Construction of splitting fields

In mathematics, a splitting field of a polynomial with coefficients in a field is an extension of that field over which the polynomial factors into linear factors. The purpose of this article is to describe an iterative process for constructing the splitting field of a given polynomial.

Motivation

Finding roots of polynomials has been an important problem since the time of the ancient Greeks. Some polynomials, however, have no roots such as $x^2+1$ over R, the real numbers. By constructing the splitting field for such a polynomial one can find the roots of the polynomial in the new field. (In this example the splitting field is C the complex numbers, where $x^2\; +\; 1\; =\; (x+i)(x-i)$.)

Construction

Let F be a field and p(x) be a polynomial in the polynomial ring F [x] of degree n. The general process for constructing K, the splitting field of p(x) over F, is to construct a sequence of fields $F=K\_0,\; K\_1,\; ...\; K\_\{r-1\},\; K\_r=K$ such that $K\_i$ is an extension of $K\_\{i-1\}$ containing a new root of p(x). Since p(x) has at most n roots the construction will require at most n extensions. The steps for constructing $K\_i$ are given as follows:

*Factorize p(x) over $K\_i$ into irreducible factors $f\_1(x)f\_2(x)...f\_k(x)$.
*Chose any irreducible factor $f(x)\; =\; f\_i(x)$
*Construct the quotient ring $K\_\{i+1\}=K\_i\; [x]\; /(f(x))$ where (f(x)) denotes the ideal in $K\_i\; [x]$ generated by f(x)
*Repeat the process for $K\_\{i+1\}$ until p(x) factorizes completely.

The irreducible factor $f\_i$ used in the quotient construction may be chosen arbitrarily. Although different choices of factors may lead to different subfield sequences the resulting splitting fields will be isomorphic.

Since f(x) is irreducible (f(x)) is a maximal ideal and hence $K\_i\; [x]\; /(f(x))$ is, in fact, a field. Moreover, if we let $pi\; :\; K\_i\; [x]\; ->\; K\_i\; [x]\; /(f\_1(x))$ be the natural projection of the ring onto its quotient then $f(pi(x))\; =\; pi(f(x))\; =\; f(x)$ mod $f(x)\; =\; 0$ so $pi(x)$ is a root of f(x) and of p(x).

The degree of a single extension [$K\_\{i+1\}\; :\; K\_i$] is equal to the degree of the irreducible factor f(x). The degree of the extension [K : F] is given by $[K\_r\; :\; K\_\{r-1\}]\; ...\; [K\_2\; :\; K\_1]\; [K\_1\; :\; F]$ and is at most n!.

The Field $K\_i\; [x]\; /(f(x))$

As mentioned above the quotient ring $K\_\{i+1\}\; =\; K\_i\; [x]\; /(f(x))$ is a field when f(x) is irreducible. Its elements are of the form $c\_\{n-1\}alpha^\{n-1\}\; +\; c\_\{n-2\}alpha^\{n-2\}\; +\; ...\; +\; c\_1alpha^1\; +\; c\_0$ where the $c\_j$ are in $K\_i$ and $alpha\; =\; pi(x)$. (If one considers $K\_\{i+1\}$ as a vector space over $K\_i$ then the powers $alpha^j$ for 1 <= j <= n-1 form a basis.)

The elements of $K\_\{i+1\}$ can be considered as polynomials in $alpha$ of degree less than n. Addition in $K\_\{i+1\}$ is given by the rules for polynomial addition and multiplication is given by polynomial multiplication modulo f(x). That is, for $g(alpha)$ and $h(alpha)$ in $K\_\{i+1\}$ the product $g(alpha)h(alpha)\; =\; r(alpha)$ where r(x) is the remainder of g(x)h(x) divided by f(x) in $K\_i\; [x]$.

The remainder r(x) can be computed through long division of polynomials, however there is also a straightforward reduction rule that can be used to compute $r(alpha)=g(alpha)h(alpha)$ directly. First let $f(x)\; =\; x^n\; +\; b\_\{n-1\}x^\{n-1\}\; +\; ...\; +\; b\_1x\; +\; b\_0$. (The polynomial is over a field so one can take f(x) to be monic without loss of generality.) $alpha$ is a root of f(x) so $alpha^n\; =\; -(b\_\{n-1\}alpha^\{n-1\}\; +\; ...\; +\; b\_1alpha\; +\; b\_0)$. If the product $g(alpha)h(alpha)$ has a term $alpha^m$ with m >= n it can be reduced as follows:

:$alpha^nalpha^\{m-n\}\; =\; (-(b\_\{n-1\}alpha^\{n-1\}\; +\; ...\; +\; b\_1alpha\; +\; b\_0))alpha^\{m-n\}\; =\; -(b\_\{n-1\}alpha^\{m-1\}\; +\; ...\; +\; b\_1alpha^\{m-n+1\}\; +\; b\_0alpha^\{m-n+1\})$.

As an example of the reduction rule, take $K\_i\; =$ Q, the rational numbers, and take $f(x)\; =\; x^7\; -\; 2$. Let $g(alpha)\; =\; alpha^5\; +\; alpha^2,\; h(alpha)\; =\; alpha^3\; +\; 1$ be two elements of Q$/(x^7-2)$. The reduction rule given by f(x) is $alpha^7\; =\; 2$ so

:$g(alpha)h(alpha)\; =\; (alpha^5\; +\; alpha^2)(alpha^3\; +\; 1)\; =\; alpha^8\; +\; 2alpha^5\; +\; alpha^2\; =\; (alpha^7)alpha\; +\; 2alpha^5\; +\; alpha^2\; =\; 2alpha^5\; +\; alpha^2\; +\; 2alpha.$

Examples

Let "f = X² + 1" in "R [X] ". Then $R\; [x]\; :=R\; [X]\; /(f)$" has:
*elements: a + bx, a, b in R;
*addition: $(a\_1\; +\; b\_1x)\; +\; (a\_2\; +\; b\_2x)\; =\; (a\_1\; +\; a\_2)\; +\; (b\_1\; +\; b\_2)x$;
*multiplication: $(a\_1\; +\; b\_1x)(a\_2\; +\; b\_2x)=\; (a\_1a\_2\; -\; b\_1b\_2)\; +\; (a\_1b\_2\; +\; a\_2b\_1)x$.We usually write "i" for "x" and "C" for "R [x] ".