Affine Algebraic Curves - Kunz - Theorem 1.3

[Math Amateur]

I am reading Ernst Kunz book, "Introduction to Plane Algebraic Curves"

I need help with some aspects of Kunz' proof of Theorem 1.3 ...

The relevant text from Kunz is as follows:http://mathhelpboards.com/attachment.php?attachmentid=4559&stc=1In the above text we read the following:

" ... ... Therefore let \(\displaystyle p \gt 0\). Since \(\displaystyle a_p\) has only finitely many zeros in \(\displaystyle K\), there are infinitely many \(\displaystyle y \in K\) with \(\displaystyle a_p(y) \neq 0\).

Then

\(\displaystyle f(X, y) = a_0(y) + a_1(y)X + \ ... \ ... \ + a_n(y)X^n\)

is a non-constant polynomial in \(\displaystyle K[X]\). ... ..."QUESTION 1

Why are we worrying that \(\displaystyle a_p(y) \neq 0\) ... I am assuming only that we want \(\displaystyle p \gt 0\) ... that is we want the polynomial to be more than a constant \(\displaystyle a_0(y)\) ... because that is the case we are considering ... is that correct?Continuing the quote from Kunz above, Kunz 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 ... "QUESTION 1I am quite confused about this statement ...

How can \(\displaystyle x \in K\) be a zero of this polynomial which although expressed as a polynomial in \(\displaystyle X\) is actually a polynomial in \(\displaystyle X\) and Y ... can someone please explain how we can get a zero considering only \(\displaystyle X\)? Surely it also matters what value we give to \(\displaystyle Y\) ...?

Further, I think Kunz is arguing that we can get a zero for \(\displaystyle f(X,y)\) by considering only x and also arguing that there are infinitely many y that go with such an x ... how does this work?

Can someone explain how this may be so ...

Peter

 

[jvicens]

ANSWER

Hello Peter,

You are correct in your understanding that we want p \gt 0 because we want the polynomial to have more than one term. This is important in proving Theorem 1.3 because we are trying to show that the curve \Gamma contains infinitely many points.

To answer your first question, we are not necessarily worrying that a_p(y) \neq 0, but rather using this fact to show that f(X,y) is a non-constant polynomial in K[X]. This is because if a_p(y) \neq 0, then f(X,y) will have at least one term with a non-zero coefficient, making it a non-constant polynomial.

Regarding your second question, x is a variable in the polynomial f(X,y), meaning that it can take on any value in K. So when we say that x is a zero of f(X,y), we mean that when we substitute x into the polynomial, it becomes equal to 0. This does not mean that x is the only variable in the polynomial, it is just the one we are considering in this particular case.

To understand why there are infinitely many points on the curve \Gamma, we can think of it in terms of a graph. Each point on the curve corresponds to a pair of values (x,y) that satisfy the equation f(X,y) = 0. So when we fix a value for x and vary y, we can find infinitely many points on the curve. And since we can do this for any value of x, there are infinitely many points on the curve overall.

I hope this helps clarify things for you. Let me know if you have any further questions or need any additional explanation.