I have a question about argumentation used in the proof of Proposition 1.7 in Fulton's book on Intersection Theory on page 18:
Proposition 1.7 Let
$\require{AMScd}$ \begin{CD} X' @>{g'}>> X\\ @Vf'VV @VVfV\\ Y' @>{g}>> Y \end{CD}
be a fibre square, with $g$ flat and f proper. Then $g'$ is flat, $f'$ is proper, and for all $\alpha \in Z_* X$,
$$ f' _{\ast} g' ^{\ast} \alpha = g^{\ast} f_{\ast} \alpha $$ in $Z_* Y'$.
some remarks on notation: the structure $Z_{\ast} X $ is explaned on page 10, after quotient out the rational equivalence the quotient becomes Chow ring $CH_*X$; compare with pages 10 - 16.
on this Chow ring there can be described two most common 'actions':
Pullback: If $f: X \to Y$ is flat, then for any subscheme $Z$ of $Y$, $$f^{\ast} [Z] = [f^{-1} (Z)].$$
in other words we obtain $f^{\ast}:CH_{\ast} Y \to CH_{\ast} X$. for details, see page 18.
Pushforward: If $g: X \to Y$ is proper, then for any subscheme $W$ of $X$, $$g_* [Z] = \deg(W/g(W))[g(W)]$$
with $\deg(W/g(W))=\deg(\operatorname{Frac}(W)/\operatorname{Frac}(g(W))$ if $\dim(W)=\dim(g(W))$, otherwise $0$; see page 11.
now we come to the proof of 1.7:
Proof Since flatness and properness are preserved by base change, we may assume $X$ and $Y$ are varieties,$f$ is surjective, and $\alpha = [X]$. Let $f^* [X] = d [Y]$. We must show that $f'_* [X'] = d [Y']$ This is a local calculation involving local rings of irreducible components, so we may assume $X = \operatorname{Spec}(L), Y = \operatorname{Spec} (K)$, with $K, L$ fields (???), $Y' = \operatorname{Spec}(A)$, with $A$ local Artinian, and $X' = \operatorname{Spec}(B)$, $B = A \otimes_K L$. Then the result follows from Lemma A.1.3. That's fine.
Q: although this is a local problem, I not understand why we can assume that $K,L$ are fields. I think that since we can argue on stalks, we can assume that $K,L$ are local rings, however why can we already consider $L,K$ as fields?
p.s.: I have already asked this at MSE without success.
