]$ such that it is impossible to have ...
4
votes
3answers
1k views

Finitely generated ideals in a Boolean ring are principal, why?

The classical book on commutative algebra Introduction to Commutative Algebra, by Atiyah and Macdonald, has the following as exercise I.11. A ring is Boolean if $x^2=x$ for any $x$ of $A$. In a ...
2
votes
0answers
261 views

does it make sense to tensor a simplicial set with a simplicial ring?

Given a simplicial commutative ring $A$ and a simplicial set $K$ does $K \otimes A$ make sense as a (commutative) simplicial ring? I'm asking as I've seen the expression $S^n \otimes A$ written down ...
1
vote
1answer
434 views

Is there a Noetherian faithful module over a non-noetherian ring?

Let $R$ be a ring which is not Noetherian. Let $M$ be an $R$-module whose annihilator is trivial. Is it possible for $M$ to be Noetherian? Intuitively, it seems like the answer should be no, as ...
7
votes
2answers
3k views

When is a product of two ideals strictly included in their intersection?

Let $I,J$ two ideals in a ring $R$. The product of ideals $IJ$ is included in $I \cap J$. For example we have equality in $\mathbb{Z}$ if generators have no common nontrival factors, in a ring $R$ ...
9
votes
1answer
298 views

Faithfully Flat Ring Homomorphism of Power Series

Let $R$ be a one-dimensional local ring and let $f:R[[x]][y] \rightarrow R[y][[x]]$ be the inclusion map. How can I show that $f$ is a faithfully flat ring homomorphism? Or can you give me a ...
2
votes
4answers
694 views

Prime ideals in polynomials rings

Let $A$ be a commutative ring, $\mathbb{q}\subset A$ an ideal of $A$, and $\mathbb{q}A[x]$ the ideal of $A[x]$ generated by $\mathbb{q}$ (consists of the polynomials with coefficients in ...
14
votes
3answers
689 views

Direct way to show: $\operatorname{Spec}(A)$ is $T_1$ $\Rightarrow$ $\operatorname{Spec}(A)$ is Hausdorff

In the book of Atiyah and MacDonald, I was doing exercise 3.11. One has to show that for a ring $A$, the following are equivalent: $A/\mathfrak{N}$ is absolute flat, where $\mathfrak{N}$ is the ...
7
votes
3answers
238 views

$\operatorname{Spec} (A)$ as a topological space satisfying the $T_0$ axiom

I have been spending a few days now proving the last bit of the following problem of Atiyah Macdonald: Prove that $X = \operatorname{Spec}(A)$ as a topological space with the Zariski Topology ...
6
votes
2answers
360 views

Intersection of algebraic sets not equal to $\{0\}$?

I was hoping to ask a small follow up to the question I asked here. Suppose $V$ is an algebraic variety over arbitrary field $k$. (For this situation, I'll take the definition $\dim\ V=\deg_k(k(x))$, ...
4
votes
1answer
325 views

Dimension of its irreducible components in Elimination Theory.

There is a small result I don't understand. To preface, for an algebraic variety $V\subset\mathbb{A}^n$ over some field $F$, one defines $\dim V=\operatorname{trdeg}(F(x)/F)$ for a generic point ...
9
votes
0answers
421 views

Trivial intersection of algebraic sets?

The question came up while reading a bit more into the Hilbert-Zariski theorem I asked about the other week. Suppose $V$ is an algebraic variety over arbitrary field $k$. (For this situation, I'll ...
1
vote
1answer
80 views

How is this lower bound of the dimension of a homogeneous algebraic variety reached?

I'm hoping to see how the following bound is reached. For an algebraic variety $V\subset\mathbb{A}^n$ over some field $F$, one defines $\dim V=\operatorname{trdeg}(F(x)/F)$ for a generic point $(x)$ ...
2
votes
1answer
219 views

fractional ideals

If $D$ is a domain and $K$ is field then for $x\in K$, $xD$ is a fractional ideals of $D$. If $xD$ and $yD$ are two fractional ideals then is true or not $xyD\subseteq xD$ ?. Thanks
4
votes
2answers
547 views

Inverse of Elements Modulo the Maximal Ideals of the Ring of Continuous functions on $[0,1]$

Let $R$ be the ring of continuous functions on $[0,1]$ under pointwise addition and multiplication, $c \in [0,1]$, and $M_c$ the ideal defined by the set of $f\in R$ that vanish at $c$. It is ...
4
votes
1answer
151 views

Given $f, g \in k[x,y]$ coprime, why can we find $u,v \in k[x,y]$ such that $uf + vg \in k[x]\backslash 0$?

