Dimension of a polynomial ring

Remarks

For the lower bound, a chain of prime ideals $\mathfrak p_0 \subsetneq \cdots \subsetneq \mathfrak p_n$ of $R$ gives the chain $\mathfrak p_0[X] \subsetneq \cdots \subsetneq \mathfrak p_n[X] \subsetneq \mathfrak p_n[X] + (X)$ in $R[X]$. For the upper bound, fix a prime $\mathfrak p$ of $R$ and let $\kappa(\mathfrak p)$ be the fraction field of $R/\mathfrak p$. The primes of $R[X]$ that contract to $\mathfrak p$ correspond to primes of $\kappa(\mathfrak p)[X]$, so at most two of them lie in one chain. In the Noetherian case one shows that a prime $\mathfrak P$ of $R[X]$ over $\mathfrak p$ has height at most $\operatorname{ht} \mathfrak p + 1$. By induction, for a field $k$ the ring $k[X_1, \dots, X_n]$ has dimension $n$, and for a Noetherian ring $R$ of finite dimension, $\dim R[X_1, \dots, X_n] = \dim R + n$. For example, $\dim \Z[X] = 2$, with the maximal chain $0 \subsetneq (2) \subsetneq (2, X)$. Without the Noetherian hypothesis the equality in clause 2 can fail: there are non-Noetherian rings with $\dim R[X] > \dim R + 1$.

Going deeper

For a student

This result is the algebraic reason why affine $n$-space over a field has dimension $n$. Together with Noether normalization it shows that a finitely generated domain over a field has dimension equal to the transcendence degree of its fraction field. The general bounds come from looking at the fibres of $\operatorname{Spec} R[X] \to \operatorname{Spec} R$, which are spectra of polynomial rings in one variable over fields.