Let $K$ be a field, and let $n\in\mathbb{N}$.
- dim$K[x_1,...,x_n] = n$.
- All maximal chains of prime ideals in $K[x_1,...,x_n]$ have length $n$.
This is Proposition 11.9 in Chapter 11 from the class notes : Gathmann - "Commutative Algebra". The proof starts by let $0=P_0\subset P_1\subset \cdots\subset P_m$ be a chain of prime ideals in $K[x_1,...,x_n]$, where the height of $P_1$ is $1$ and $P_m$ is maximal. Then $P_1=(f)$ for some nonzero polynomial $f\in K[x_1,...,x_n]$, but I don't understand why we can assume that $f$ is monic in $x_n$. I know that there are $a_1,...,a_{n-1}\in\mathbb{N}$ and set \begin{equation} y_1=x_1-x_n^{a_1},y_2=x_2-x_n^{a_2},...,y_{n-1}=x_{n-1}-x_n^{a_{n-1}}, y_n=x_n. \end{equation} Then $f(y_1+y_n^{a_1},...y_{n-1}+y_n^{a_{n-1}},y_n)$ is monic in $y_n$. So we will look at $K[y_1,...,y_n]$ as the ring of the polynomial with indeterminates $y_1,...,y_n$ instead of $K[x_1,...,x_n]$ ? If yes, why can $K[y_1,...,y_n]$ be viewed as a ring of a polynomial? I try to prove that the map $f(x_1,...,x_n)\longrightarrow f(y_1+y_n^{a_1},...y_{n-1}+y_n^{a_{n-1}},y_n)$ is an isomorphism, but I can't prove that it is one-to-one.
The map \begin{align}k[y_1,\dotsc ,y_n] &\to k[x_1,\dotsc,x_n]\\ y_i&\mapsto x_i+x_n^{a_i}&i<n\\ y_n&\mapsto x_n\end{align} is an inverse to the map you wrote.