Let $k$ be a field. Given $f, g \in k[x,y]$ coprime, why can we find $u,v \in k[x,y]$ such that $uf + vg \in k[x]\backslash 0$? I can do it for specific polynomials, but I'm struggling to structure a ...
5
votes
1answer
546 views

Structure Sheaf of the Spectrum of a Ring

Let $A$ be a ring and $X$ be the spectrum of $A$ with the Zariski topology. For an element $f\in A$ let $X_f:=\{p\subset A\text{ prime ideal }\,|\,f\notin p\}$; the $X_f$ form a basis of the topology ...
5
votes
2answers
1k views

Finding generators of an ideal / showing an ideal is prime

Suppose I have an ideal like $\mathrm{rad}(\langle y - x^2, z - x^3 \rangle)$. How do I go about finding generators for the ideal? If I could show that $\langle y - x^2, z - x^3 \rangle$ is a ...
-1
votes
1answer
520 views

Prime ideals in a polynomial ring

Let $R$ be a Noetherian ring, and let $\mathfrak p$ be a prime of $R$ of codimension $d$. Suppose that $P\subset R[X]$, $P$ prime, intersects $R$ in $\mathfrak p$. Prove that if $P\neq\mathfrak ...
14
votes
1answer
1k views

How to compute localizations of quotients of polynomial rings

At the moment I'm trying to understand the concept of localizations of rings / modules. I have done some exercises (using the book of Atiyah / MacDonald) and I will do some more, but a more practical ...
1
vote
1answer
81 views

How to show a given algebra is not generated by one element

Suppose I have a $k$-algebra $k[x,y]/\langle f\rangle$, where $f$ is a (given, fixed) irreducible polynomial. What are the strategies for showing that this isn't generated by one element as a ...
3
votes
1answer
150 views

Hilbert series and initial ideals

Suppose $S$ is a polynomial ring, $I$ an ideal and $<$ some term order. Why is the Hilbert series of $S/I$ the same as the Hilbert series of $S/\mathrm{in}_<(I)$? I truly suspect the answer ...
4
votes
4answers
201 views

How to directly prove that $M$ is maximal ideal of $A$ iff $A/M$ is a field?

An ideal $M$ of a commutative ring $A$ (with unity) is maximal iff $A/M$ is a field. This is easy with the correspondence of ideals of $A/I$ with ideals of $A$ containing $I$, but how can you prove ...
3
votes
1answer
195 views

subalgebras of a polynomial ring

If $k$ is a field, I know that any subalgebra $A \subset k[x]$ is finitely generated, but I wonder if there is a good algorithm to find a set of generators for $A$. In particular, if $(f) \subset ...
7
votes
1answer
289 views

(Possibly) alternative statement of Hilbert's Nullstellensatz

In my notes, there is a statement entitled "Nulstellensatz version 2": If $k = \bar{k}$, and $\mathfrak{m} \subseteq k[x_1,\ldots,x_n]$ is a maximal ideal, then $k[x_1,\ldots,x_n]/\mathfrak{m} ...
6
votes
1answer
253 views

Equivalent condition for flatness of an A-module (Atiyah-MacDonald ex. 2.26)

I would like to solve the following exercise (2.26) from Atiyah & MacDonald's "Introduction to Commutative Algebra": If $M$ is an $A$-module (where $A$ is a commutative ring), then: $$M \text{ is ...
3
votes
1answer
95 views

Interpreting an $S^{-1}R$-module as an $R$-module.

Can one do it? I'm trying to prove that $S^{-1}I$ is an injective $S^{-1}R$-module whenever $I$ is an injective $R$-module. So I need to start with a situation where I have: (i) $S^{-1}R$-modules ...
17
votes
3answers
2k views

$A^m\hookrightarrow A^n$ implies $m\leq n$ for a ring $A\neq 0$

I'm trying to prove that if $A\neq 0$ is a commutative ring and there is an injective $A$-module homomorphism $A^m\hookrightarrow A^n$ then $m\leq n$ must necessarily hold. This is exercise 2.11 ...
4
votes
2answers
329 views

$A$ an absolutely flat ring $\Rightarrow$ $S^{-1}A$ is absolutely flat

I was doing some exercises in the book of Atiyah / MacDonald on Commutative Algebra, and I'm a little "stuck" with number 3.10 (i): If $A$ is an absolutely flat ring and $S\subseteq A$ a ...
19
votes
2answers
774 views

Is the radical of an irreducible ideal irreducible?

