Krull dimension of affine algebra is equal to maximum of transcendence degrees

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I'm going through some introductory books on commutative algebra and I'm struggling with the following problem:

Let $A$ be a non-trivial affine algebra over the field $K$. Since $A$ is noetherian, we know that the zero ideal of $A$ has only finitely many minimal prime ideals: let's call them $P_1,...,P_n.$ For each $i = 1,...,n$, let $L_i = \mathrm{Quot}(A/P_i)$ where $A/P_i$ are integral domains. So we can look at $L_i$ as an extension field of $K$.

Now I have to show the following:

i) $\dim(A) = \max\{\mathrm{tr.deg}_KL_i: 1 \leq i \leq n\}$.

ii) If $\dim(A) = \mathrm{tr.deg }_KL_i$ for all $ i = 1,...,n$, then $\mathrm{ht}(P) +\dim(A/P) = \dim(A)$ for all $P \in \mathrm{Spec}(A)$.

What I know so far, are the following properties:

  • $\dim(A) \geq \mathrm{ht}(P_i) + \dim(A/P_i)$ for $i = 1,...,n$.
  • I think that $\mathrm{ht}(P_i) \leq 1$ since these are minimal prime ideals of the zero ideal.
  • if an integral domain $E$ is an affine algebra over $K$ with $L=\mathrm{Quot}(K)$ then $\dim(E) = \mathrm{tr.deg}_KL$.

I'm not sure if these properties are enough to solve the problem. Is it possible to view the integral domains $A/P_i$ as affine algebras over $K$, so we can use the 3rd property?