Editing Splitting field (section)

==Constructing splitting fields==
===Motivation===
Finding [[root of a polynomial|roots of polynomials]] has been an important problem since the time of the ancient Greeks. Some polynomials, however, such as {{math|''x''<sup>2</sup> + 1}} over {{math|'''R'''}}, the [[real number]]s, have no roots. By constructing the splitting field for such a polynomial one can find the roots of the polynomial in the new field.

===The construction===
Let ''F'' be a field and ''p''(''X'') be a polynomial in the [[polynomial ring]] ''F''[''X''] of [[degree of a polynomial|degree]] ''n''. The general process for constructing ''K'', the splitting field of ''p''(''X'') over ''F'', is to construct a [[chain (ordered set)|chain]] of fields <math>F=K_0 \subseteq K_1 \subseteq \cdots \subseteq K_{r-1} \subseteq K_r=K</math> such that ''K<sub>i</sub>'' is an extension of ''K''<sub>''i''−1</sub> 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<sub>i</sub>'' are given as follows:
* [[Factorization of polynomials#Factoring over algebraic extensions (Trager's method)|Factorize]] ''p''(''X'') over ''K<sub>i</sub>'' into [[irreducible polynomial|irreducible]] factors <math>f_1(X)f_2(X) \cdots f_k(X)</math>.
* Choose any nonlinear irreducible factor ''f''(''X'').
* Construct the [[field extension]] ''K''<sub>''i''+1</sub> of ''K<sub>i</sub>'' as the [[quotient ring]] ''K''<sub>''i''+1</sub> = ''K''<sub>''i''</sub>[''X''] / (''f''(''X'')) where (''f''(''X'')) denotes the [[ideal (ring theory)|ideal]] in ''K''<sub>''i''</sub>[''X''] generated by ''f''(''X'').
* Repeat the process for ''K''<sub>''i''+1</sub> until ''p''(''X'') completely factors.

The irreducible factor ''f''(''X'') 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]] of ''K''<sub>''i''</sub>[''X''] and ''K''<sub>''i''</sub>[''X''] / (''f''(''X'')) is, in fact, a field, the [[residue field]] for that maximal ideal. Moreover, if we let <math>\pi : K_i[X] \to K_i[X]/(f(X))</math> be the natural projection of the [[ring (mathematics)|ring]] onto its quotient then
:<math>f(\pi(X)) = \pi(f(X)) = f(X)\ \bmod\ f(X) = 0</math>
so ''π''(''X'') is a root of ''f''(''X'') and of ''p''(''X'').

The degree of a single extension <math>[K_{i+1} : K_i]</math> is equal to the degree of the irreducible factor ''f''(''X''). The degree of the extension [''K'' : ''F''] is given by <math>[K_r : K_{r-1}] \cdots [K_2 : K_1] [K_1 : F]</math> and is at most ''n''!.

=== The field ''K''<sub>''i''</sub>[''X'']/(''f''(''X'')) ===
As mentioned above, the quotient ring ''K''<sub>''i''+1</sub> = ''K''<sub>''i''</sub>[''X'']/(''f''(''X'')) is a field when ''f''(''X'') is irreducible. Its elements are of the form

:<math>c_{n-1}\alpha^{n-1} + c_{n-2}\alpha^{n-2} + \cdots + c_1\alpha + c_0</math>

where the ''c<sub>j</sub>'' are in ''K<sub>i</sub>'' and ''α'' = ''π''(''X''). (If one considers ''K''<sub>''i''+1</sub> as a [[vector space]] over ''K<sub>i</sub>'' then the powers ''α''<sup>&thinsp;''j''</sup> for {{nowrap|0 ≤ ''j'' ≤ ''n''−1}} form a [[Basis (linear algebra)|basis]].)

The elements of ''K''<sub>''i''+1</sub> can be considered as polynomials in ''α'' of degree less than ''n''. Addition in ''K''<sub>''i''+1</sub> is given by the rules for polynomial addition, and multiplication is given by polynomial multiplication modulo ''f''(''X''). That is, for ''g''(''α'') and ''h''(''α'') in ''K''<sub>''i''+1</sub> their product is ''g''(''α'')''h''(''α'') = ''r''(α) where ''r''(''X'') is the remainder of ''g''(''X'')''h''(''X'') when divided by ''f''(''X'') in ''K''<sub>''i''</sub>[''X''].

The remainder ''r''(''X'') can be computed through [[polynomial long division]]; however there is also a straightforward reduction rule that can be used to compute ''r''(''α'') = ''g''(''α'')''h''(''α'') directly. First let

:<math>f(X) = X^n + b_{n-1} X^{n-1} + \cdots + b_1 X + b_0.</math>

The polynomial is over a field so one can take ''f''(''X'') to be [[monic polynomial|monic]] [[without loss of generality]]. Now ''α'' is a root of ''f''(''X''), so

:<math>\alpha^n = -(b_{n-1} \alpha^{n-1} + \cdots + b_1 \alpha + b_0).</math>

If the product ''g''(''α'')''h''(''α'') has a term ''α''<sup>''m''</sup> with {{nowrap|''m'' ≥ ''n''}} it can be reduced as follows:

:<math>\alpha^n\alpha^{m-n} = -(b_{n-1} \alpha^{n-1} + \cdots + b_1 \alpha + b_0) \alpha^{m-n} = -(b_{n-1} \alpha^{m-1} + \cdots + b_1 \alpha^{m-n+1} + b_0 \alpha^{m-n})</math>.

As an example of the reduction rule, take ''K<sub>i</sub>'' = '''Q'''[''X''], the ring of polynomials with [[rational number|rational]] coefficients, and take ''f''(''X'') = ''X''<sup>&thinsp;7</sup> − 2. Let <math>g(\alpha) = \alpha^5 + \alpha^2</math> and ''h''(''α'') = ''α''<sup>3</sup> +1 be two elements of '''Q'''[''X'']/(''X''<sup>&thinsp;7</sup> − 2). The reduction rule given by ''f''(''X'') is ''α''<sup>7</sup> = 2 so

:<math>g(\alpha)h(\alpha) = (\alpha^5 + \alpha^2)(\alpha^3 + 1) = \alpha^8 + 2 \alpha^5 + \alpha^2 = (\alpha^7)\alpha + 2\alpha^5 + \alpha^2 = 2 \alpha^5 + \alpha^2 + 2\alpha.</math>