Remark 2

Suppose that the extension k' of k is radicial. Then the canonical mapping $f : \mathrm{Spec}(A') \to \mathrm{Spec}(A)$ is a homeomorphism.

Let indeed $p \in \mathrm{Spec}(A)$. By Lemma 1 of No. 1, the space $f^{-1}(\{p\})$ is homeomorphic to $\mathrm{Spec}(\kappa(p) \otimes_k k')$. Now the set a of nilpotent elements of $\kappa(p) \otimes_k k'$ is a prime ideal (A, V, p. 134, corollary) and the quotient ring $(\kappa(p) \otimes_k k')/a$, integral and finite over $\kappa(p)$, is a field (A, V, p. 16, cor. 1 and p. 10, prop. 1). Consequently $f^{-1}(\{p\})$ is reduced to one element. It follows that the mapping $f$ is an increasing bijection of $\mathrm{Spec}(A')$ onto $\mathrm{Spec}(A)$, these two sets being ordered by inclusion of the prime ideals, and hence induces a bijection between the irreducible closed subsets of $\mathrm{Spec}(A)$ and those of $\mathrm{Spec}(A')$. Since the closed subsets of $\mathrm{Spec}(A)$ (resp. $\mathrm{Spec}(A')$) are finite unions of irreducible closed subsets, $f$ is a homeomorphism.

For a generalization, see exerc. 24, p. 98.