Understand Theorem 1: Weak Nullstellensatz Proof by Cox et al - Exercise 3(a)

[Math Amateur]

I am reading the undergraduate introduction to algebraic geometry entitled "Ideals, Varieties and Algorithms: An introduction to Computational Algebraic Geometry and Commutative Algebra (Third Edition) by David Cox, John Little and Donal O'Shea ... ...

I am currently focused on Chapter 4, Section 1: Hilbert's Nullstellensatz ... ... and need help with the aspects of Cox et al's interesting proof of the Weak Nullstellensatz as outlined in Exercise 3 ...

Exercise 3 (Chapter 4, Section 1) reads as follows:
https://www.physicsforums.com/attachments/5676As Exercise 3 refers to aspects of the proof of Theorem 1: The Weak Nullstellensatz, I am providing the first part of the proof of that Theorem as follows:
https://www.physicsforums.com/attachments/5677
View attachment 5678
My question is as follows:

How do we formulate and state a formal and rigorous proof of 3(a) ... ... that is a proof of the proposition that the coefficient \(\displaystyle c(a_1, a_2, \ ... \ ... \ , a_n )\) of \(\displaystyle \tilde{x}^N_1\) in \(\displaystyle f\) is \(\displaystyle h_N (1, a_2, \ ... \ ... \ , a_n )\) ... ...I can see that the proposition is likely true from the following simple example ... ...

Consider \(\displaystyle f_1 = 3 x_1^2 x_2^2 x_3 + 2 x_2^2 x_3^2\)The total degree of \(\displaystyle f_1\) is \(\displaystyle N= 5\), determined by the first term, namely \(\displaystyle 3 x_1^2 x_2^2 x_3\) ... ...

Also note that \(\displaystyle h_N(x_1, x_2, x_3) = h_5(x_1, x_2, x_3) = 3 x_1^2 x_2^2 x_3\)

and that \(\displaystyle h_5( 1, a_2, a_3) = 3 a_2^2 a_3\)

... ... ...

Consider now the transformation f_1 \mapsto \tilde{f_1} given by:

\(\displaystyle x_1 = \tilde{x}_1\)

\(\displaystyle x_2 = \tilde{x}_2 + a_2 \tilde{x}_1\)

... ...

... ...

\(\displaystyle x_n = \tilde{x}_n + a_n \tilde{x}_1\)Considering the above transformation, it is clear that the term \(\displaystyle 3 x_1^2 x_2^2 x_3\) will give rise the the coefficient of \(\displaystyle \tilde{x}^N_1\) ... ... so we examine this term under the transformation ... ...... so then ...

\(\displaystyle 3 x_1^2 x_2^2 x_3 = 3 \tilde{x}_1^2 ( \tilde{x}_2^2 + 2 a_2 \tilde{x}_1 \tilde{x}_2 + a_2^2 \tilde{x}_1^2 ) ( \tilde{x}_3 + a_3 \tilde{x}_1 )\)\(\displaystyle = ( 3 \tilde{x}_1^2 \tilde{x}_2^2 + 6 a_2 \tilde{x}_1^3 \tilde{x}_2 + 3 a_2^2 \tilde{x}_1^4 ) ( \tilde{x}_3 + a_3 \tilde{x}_1 )\)
Clearly, the term involving \(\displaystyle \tilde{x}_1^N = \tilde{x}_1^5\) will be

\(\displaystyle 3 a_2^2 a_3 \tilde{x}_1^5\)
So we have that \(\displaystyle h_N( \tilde{x}_1, \tilde{x}_2, \tilde{x}_3 )\) in \(\displaystyle \tilde{f}_1\) is \(\displaystyle h_5( \tilde{x}_1, \tilde{x}_2, \tilde{x}_3 ) = 3 a_2^2 a_3 \tilde{x}_1^5
\)

so we have that

\(\displaystyle h_5( 1, a_2, a_3 ) = 3 a_2^2 a_3\)

as required ... ... BUT ...

... ... how do we formulate and state a formal proof of the general proposition that the coefficient \(\displaystyle c(a_1, a_2, \ ... \ ... \ , a_n )\) of \(\displaystyle \tilde{x}^N_1\) in \(\displaystyle f\) is \(\displaystyle h_N (1, a_2, \ ... \ ... \ , a_n )\) ... ... ?Does the proof just describe the computation process in general terms ... ?Hope someone can help ... ...

Help will be appreciated ...

Peter

 

[jvicens]

Dear Peter,

Thank you for reaching out for help with your question on the proof of Exercise 3 in Chapter 4, Section 1 of Cox et al's "Ideals, Varieties and Algorithms." The proposition that you are trying to prove is a fundamental result in algebraic geometry known as the Weak Nullstellensatz. It states that for a polynomial f in n variables, if f has no zeros in the algebraically closed field K, then there exist polynomials g_1, ..., g_n in K[x_1, ..., x_n] such that 1 = f g_1 + ... + f_n g_n.

To prove part 3(a) of Exercise 3, we need to show that the coefficient c(a_1, ..., a_n) of \tilde{x}^N_1 in f is equal to h_N(1, a_2, ..., a_n). To do this, we will use the transformation f \mapsto \tilde{f} defined in the exercise, and show that the term involving \tilde{x}^N_1 in \tilde{f} is equal to h_N(1, a_2, ..., a_n).

Let's begin by considering the term c(a_1, ..., a_n) in \tilde{f}. This term comes from the term c(x_1, ..., x_n) in f under the transformation x_i = \tilde{x}_i + a_i \tilde{x}_1 for i = 2, ..., n. Thus, we can write c(a_1, ..., a_n) as c(\tilde{x}_1 + a_1 \tilde{x}_1, ..., \tilde{x}_n + a_n \tilde{x}_1). Using the binomial theorem, we can expand this as a sum of terms of the form c_i \tilde{x}_1^i for some constants c_i. The term c_N \tilde{x}_1^N is the one we are interested in, as it will give us the coefficient of \tilde{x}^N_1 in \tilde{f}.

Next, let's look at the term h_N(1, a_2, ..., a_n) in \tilde{f}. This term comes from the term h_N(x_1, ..., x_n) in f under the transformation x_i