The following proof is taken from "Algebraic Varieties" by Kempf.
There seems to be way too many $n$'s in the proof. The projective variety is $\mathbb{P}^n$, the number of polynomials is $n$, there seems to be a linear map $\beta_n$ for each $n$. How does one clean this up?
Additionally, Kempf says $P_j$ is the space of homogenous polynomials of degree $a_j$, then he write $P_{n-a_j}$, does he mean the space of homogenous polynomials of degree $n-a_j$? And does $P_n$ refer to polynomials of degree $n$ or degree $a_n$?
Do you also find this proof to be highly confusing?
This proof works fine as-is through the start of the paragraph beginning "We will show that $\pi_Y(Z)$...". We'll start making some alterations there, which we'll note in $\color{#F00}{\text{red}}$ when possible.
We will show that $\pi_Y(Z)$ is closed by writing it as the intersection of determinantal subvarieties. Let $a_j$ be the degree of $f_j$ in the $X_j$'s. Let $P_{\color{#F00}{d}}$ be the vector space of all homogeneous polynomials in $k[X_0,\cdots,X_n]$ of degree $\color{#F00}{d}$. Then consider $k[X_0,\cdots,X_n]$-linear combinations of the $f_j$. They define a $k[Y]$-linear mapping $\beta_{\color{#F00}{m}}:\bigoplus_j P_{\color{#F00}{m-a_j}} \otimes_k k[Y] \to P_{\color{#F00}{m}}\otimes_k k[Y]$ for all $\color{#F00}{m}$. Thus in terms of a basis of the $P_{\color{#F00}{d}}$, $\beta_{\color{#F00}{m}}$ is given by a matrix $\alpha_{\color{#F00}{m}}$ with coefficients in $k[Y]$. Let $E_{\color{#F00}{m}}$ be the determinantal subvariety $\{\operatorname{rank} \alpha_{\color{#F00}{m}}(y) \leq \dim P_{\color{#F00}{m}} - 1\}$. Then $y$ is in $E_{\color{#F00}{m}}$ if and only if $\beta_{\color{#F00}{m}}(y)$ is not surjective.
Explanation of changes: we redefined $P_d$, the space of polynomials, fixing an error where the degree was not defined correctly. We replaced the index on $\beta$ and $E$ with $m$ to avoid a naming collision with the dimension of $\Bbb P^n$.
Next paragraph:
The projective nullstellensatz says that $y\in\pi_Y(Z)$ $\Leftrightarrow$ $\pi_Y^{-1}(y)\neq \emptyset$ $\Leftrightarrow$ $\color{#F00}{f_j(y,X)=0}$ for all $\color{#F00}{j}$ is a non-empty subset $\Leftrightarrow$ $\beta_{\color{#F00}{m}}(y)$ is not surjective for all $\color{#F00}{m}$ $\Leftrightarrow$ $y\in\bigcap_\color{#F00}{m} E_\color{#F00}{m}$ for all $\color{#F00}{m}$. Thus $\pi_Y(Z)$ is the intersection of closed determinantal subvarieties. Hence $\pi_Y(Z)$ is closed.
Explanation of changes: we replaced the expression "$f_1(y,X)=\ldots=f_n(y,X)=0$" with the more appropriately quantified "$\color{#F00}{f_j(y,X)=0}$ for all $\color{#F00}{j}$". We again swapped the index on $\beta$ and $E$ to $m$ in order to avoid a naming collision with the dimension of $\Bbb P^n$.
Now that we've got a proof that makes more sense (or is at least readable without worrying about duplicate variable names), let's examine the logic. The goal is to come up with elements of $k[Y]$ which vanish at $y\in Y$ exactly when there's a point of $Z$ over $y$.
Let's start with the case when $Y$ is a single point itself. Then we're asking for when a set of homogeneous polynomials $\{f_j\}$ in $k[X_0,\cdots,X_n]$ has no common zero in $\Bbb P^n$. There are two separate ways this can happen: either $I$, the ideal generated by the $f_j$, has $I=(1)$, or $\sqrt{I}=(X_0,\cdots,X_n)$. If neither of those happen, then $\sqrt{I}$ is properly contained in $(X_0,\cdots,X_n)$ and therefore determines a nonempty projective variety. From the definition of $\sqrt{I}=\{ \gamma \in k[X_0,\cdots,X_n] \mid \exists e\in\Bbb Z_{>0} \text{ such that } \gamma^e\in I\}$, we see that if $I=(1)$ or $\sqrt{I}=(X_0,\cdots,X_n)$, then there must be some $l>0$ so that $X_0^l,X_1^l,\cdots,X_n^l$ are all in $I$. Therefore $V(I)\subset\Bbb P^n$ empty is equivalent to the existence of some $l>0$ so that $I\cap k[X_0,\cdots,X_n]_l=k[X_0,\cdots,X_n]_l$.
Let's reinterpret this in a fashion closer to what's used in the proof. What's $I\cap k[X_0,\cdots,X_n]_m$? It's the vector space of sums of polynomials $p\cdot f_j$, where $p$ has degree $m-a_j$. So it can be described as the image of $\beta_m:\bigoplus_j P_{m-a_j}\to P_m$. Saying that it's not all of $P_m$ means that the rank of $\beta_m$ is strictly less than $\dim P_m$, which we can measure via choosing bases, getting a matrix, then looking at some determinants: a linear map of vector spaces fails to be of rank $r$ iff all the determinants of $r\times r$ minors of a matrix $\alpha_m$ representing it vanish. Therefore the condition that $\beta_m$ fails to be surjective is described by the vanishing of some polynomials in the entries of $\alpha_m$ (*). Since $V(I)$ nonempty is equivalent to $\beta_m$ not surjective for any $m$, we see that $V(I)$ nonempty is equivalent to the vanishing of all the $(\dim P_m-1)\times(\dim P_m-1)$ minors of $\alpha_m$ over all $m$.
Now let's upgrade everything from the case of $Y$ a single point to a general variety. This will go quickly: we upgrade $\beta_m$ from a map $\bigoplus_j P_{m-a_j}\to P_m$ (aka $\bigoplus_j P_{m-a_j}\otimes_k k\to P_m\otimes_k k$) to $\bigoplus_j P_{m-a_j}\otimes_k k[Y]\to P_m \otimes_k k[Y]$, which turns our matrix $\alpha_m$ to having coefficients in $k[Y]$ instead of just $k$. Therefore the determinants of the minors are actually polynomials in elements of $k[Y]$, and therefore elements of $k[Y]$. These determinants are the equations cutting out $E_m$. Evaluating at some point $y\in Y$, we recover the statements we proved when $Y$ was a single point: the equations for $E_m$ turn in to the polynomials mentioned at (*), and so a point $y\in Y$ is in $E_m$ iff $\beta_m(y)$ fails to be surjective, where $\beta_m(y)$ means evaluating at $y$. Therefore there's a point of $Z$ over $y$ exactly when $y\in E_m$ for all $m$, or $y\in \bigcap_m E_m$.