Is It Possible to Write Multivariable Polynomials as Products of Linear Combinations in an Algebraically Closed Field?

In the context of algebraically closed fields, a prominent example is the field of complex numbers. This article explores the question of whether multivariable polynomials can be uniquely represented as products of linear combinations. Specifically, we examine the polynomial fxy x2y2 - 1, and demonstrate why this is not possible through a detailed analysis.

Introduction

Consider the polynomial fxy x2y2 - 1 in the context of algebraically closed fields. An algebraically closed field is one where every non-constant polynomial has a root. The complex numbers (( mathbb{C} )) provide an excellent example of such a field. We aim to determine if this polynomial can be uniquely represented as a product of linear combinations of the variables x and y.

Exploring the Polynomial fxy

Begin with the polynomial:

[mathbf{f_{xy} x^2y^2 - 1}]

Assume for the sake of contradiction that:

[mathbf{f_{xy} a x^a y^b (c x d y) (e x f y)}]

Expanding the right-hand side, we get:

[mathbf{a x^a y^b (c x d y) (e x f y) a x^a y^b (ce x^2 (cf de)xy df y^2)}]

Equating the coefficients of the expanded form with the given polynomial, we obtain the system of equations:

[begin{cases} a^2 c e 1 a b (c f d e) 0 a b d f 0 a b (c f d e) 0 a b d f -1 a b (c e) 0 end{cases}]

Now, let's analyze these equations step by step.

Equation Analysis

Starting with the third and fifth equations:

[mathbf{a b d f 0} quad text{and} quad mathbf{a b d f -1}]

These equations directly contradict each other. For the product (a b d f) to be zero and simultaneously equal to -1 is impossible. Hence, this assumption leads to a contradiction.

Further Analysis

Consider the second and fourth equations:

[mathbf{a b (c f d e) 0}]

This equation implies that either (a 0), (b 0), (c f d e 0), or all of these factors are zero. If (a 0) or (b 0), then the original polynomial cannot be expressed in the desired form as the leading term would vanish. Therefore, we must have (c f d e 0).

Linear Combination Analysis

Assume without loss of generality that:

[mathbf{c 1} quad text{and} quad mathbf{d -i}]

Then, the equation (c f d e 0) becomes:

[mathbf{f (-i) e 0 quad text{or} quad f i e}]

Now, consider the first and second equations:

[mathbf{a^2 c e 1} quad text{and} quad mathbf{a b (c f d e) 0}]

Substituting (c 1) and (d -i), and (f i e), we get:

[mathbf{a^2 e 1} quad text{and} quad mathbf{a b e 0}]

The equation (a b e 0) implies that either (a 0), (b 0), or (e 0). If (a 0) or (b 0), the polynomial would again vanish, which is not the case. Thus, we must have (e 0), but this leads to:

[mathbf{a^2 1 quad text{and} quad c f 0}]

From (c f 0), we get (f 0) (since (c 1)). This implies that (e f 0), which contradicts the requirement that (a b d f -1).

Conclusion

In conclusion, it is not possible to represent the polynomial (f_{xy} x^2y^2 - 1) as a product of linear combinations of (x) and (y) in the field of complex numbers. This analysis highlights the limitations on the unique representation of multivariable polynomials under certain conditions, particularly within algebraically closed fields like the complex numbers.

Related Keywords

multivariable polynomials, linear combinations, algebraically closed fields, complex numbers, unique representation