# Splitting Fields ## Polynomials and Their Roots A central problem in algebra is to determine where a polynomial factors completely into linear terms. Consider the polynomial $$ f(x)=x^2-2. $$ Over the rational numbers $\mathbb{Q}$, this polynomial cannot be factored further, since $\sqrt{2}\notin\mathbb{Q}$. Over the real numbers $\mathbb{R}$, however, $$ x^2-2=(x-\sqrt{2})(x+\sqrt{2}). $$ Thus the factorization of a polynomial depends on the field in which it is considered. In general, a polynomial may fail to have all its roots in a given field. By enlarging the field appropriately, one can construct a field in which the polynomial decomposes completely into linear factors. This leads to the notion of a splitting field. ## Definition of a Splitting Field Let $K$ be a field and let $$ f(x)\in K[x] $$ be a nonconstant polynomial. **Definition.** A splitting field of $f(x)$ over $K$ is a field extension $L/K$ satisfying the following conditions: 1. The polynomial $f(x)$ factors completely into linear factors over $L$. 2. The field $L$ is generated over $K$ by the roots of $f(x)$. Equivalently, $$ L=K(\alpha_1,\alpha_2,\ldots,\alpha_n), $$ where $\alpha_1,\ldots,\alpha_n$ are all roots of $f(x)$. The second condition expresses minimality. The splitting field contains no unnecessary elements beyond those needed to include all roots of the polynomial. ## First Examples Consider $$ f(x)=x^2-2 $$ over $\mathbb{Q}$. Its roots are $$ \pm\sqrt{2}. $$ The field generated by these roots is $$ \mathbb{Q}(\sqrt{2}). $$ Since both roots lie in this field and the polynomial factors as $$ x^2-2=(x-\sqrt{2})(x+\sqrt{2}), $$ the splitting field is $$ \mathbb{Q}(\sqrt{2}). $$ The extension has degree $$ [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2. $$ Now consider $$ f(x)=x^2+1 $$ over $\mathbb{R}$. The roots are $$ \pm i. $$ The splitting field is therefore $$ \mathbb{R}(i)=\mathbb{C}. $$ Again the extension degree equals $2$. More complicated examples require adjoining several roots successively. ## Splitting Field of $x^3-2$ Consider the polynomial $$ f(x)=x^3-2 $$ over $\mathbb{Q}$. One root is the real number $$ \sqrt[3]{2}. $$ However, the polynomial has three complex roots: $$ \sqrt[3]{2}, \qquad \omega\sqrt[3]{2}, \qquad \omega^2\sqrt[3]{2}, $$ where $\omega$ is a primitive cube root of unity satisfying $$ \omega^3=1, \qquad \omega\neq 1. $$ The roots of $$ x^3-1 $$ are $1,\omega,\omega^2$, and one computes that $$ \omega=\frac{-1+\sqrt{-3}}{2}. $$ Thus the splitting field of $x^3-2$ over $\mathbb{Q}$ is $$ \mathbb{Q}(\sqrt[3]{2},\omega). $$ The polynomial does not split in $\mathbb{Q}(\sqrt[3]{2})$ alone because the complex roots are missing. This example shows that splitting fields often require adjoining algebraic numbers of different types simultaneously. ## Existence of Splitting Fields Every polynomial over a field possesses a splitting field. **Theorem.** Let $K$ be a field and let $$ f(x)\in K[x] $$ be nonconstant. Then there exists a field extension $L/K$ in which $f(x)$ splits completely. The construction proceeds iteratively. If $f(x)$ does not already have a root in $K$, one adjoins a root $\alpha$ to form $$ K(\alpha). $$ The polynomial then factors partially over this larger field. Repeating the process eventually produces a field containing all roots. Thus splitting fields always exist. ## Uniqueness up to Isomorphism Although a splitting field may be constructed in different ways, its algebraic structure is essentially unique. **Theorem.** If $L_1$ and $L_2$ are splitting fields of the same polynomial over $K$, then there exists a field isomorphism $$ L_1\cong L_2 $$ that fixes every element of $K$. This means that all splitting fields of a polynomial are identical from the viewpoint of field theory, even if they appear differently as concrete subsets of larger fields. For this reason one usually speaks of “the” splitting field. ## Repeated Roots and Separability A polynomial may possess repeated roots. For example, $$ f(x)=x^2-2x+1=(x-1)^2. $$ Here the root $1$ occurs twice. Repeated roots are detected using derivatives. A polynomial $f(x)$ has a repeated root precisely when it shares a nontrivial common factor with its derivative $f'(x)$. In characteristic zero, irreducible polynomials never have repeated roots. Consequently, extensions of fields such as $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ behave especially well. This observation becomes important in Galois theory. ## Splitting Fields and Symmetry The splitting field contains all algebraic information associated with the roots of a polynomial. Once all roots lie in a common field, one may study how the roots relate to one another. For example, the polynomial $$ x^2-2 $$ has roots $\sqrt{2}$ and $-\sqrt{2}$. There is a symmetry exchanging these two roots while preserving all rational numbers: $$ \sqrt{2}\mapsto -\sqrt{2}. $$ Similarly, the roots of $$ x^3-2 $$ can be permuted in more complicated ways. The collection of such symmetries forms the Galois group of the polynomial. This idea connects algebraic equations with group theory and lies at the center of modern algebraic number theory. ## Splitting Fields in Number Theory Splitting fields appear naturally throughout number theory. Cyclotomic fields arise as splitting fields of $$ x^n-1. $$ Quadratic fields arise from polynomials of degree two. Finite fields are constructed as splitting fields of irreducible polynomials over prime fields. Many reciprocity laws are most naturally expressed in terms of how primes behave inside splitting fields. The arithmetic of a polynomial is therefore deeply connected to the arithmetic of the field generated by its roots.