1
vote
0answers
53 views

Krull dimension of localization

If $R$ is a commutative ring and $m$ a maximal ideal therein, then what are the conditions for the Krull dimension of $R$ equaling to the Krull dimension of $R_m$?
0
votes
0answers
51 views

prove $\dim\mathbb{Z}[X_1,X_2]=3$ from first principles

Since $\dim R[X] =\dim R+1$ for any Noetherian ring $R$, the ring $\mathbb{Z}[X_1,X_2]$ must have dimension 3. But how can this be proved 'from first principles', i.e. without using any big theorems ...
3
votes
1answer
69 views

Principal ideal domain is universally catenary …

... actually, even more general statement is true: Theorem. Every regular ring is universally catenary. (see for example Algebraic Geometry by Qing Liu, Corollary 2.16, Chapter 8) Though, the ...
5
votes
2answers
130 views

Equivalence of definitions of Krull dimension of a module

I've seen two definitions of Krull dimension of a module $M$ over a (commutative) ring $R$, and their equivalance does not seem obvious: Matsumura on page 31 of his book Commutative Ring Theory ...
3
votes
1answer
199 views

Proof details of Theorem 11.1 in Atiyah-Macdonald

I have some trouble filling in the details of this proof from Atiyah-Macdonald. In this result, the authors assume what follows: 1) $A = \oplus_{n=0}^\infty A_n$ is a Noetherian graded ring, and ...
4
votes
3answers
100 views

Converse of a dimension lemma

Consider the following lemma. It comes from the Stacks Project. Lemma 9.59.11. Suppose that $R$ is a Noetherian local ring and $x\in\mathfrak m$ an element of its maximal ideal. Then $\dim R\le ...
3
votes
1answer
126 views

Krull dimension of a $\mathbb Q$-algebra

I'm trying to find the Krull dimension of $\mathbb{Q}[X,Y,Z]/(X^{2}-Y,Z^{2})$. My professor said that I have to consider that $\mathbb{Q}[X,Y,Z]/(X^{2}-Y,Z^{2})$ is a $\mathbb{Q}$-algebra but I ...
0
votes
0answers
78 views

How to show an ideal is zero-dimensional? [duplicate]

Let $J$ denote the ideal in $\mathbb{Q}[x,y,z]$ generated by $\{y^2-xy-2xz,y^3+z^2+1, x^2yz-yz\}$. Show that $J$ is zero-dimensional. How do I go about showing this?
72
votes
0answers
3k views

A short proof for $\dim(R[T])=\dim(R)+1$?

If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$. For noetherian $R$, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial ...
2
votes
1answer
111 views

Poles of formal power series (Hilbert-Poincaré series)

How are poles and orders of poles of formal power series defined? The particular case, I am interested in, is the following definition from [Atiyah-Macdonald, Introduction to commutative algebra, ...
1
vote
0answers
157 views

Hilbert (polynomial) dimension and dimension of a support of a module

$\newcommand{\Supp}{\mathrm{Supp}}$ $\newcommand{\Ann}{\mathrm{Ann}}$ Let $X$ be an affine algebraic variety (over a field $K$, can assume it is algebraicaly closed), $M$ a finitely generated ...
2
votes
1answer
105 views

Injective endomorphism and Hilbert dimension

Let $\mathcal{O}$ be a finitely-generated $K$-algebra where $K$ is a field and let $M$ be a finitely-generated $\mathcal{O}$-module. For every good filtration $0 = M_0 \subset M_1 \subset M_2 \subset ...
2
votes
1answer
158 views

Transcendence degree of $K[X_1,X_2,\ldots,X_n]$

Let $K$ be field. How do I proof that transcendence degree of $K[X_1,X_2,\ldots,X_n]$ is $n$? The set $\{X_1,X_2,\ldots,X_n\}$ is algebraically independent over $K$. So, I have to show that every ...
3
votes
2answers
179 views

Krull dimension of the injective hull of residue field

Let $(R,\mathfrak{m})$ be a noetherian local ring, and $E=E_R(R/\mathfrak{m})$ the injective hull of $R/\mathfrak{m}$. What do we know about the Krull dimension of $E$? Thank you.
3
votes
1answer
89 views

Prime ideals of height less than the dimension

Let $A$ be a noetherian local ring with maximal ideal $\mathfrak{m}$ of height $d$, and suppose $\mathfrak{p}_1, \ldots, \mathfrak{p}_s$ are prime ideals of height $i - 1 < d$. It's quite clear ...
7
votes
1answer
185 views

Which definition of dimension came first?

In my algebraic geometry class, the dimension of an affine variety $X=V(I)$ was defined as the supremum of the length of chains of prime ideals in the coordinate ring $R=k[x_1,\ldots,x_n]/\sqrt{I}$, ...
4
votes
1answer
143 views

If the special fiber of a flat morphism is reduced, then any other fiber is reduced?

Suppose $R=\mathbb{C}[x_1,\ldots, x_n]$ is a polynomial ring with $I$ being an ideal of $R$. Let $I'$ be an ideal of $R[t]$. If $R[t]/I'$ is flat as a $\mathbb{C}[t]$-module and over $0$, ...
9
votes
3answers
798 views

Krull dimension of $\mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$

Krull dimension of a ring $R$ is the supremum of the number of strict inclusions in a chain of prime ideals. Question 1. Considering $R = \mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 ...
2
votes
1answer
173 views

On Krull dimension of $M/(0 :_{M} \mathfrak{m}^t)$ module

Let $(R,\mathfrak{m})$ be a commutative Noetherian local ring and $M$ is an $R$-module. There is an non-negative integer $t$ such that $M/(0 :_{M} \mathfrak{m}^t)$ is finitely generated. Then $$\dim ...
3
votes
1answer
147 views

Hilbert function on ideal generated by linear forms.

This is a slight extension of a remark a read a few days ago. Let $K$ be a field, and let $A=K[X_0,\dots,X_N]$ be a polynomial ring, which is graded in the standard way (the elements of degree $n$ ...
11
votes
2answers
521 views

$\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$

Let $A$ be an integral domain of finite Krull dimension. Let $\mathfrak{p}$ be a prime ideal. Is it true that $$\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$$ where $\dim$ ...