TITLE: What is the Krull dimension of $\mathbb{Q}[x^2+y+z,\ x+y^2+z,\ x+y+z^2,\ x^3+y^3,\ y^4+z^4]$?
QUESTION [6 upvotes]: I am studying commutative algebra and saw the following question in one of the tests: 

What is the Krull dimension of $R=\mathbb{Q}[x^2+y+z,\ x+y^2+z,\ x+y+z^2,\ x^3+y^3,\ y^4+z^4]?$ 

I know that $\dim R \leq \dim\mathbb{Q}[x,\ y,\ z]=3$ and $\dim R>0$ since $R$ is not a field, but this is not very helpful. 
I guess I should find a maximal chain of prime ideals, or maybe use $\dim R= \dim(R/P)+\operatorname{height}(P)$ for some prime ideal $P$, but I couldn't think of anything.. 
I would be grateful for any help!

REPLY [5 votes]: Tensor with $\mathbb{C}$ and the dimension won't change. 
$\dim \mathbb{C}[x^2+y+z,\ x+y^2+z,\ x+y+z^2,\ x^3+y^3,\ y^4+z^4]$ is the dimension of the (closure of the) image of $\mathbb{C}^3$ under the map
$$\mathbb{C}^3 \ni (x,y,z) \mapsto (x^2+y+z,\ x+y^2+z,\ x+y+z^2,\ x^3+y^3,\ y^4+z^4) \in \mathbb{C}^5$$
Look at the jacobian matrix and notice that it's an imersion at almost all points. Hence the dimension is $3$.