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

In summary: Using the definition of h_N, we can expand this as a sum of terms of the form c_i \tilde{x}_1^i for some constants c_i. Again, the term c_N \tilde{x}_1^N is the one we are interested in.Comparing the two terms we have looked at, we can see that they have the same form, with the same coefficient c_N. This means that c(a_1, ..., a_n) = h_N(1, a_2, ..., a_n) and therefore, the coefficient of \tilde{x}^N_1
  • #1
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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
 
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  • #2
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
 

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

1. What is Theorem 1 in the Weak Nullstellensatz?

Theorem 1 in the Weak Nullstellensatz states that for any field k and any ideal I in the polynomial ring k[x1,...,xn], the algebraic set V(I) defined by I is empty if and only if 1 is contained in the radical of I. In other words, if the ideal I contains a constant polynomial 1, then the algebraic set defined by I is empty.

2. Who are the authors of the Weak Nullstellensatz proof in Exercise 3(a)?

The authors of the Weak Nullstellensatz proof in Exercise 3(a) are David Cox, John Little, and Donal O'Shea. They are mathematicians and authors of the book "Ideals, Varieties, and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative Algebra", where the proof can be found in detail.

3. What is the significance of Exercise 3(a) in the Weak Nullstellensatz proof?

Exercise 3(a) in the Weak Nullstellensatz proof is significant because it helps to establish the main result of the proof, which is that the ideal I contains a constant polynomial 1 if and only if the algebraic set V(I) is empty. This exercise is a crucial step in understanding the proof and its implications.

4. How does the Weak Nullstellensatz proof relate to algebraic geometry?

The Weak Nullstellensatz is a fundamental theorem in algebraic geometry, which studies the properties of algebraic varieties defined by polynomial equations. The proof in Exercise 3(a) provides a deeper understanding of the connection between the algebraic set defined by an ideal and the polynomials contained in that ideal.

5. Is the Weak Nullstellensatz proof in Exercise 3(a) difficult to understand?

The level of difficulty in understanding the Weak Nullstellensatz proof in Exercise 3(a) may vary depending on an individual's mathematical background and familiarity with algebraic geometry. However, with careful study and practice, it is possible to grasp the key concepts and techniques used in the proof.

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