Let $A$ be a Noetherian integral domain, $B$ a ring extension of $A$ that is an integral domain, $P \in \operatorname{Spec} B, \, p = P \cap A$. Denote by $\kappa(p),\ \kappa(P)$ the residue fields of the two prime ideals. Then we see that there is a field extension $\kappa(p) \hookrightarrow \kappa(P)$. Let $t$ be a non-negative integer such that $t \le {\rm tr}.\deg_{\kappa(p)} \kappa(P)$.
Matsumura in his Commutative Ring Theory, proof of Theorem 15.5, says "let $c_1,\dots,c_t \in B$ such that their images modulo $P$ are algebraically independent over $A/p$."
Question: Why would such elements exist?
(Edited)
Remark: Matsumura wants to prove that $\operatorname{ht}(P)+\operatorname{tr.deg}_{\kappa(p)} \kappa(P) \le \operatorname{ht}(p)+ \operatorname{tr.deg}_A B$ and he starts by proving that we can assume that $B$ is a finitely generated $A$-algebra. Specifically, he is showing that we can construct a subring $C$ of $B$ that is a finitely-generated $A$-algebra, and that if the theorem is true for $C$, then it is true of $B$. My question relates to his argument of "why we can assume that". Consequently, in answering my question, we can not make the assumption that $B$ is a finitely generated $A$-algebra.
Answer
Take $x_1,\dots,x_t$ to be algebraically independent over $k(p)$ in $k(P)$. Then, write each $x_i$ as a fraction. Then, you get $2t$ elements of $B/P$. I claim that there are $t$ algebraically independent among the $2t$ elements. Why? What happens if there is only $m
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