Bourbaki

Exercise 14, § 2

Commutative Algebra, VIII, p. 85

Let AA be a ring which is an integral domain, with field of fractions KK, and let m∈Nm \in \mathbf{N}. Now prove that one has

dim⁡(A[X1,...,Xm])=m+sup⁡{dim⁡(A[t1,...,tm])∣ti∈K,1≤i≤m}. \dim(A[X_1, ..., X_m]) = m + \sup \left\{ \dim(A[t_1, ..., t_m]) | t_i \in K, 1 \leq i \leq m \right\}.
(One will use Exercise 10). More generally if FF is a field, extension of KK, prove that one has
dim⁡(A[X1,...,Xm])=sup⁡{dim⁡(A[t1,...,tm])∣ti∈F,1≤i≤m}+sup⁡(0,m−deg⁡.tr⁡K(F)). \dim(A[X_1, ..., X_m]) = \sup \left\{ \dim(A[t_1, ..., t_m]) | t_i \in F, 1 \leq i \leq m \right\} + \sup(0, m - \deg. \operatorname{tr}_K(F)).

Solution

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Machine translation of the English, by gpt-5-6. Not checked by a person.