Question

Let \(F\) be a field, and consider the ring of bivariate polynomials \(F[X, Y]\). Show that in this ring, the polynomial \(X^{2}+Y^{2}-1\) is irreducible, provided \(F\) does not have characteristic 2 . What happens if \(F\) has characteristic \(2 ?\)

Short answer

#Short Answer# When \(F\) does not have characteristic 2, the polynomial \(X^2 + Y^2 - 1\) is irreducible in \(F[X, Y]\). We showed this by assuming it can be factored into \(f(X, Y)g(X, Y)\) and reaching a contradiction. However, when \(F\) has characteristic 2, the polynomial may factor. For example, in \(F = \mathbb{Z}_2\), \(X^2 + Y^2 - 1 = (X + Y + 1)^2\), which demonstrates that the polynomial is reducible under this condition.

Step-by-step solution

Recall the definition of irreducibility

A polynomial \(f(X, Y)\) is irreducible in the ring of bivariate polynomials \(F[X, Y]\) if it cannot be factored into the product of two non-constant polynomials in \(F[X, Y]\).

Assume the polynomial can be factored

Let's assume that the polynomial \(X^2 + Y^2 - 1\) can be factored as \(f(X, Y)g(X, Y)\) with \(f, g \in F[X, Y]\) are both non-constant. We are aiming for a contradiction.

Examine properties of the polynomial in characteristic not 2

Note that the polynomial \(X^2 + Y^2 - 1 = 0\) represents a circle of radius 1 centered at the origin. The coefficients of the polynomial are all either 1 or -1. When \(F\) does not have characteristic 2, the polynomial does not factor, as its solutions form a circle that does not intersect any axis and cross each other.

Prove the polynomial is irreducible when \(F\) does not have characteristic 2

Suppose \(X^2 + Y^2 - 1\) is reducible in \(F[X, Y]\) with \(F\) not having characteristic 2. That means \(f(X, Y)g(X, Y) = X^2 + Y^2 - 1\). Without loss of generality, assume \(f(X, Y) = X - a\) and \(g(X, Y) = X^2 + Y^2 - 1 - (X-a)\) with \(a \in F\). Then, \((X - a)(X^2 + Y^2 - 1 - (X - a)) = X^2 + Y^2 - 1\). Expanding it, we obtain \(X^3 - aX^2 + (a - 1)X - aY^2 + 1 = X^2 + Y^2 - 1\). Matching the coefficients, we have \(-a = 1\) and \(a-1 = 0\), which is impossible, as they imply \(a = 1\) and \(a = -1\) simultaneously. Thus, \(X^2 + Y^2 - 1\) is irreducible in \(F[X, Y]\).

Examine the case when \(F\) has characteristic 2

In the case when \(F\) has characteristic 2, the polynomial does factor. For instance, when \(F = \mathbb{Z}_2\), we have \(X^2 + Y^2 - 1 = X^2 + Y^2 + 1 = (X + Y + 1)^2\). Notice that the polynomial becomes reducible under this condition.

Key concepts

Field Characteristics

In the world of algebra, fields can have different characteristics, usually denoted by a number. The characteristic of a field \(F\) is the smallest positive number \(n\) such that repeatedly adding the field's "one" element to itself \(n\) times yields zero. If no such \(n\) exists, the characteristic is said to be zero. This characteristic plays a crucial role in the behavior of polynomials.

For instance, if \(F\) does not have characteristic 2, it means adding "1" to itself never results in zero. This affects how polynomials like \(X^2 + Y^2 - 1\) behave. In particular, because the coefficients of this polynomial are 1 or -1, these elements won't simplify to zero, affecting factorizability.

When \(F\) has characteristic 2, an interesting thing happens. Here, \(1 + 1 = 0\). This change means expressions involving "-1" may effectively become zero, allowing different factorization possibilities. It's these subtle shifts that affect polynomial behavior and factorizability.

Bivariate Polynomials

Bivariate polynomials are polynomials that involve two variables, typically \(X\) and \(Y\). The ring \(F[X, Y]\) consists of all polynomials with these two variables and coefficients from the field \(F\).

Bivariate polynomials like \(X^2 + Y^2 - 1\) have geometric interpretations. For instance, \(X^2 + Y^2 - 1 = 0\) represents a circle with radius 1 centered at the origin in a Cartesian plane. These geometric interpretations can give insights into their properties.

Analyzing these polynomials involves checking whether they can be expressed as the product of simpler non-constant polynomials, considering their interactions within \(F[X, Y]\). The presence of two variables adds complexity, but it also allows representation of more complex shapes and forms.

Polynomial Factorization

Factorization of polynomials involves expressing a given polynomial as a product of simpler polynomials. A polynomial is considered irreducible if it cannot be factored into non-constant polynomials over a given ring.

The polynomial \(X^2 + Y^2 - 1\) is shown to be irreducible in \(F[X, Y]\) when \(F\) does not have characteristic 2. This conclusion comes from trying hypothetical factorizations and finding contradictions, leading to the realization that no such factorization exists over the ring.

However, if \(F\) has characteristic 2, such as \(\mathbb{Z}_2\), the polynomial becomes reducible. It can be factored as \((X + Y + 1)^2\), indicating a different arithmetic context where factorization becomes possible. This highlights how characteristics of fields influence polynomial behavior extensively.