Kunz - Plane Algebraic Curves - Theorem 1.3 - basic question

[Math Amateur]

I am reading the book, "Introduction to Plane Algebraic Curves" by Ernst Kunz - which the author claims gives a basic introduction to the elements of algebraic geometry.

I need help with an apparently simple statement that I find confusing and puzzling.

Theorem 1.3 and its proof reads as follows:

View attachment 2683In the above, Kunz writes:

"Since \(\displaystyle a_p \) has only finitely many zeros in K, there are infinitely many y with \(\displaystyle a_p (y) \ne 0 \)."

I am puzzled by the statement "Since \(\displaystyle a_p \) has only finitely many zeros in K" - what does Kunz mean by this, since \(\displaystyle a_p \) is just a coefficient?

Can someone please clarify this issue for me?

Peter***EDIT-1*** Apologies to MHB members - I should have read Kunz' proof more carefully ... \(\displaystyle a_p \in K[Y] \) and hence \(\displaystyle a_p \) is a polynomial of the form \(\displaystyle b_0 + b_1y + ... \ ... + b_n x_n \) which has at most n roots i.e. a finite number of roots ... ... BUT ... it still leaves an issue ... ... how do we then conclude as Kunz does that therefore "there are infinitely many \(\displaystyle y \in K \) with \(\displaystyle a_p (y) \ne 0 \)?*** Possible answer to EDIT-1:

Suppose \(\displaystyle a_p(Y) = b_0 + b_1 Y + b_2 Y^2 + ... \ ... + b_n Y^n \)

Then there are a finite set of values y (at most n) in K such that \(\displaystyle a_p (Y = y ) = 0 \)

But K is infinite (since in the first line of his text Kunz declares that "we will study algebraic curves over an arbitrary algebraically closed field K" and we know that arbitrary algebraically closed fields are infinite) so the set of points at which \(\displaystyle a_p ( y ) \ne 0 \) is infinite (i.e. infinite set of points minus a finite set of points leaves an infinite set of points)

Can someone please confirm that the above analysis is correct?
***EDIT-2*** I now have a further question:

Kunz writes:

""Since \(\displaystyle a_p \) has only finitely many zeros in K, there are infinitely many y with \(\displaystyle a_p (y) \ne 0 \).

Then \(\displaystyle f(X,y) = a_0 (y) + a_1 (y)X + ... \ ... a_p (y)X^p \)

is a non-constant polynomial in \(\displaystyle K[X] \)."

I note that in the above X is a capital letter and y is lower case - which seems to me to indicate X is an indeterminate but that y is a particular (unspecified) value of the indeterminate y. Is that what Kunz wants us to assume? Further, in thinking about y in the above text, do we think of y ranging over a possible infinity of values of Y in K (assuming K is infinite of course, which (just thinking about it) it may not be?
***EDIT-3***

As noted above Kunz writes:

""Since \(\displaystyle a_p \) has only finitely many zeros in K, there are infinitely many y with \(\displaystyle a_p (y) \ne 0 \).

Then \(\displaystyle f(X,y) = a_0 (y) + a_1 (y)X + ... \ ... a_p (y)X^p \)

is a non-constant polynomial in \(\displaystyle K[X] \)."

Then he writes:

"If \(\displaystyle x \in K \) is a zero of this polynomial, then \(\displaystyle (x,y) \in \Gamma \); therefore \(\displaystyle \Gamma \) contains infinitely many points."

How does the statement in the last sentence follow?
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[jvicens]

The statement in the last sentence follows because if (x,y) is a point on the curve, then for each value of x, there are infinitely many values of y that satisfy the equation f(x,y) = 0. This is because f(x,y) is a polynomial in y and, as established in the previous statement, a_p(y) has infinitely many non-zero values. Therefore, for each x, there are infinitely many points (x,y) on the curve, making the curve contain infinitely many points.