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

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!

Answer 1

Macaulay2 shows that a Gröbner basis for the ideal of $S=\mathbb Q[x,y,z]$ generated by $x^2+y+z,\ x+y^2+z,\ x+y+z^2,\ x^3+y^3,\ y^4+z^4$ is $x,y,z$.

Let $m$ be the ideal of $R$ generated by $x^2+y+z,\ x+y^2+z,\ x+y+z^2,\ x^3+y^3,\ y^4+z^4$, and $M$ the ideal of $S$ generated by $x,y,z$. Then $mS=M$, so $m\subseteq M\cap R$. By the dimension inequality (see Matsumura, CRT, Theorem 15.1(i)) we have $3=\operatorname{ht}M\le\operatorname{ht}(M\cap R)+\dim S_M/(M\cap R)S_M$. But $S_M/(M\cap R)S_M=S_M/MS_M$, so $\dim S_M/(M\cap R)S_M=0$. Thus $3\le\operatorname{ht}(M\cap R)$ and since $\dim R\le\dim S=3$ we have equality.

Answer 2

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 $3$.