0
votes
0answers
44 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 ...
4
votes
3answers
98 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 ...
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?
57
votes
0answers
2k 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
135 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 ...