Proof of Zariski's Lemma to Nullstellensatz (Fulton)

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This proof of Zariski's Lemma is from Fulton's Algebraic Curves.

Proposition. If a field L is ring-finite over a subfield K, then L is module-finite (and hence algebraic) over $K$.

Proof. Suppose $L = K[v_1,...,v_n]$. The case n = 1 is taken care of by the above discussion, so we assume the result for all extensions generated by n − 1 elements. Let $K_1 = K(v_1)$. By induction, $L = K_1[v_2,...,v_n]$ is module-finite over $K_1$. We may assume $v_1$ is not algebraic over K (otherwise Problem 1.45(a) finishes the proof). Each $v_i$ satisfies an equation $v_i^{n_i} +a_{i1}v_i^{n_i−1}+···=0,a_{ij} ∈K_1$. If we take $a∈K[v_1]$ that is a multiple of all the denominators of the $a_{ij}$, we get equations $(av_i)^{n_i} +aa_{i1}(av_i)^{n_i−1}+···=0$. It follows from the Corollary in §1.9 that for any $z ∈ L = K[v_1,...,v_n]$, there is an $N$ such that $a^Nz$ is integral over $K[v_1]$. In particular this must hold for $z ∈ K (v_1)$. But since $K (v_1)$ is isomorphic to the field of rational functions in one variable over $K$ , this is impossible (Problem 1.49(b)).

The corollary refers to this statement: Let S be a domain. The set of elements of $S$ that are integral over $R$ is a subring of $S$ containing $R$.

I have two questions:

How does "for any $z ∈ L = K[v_1,...,v_n]$, there is an $N$ such that $a^Nz$ is integral over $K[v_1]$." follow from the Corollary?

Why is $z ∈ K (v_1)$ a particular case for this statement? It does not appear to me that $K (v_1)$ is a subset of $K[v_1,...,v_n]$

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As per my professor:

For the first question, we have that for every $i=1,\dots,n$, we have that $av_i$ is integral over $K[v_1]$. Further $K$ is obviously integral over $K[v_1]$. By the corollary, since (finite) multiplication and addition of integral elements is integral, we must have that $K[av_1,\dots,av_n]$ is integral over $K[v_1]$. Therefore for any element $z\in K[v_1,\dots,v_n]$ has some larger enough N such that $a^Nz$ is integral over K[v_1].

For the second question, since $K[v_1,\dots,v_n]$ is a field, we have that $K(v_1)$ is one of its subfields.