This is a rather specific question stemming from the proof of lemma 5.6 in Gregor Kemper's A Course in Commutative Algebra. The lemma is as follows:
Let $A$ be an algebra over a field $K$, and let $S\subseteq A$ be a subset that generates $A$ as an algebra. Then $$\dim(A)\leq \sup\{|T| \mid T\subseteq S ~\text{is finite and algebraically independent} \}.$$
Calling the supremum on the right side $n$, Kemper says that we need to show $\text{dim}(A/P)\leq n$ for all $P\in\text{Spec}(A)$. However, I don't see why it couldn't be that $\text{dim}(A/P)<\text{dim}(A)$ for every $P\in\text{Spec}(A)$ which would lead to the proof being insufficient as I see it. The dimension used here is the Krull dimension.
If $\dim A < \infty$, then $\dim A = \dim A/P$ for some minimal prime ideal $P$.
Proof for this: Let $\dim A=n$. By the definition of the dimension, there is a chain of prime ideals $P_0 \subset P_1 \subset \dotsc \subset P_n$ with all inclusions strict. Then $\dim A/P_0=n$, because we have the chain $0 \subset P_1/P_0 \subset \dotsc \subset P_n/P_0$, thus $\dim A/P_0 \geq n$. The other inequality holds for any quotient.
If $\dim A = \infty$, then $\sup\limits_{P} \dim A/P = \infty$, i.e. if we show $\dim A/P \leq n$ for any $P$, we have automatically shown the desired $n=\infty$.