Fix a commutative ring $R$. Recall that an ideal $I$ of $R$ is irreducible if $I = J_1 \cap J_2$ for ideals $J_1$ and $J_2$ only when either $I = J_1$ or $I = J_2$. Question : Assume that $I$ is an ...
6
votes
1answer
2k views

An ideal is homogeneous if and only if it can be generated by homogeneous elements

Let $S$ be a graded ring with decomposition $S = \bigoplus_{d \geq 1} S_d$, where the $S_d$ are additive abelian groups such that $S_d S_e \subseteq S_{d+e}$ for $e,d \geq 1$. An element in $S_d$ is ...
10
votes
2answers
820 views

Rings whose spectrum is Hausdorff

Let $A$ be a commutative ring with $1$ and consider the Zariski topology on $\operatorname{Spec}(A)$. When will $\operatorname{Spec}(A)$ be a Hausdorff space? If $A$ has positive or infinite ...
2
votes
1answer
69 views

What is the length of the following local ring

Let $f:Y\to X$ be a finite etale cover of smooth projective connected varieties. (Or, just a finite degree connected topological cover of connected Riemann surfaces.) Let $y\in Y$ and let $x=f(y)$. ...
9
votes
1answer
121 views

Curious about Hilbert-Zariski theorem involving homogeneous variety and set of zeroes.

I got myself in a confusing situation the other week while trying to read a bit of algebraic geometry. I'm hoping someone can pull me out. Suppose $k$ is a field, and $V$ a homogeneous variety with ...
2
votes
1answer
164 views

how does one intersect the diagonal with a graph on the surface $X\times X$

I want to do a concrete example of an intersection product for myself. Consider the endomorphism $f:\mathbf{P}^1_k\to \mathbf{P}^1_k$ given by $(x:y)\to (y:x)$. It has precisely two fixed points: ...
12
votes
2answers
404 views

Integral domain with fraction field equal to $\mathbb{R}$

I wonder if there is an integral domain $A\subseteq \mathbb{R}$ which is not a field, and such that the field of fractions of $A$ is equal to $\mathbb{R}$? Edit: here as a possible direction: it is ...
8
votes
1answer
148 views

Generators of a certain ideal

Crossposted on MathOverflow. The MathOverflow version of the question has been rewritten. For the sake of completeness, I pasted it here in a condensed form. I also deleted the old version. Let $K$ ...
5
votes
1answer
138 views

Three maximal ideals lying over $3\mathbb{Z}$?

A few weeks ago I asked a question about finding the number of maximal ideals lying above $3\mathbb{Z}$ in $B$, where $B$ is the integral closure of $\mathbb{Z}$ in a splitting extension ...
5
votes
1answer
132 views

Mod-$R$, Mod-$S$ and Mod-$R \otimes S$

Let $R,S,T$ be commutative rings and assume that $R,S$ are $T$-algebras. In an answer to this question, Pierre-Yves Gaillard gives an example of an $R \otimes_T S$-module that cannot be written as ...
16
votes
2answers
1k views

Compactness of $\operatorname{Spec}(A)$

In an exercise in Atiyah-Macdonald it asks to prove that the prime spectrum $\operatorname{Spec}(A)$ of a commutative ring $A$ as a topological space $X$ (with the Zariski Topology) is compact. Now ...
8
votes
1answer
281 views

Modules over a tensor product

Let $k$ be a field. Suppose $A$ and $B$ are two commutative $k$-algebras. Let $M$ be a finite $A\otimes_k B$-module. Can one find a finite $A$-module $N$ and a finite $B$-module $L$ such that $M ...
5
votes
1answer
110 views

Does the regularity of $A$ imply the regularity of $A[X]$?

Let $A$ be a commutative Noetherian ring. We say it is regular if its localization at every prime ideal is a regular local ring. If this is the case, is it true that $A[X]$ is regular?
5
votes
1answer
459 views

Tensor products: proving that $I \otimes_R M \cong IM$

Assume it if it´s neccesarly that the ring has an 1 or is commutative ( I´m not sure if it´s needed) Given a ring $R$ an ideal $I$ of $R$, and a $R$ module $M$ , prove that: $ I \otimes _R M \cong ...
2
votes
2answers
133 views

Noetherian and local Laurent polynomials

Let $R$ be a commutative ring. I am interested in the following two questions: 1) If $R$ is Noetherian, then is $R[u,u^{-1}]$ Noetherian? In fact, I know that this is true (the Hilbert basis theorem ...
3
votes
1answer
183 views

'Divides means contains' for Dedekind domains, by treating PIDs and localising

I'm trying to solve the following problem: Suppose $R$ is a Dedekind domain which contains (nonzero) ideals $\mathfrak{a}$ and $\mathfrak{b}$. By first dealing with the case where $R$ is a PID and ...
6
votes
2answers
1k views

A slick proof that a field which is finitely generated as a ring is finite

It is a known fact that if $k$ is a field that is finitely generated as a ring, which is the same as having a surjective ring homomorphism $f:\mathbb{Z}[x_1,\dots,x_n]\to k$ for some $n\in ...
2
votes
2answers
2k views

Showing the factor ring R/I is a field

Morning everybody! I have the following question on ring theory, I would like somebody to help me: Let $R =\mathbb Z_2[x]$, and consider the ideal $I$ of $R$ generated by the irreducible ...