Question

Let \(K\) be a field, \(F \in K[X] .\) Show that there is a field \(L\) containing \(K\) such that \(F=\prod_{i=1}^{n}\left(X-x_{i}\right) \in L[X] .\) (Hint: Use Problem \(1.51(\mathrm{~d})\) and induction on the degree.) \(L\) is called a splitting field of \(F\).

Short answer

The main goal of the exercise is to show that there is a bigger field \(L\) containing \(K\) such that the polynomial \(F\) can be written as the product of linear factors in \(L[X]\).

Step-by-step solution

Basis step: Degree 1 polynomial

For a polynomial of degree 1, let \(F(X) = ax+b\) with \(a,b \in K\) and \(a \neq 0\). Then, we can rewrite \(F(X) = a \cdot (X-\frac{-b}{a})\) as a product of linear factors in \(K[X]\) itself. Therefore, the statement is true for a polynomial of degree 1.

Inductive step

Now, let the statement be true for any polynomial of degree \(n\), i.e., there exists a field extension \(L \supseteq K\) such that \(F = \prod_{i=1}^{n}(X-x_i) \in L[X]\) for some \(x_i \in L\). We want to show that the statement is also true for a polynomial of degree \(n+1\). Let \(G \in K[X]\) be a polynomial of degree \(n+1\). Let \(p(x)\) be an irreducible factor of \(G(x)\) in \(K[X]\). By problem 1.51(d), there exists a bigger field \(\widetilde{K}\) containing \(K\) where \(p(x)\) has a root, say \(c_1\). Therefore, \(p(x)\) can be factored as \((X-c_1)Q(x)\) in \(\widetilde{K}[X]\). Now, \(\widetilde{K}[X]\) also contains the polynomial \(G\). In \(\widetilde{K}[X]\), we can write \(G\) as \((X-c_1)Q(x)H(x)\) for some polynomial \(H(x)\) with degree \(n\). Since the statement is true for polynomials of degree \(n\), there exists a field \(L \supseteq \widetilde{K}\) such that \(H=\prod_{i=2}^{n+1}(X-x_i) \in L[X]\). Thus, in the field \(L[X]\), we can write \(G\) as \(G=\prod_{i=1}^{n+1}(X-x_i)\). Since \(L \supseteq \widetilde{K} \supseteq K\), this shows that the statement is true for the polynomial \(G\) of degree \(n+1\), and the induction step is complete. Hence, we have proved that for any polynomial \(F \in K[X]\), there exists a field \(L\) containing \(K\) such that \(F = \prod_{i=1}^{n}(X-x_i) \in L[X]\), making \(L\) the splitting field of \(F\).

Key concepts

Field Extension

Understanding field extension is key to grasping the concept of splitting fields. A field extension is a pair of fields, typically denoted as \(L/K\), where \(K\) is a subfield of \(L\). Essentially, \(L\) contains all of the elements and operations that \(K\) has, plus possibly more.

Field extensions are crucial when we're dealing with polynomials that do not factor completely over the original field \(K\). A classic example is the extension \(\mathbb{C}/\mathbb{R}\), where the field of complex numbers, \(\mathbb{C}\), extends the real numbers, \(\mathbb{R}\), to include the imaginary unit \(i\), which is necessary to factor polynomials like \(X^2 + 1\).

In the context of our original exercise, finding a field \(L\) that contains \(K\) so that our polynomial \(F\) factors completely into linear terms is the essence of creating a splitting field. This transformation, realized by the field extension, allows for the polynomial's roots to exist within the field \(L\).

Polynomial Factorization

Factorization of polynomials is the process of breaking down a polynomial into a product of simpler polynomials, typically of lower degree. In simpler terms, you break the polynomial down into parts that can be multiplied to recreate the original polynomial.

When dealing with polynomials over a field, the factorization depends on the elements available within that field. For a polynomial to factor completely into linear factors, like \(F = \prod_{i=1}^{n}(X-x_{i})\), the coefficients of these factors must reside within the field. This is not always possible in the original field \(K\), and hence a field extension \(L\) is necessary.

For example, the polynomial \(X^2 - 2\) cannot be factored into real coefficients, but it can be over the field of real numbers extended by the square root of 2, \(\mathbb{R}[\sqrt{2}]\), allowing us to express it as \((X - \sqrt{2})(X + \sqrt{2})\). Solving the polynomial factorization problem often requires finding such extended fields where the factorization becomes straightforward.

Inductive Proof

Induction is a powerful proof technique frequently utilized in mathematics. It consists of two steps: the base case and the inductive step. In essence, to prove a statement for all natural numbers, you start by proving it for the simplest case, often \(n = 1\), which is the base case. Then, assuming the statement holds for an arbitrary natural number \(n\) (the inductive hypothesis), you show it must also hold for \(n + 1\).

Inductive proofs are invaluable when dealing with polynomial degrees, as they allow us to extend a property from one degree to the next. In the given exercise, the base case confirms the property for degree 1 polynomials. The inductive step extends it to higher degrees, showing that if our statement is true for any polynomial of degree \(n\), then it must also be true for polynomials of degree \(n+1\).

This structured approach ensures you can confidently assert the property holds for all polynomials, effectively leveraging the principle of mathematical induction.

Algebraic Geometry

Algebraic geometry is a branch of mathematics that studies the solutions of polynomial equations using geometric methods. While it deals with very high-level concepts, at its core, it provides a way to visualize and comprehend solutions in terms of shapes and sizes.

A central theme in algebraic geometry is the idea of viewing polynomials as defining geometric objects, such as curves or surfaces. For instance, a simple linear polynomial can be seen as defining a line, whereas a higher degree polynomial may correspond to a curve or more complex shape.

While the immediate application of algebraic geometry might not be apparent in the realm of splitting fields, understanding that fields can provide a 'space' for roots to exist connects these fields to geometric interpretations. The construction of a splitting field, for example, allows for a complete understanding of the roots of a polynomial, much like how geometric figures can help visualize mathematical problems. This perspective enriches the way we consider polynomial solutions, offering deeper insights into their nature and behavior.