Editing Splitting field

{{Short description|Field generated by all rupture-fields of a polynomial over a field}}
{{about|the splitting field of a polynomial| the splitting field of a CSA|central simple algebra}}
In [[abstract algebra]], a '''splitting field''' of a [[polynomial]] with [[coefficient]]s in a [[field (mathematics)|field]] is the smallest [[field extension]] of that field over which the polynomial ''splits'', i.e., decomposes into [[linear polynomial|linear]] factors.

==Definition==
A '''splitting field''' of a polynomial ''p''(''X'') over a field ''K'' is a field extension ''L'' of ''K'' over which ''p'' factors into linear factors

:<math>p(X) = c \prod_{i=1}^{\deg p} (X - a_i)</math> 

where <math>c \in K</math> and for each <math>i</math> we have <math>X - a_i \in L[X]</math> with ''a<sub>i</sub>'' not necessarily distinct and such that the [[root of a polynomial|roots]] ''a<sub>i</sub>'' generate ''L'' over ''K''. The extension ''L'' is then an extension of minimal [[Degree of a field extension|degree]] over ''K'' in which ''p'' splits. It can be shown that such splitting fields exist and are unique [[up to]] [[isomorphism]]. The amount of freedom in that isomorphism is known as the [[Galois group]] of ''p'' (if we assume it is [[separable polynomial|separable]]).

A splitting field of a set ''P'' of polynomials is the smallest field over which each of the polynomials in ''P'' splits.

==Properties==
An extension ''L'' that is a [[Algebraic closure#Existence of an algebraic closure and splitting fields|splitting field for a set of polynomials]] ''p''(''X'') over ''K'' is called a [[Field extension#Normal, separable and Galois extensions|normal extension]] of ''K''.

Given an [[algebraically closed field]] ''A'' containing ''K'', there is a unique splitting field ''L'' of ''p'' between ''K'' and ''A'', generated by the roots of ''p''. If ''K'' is a [[field extension|subfield]] of the [[complex number]]s, the existence is immediate. On the other hand, the existence of [[algebraic closure]]s in general is often [[mathematical proof|proved]] by 'passing to the limit' from the splitting field result, which therefore requires an independent proof to avoid [[circular definition|circular reasoning]].

Given a [[separable extension]] ''K''′ of ''K'', a '''Galois closure''' ''L'' of ''K''′ is a type of splitting field, and also a [[Galois extension]] of ''K'' containing ''K''′ that is minimal, in an obvious sense. Such a Galois closure should contain a splitting field for all the polynomials ''p'' over ''K'' that are [[Minimal polynomial (field theory)|minimal polynomials]] over ''K'' of elements of ''K''′.

==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>

==Examples==
=== The complex numbers ===
Consider the [[polynomial ring]] '''R'''[''x''], and the [[irreducible polynomial]] {{nowrap|1=''x''<sup>2</sup> + 1.}} The [[quotient ring]] {{nowrap|1='''R'''[''x'']&thinsp;/&thinsp;(''x''<sup>2</sup> + 1)}} is given by the [[Congruence relation|congruence]] {{nowrap|1=''x''<sup>2</sup> ≡ −1.}} As a result, the elements (or [[equivalence class]]es) of {{nowrap|1='''R'''[''x'']&thinsp;/&thinsp;(''x''<sup>2</sup> + 1)}} are of the form {{nowrap|1=''a'' + ''bx''}} where ''a'' and ''b'' belong to '''R'''. To see this, note that since {{nowrap|1=''x''<sup>2</sup> ≡ −1}} it follows that {{nowrap|1=''x''<sup>3</sup> ≡ −''x''}}, {{nowrap|1=''x''<sup>4</sup> ≡ 1}}, {{nowrap|1=''x''<sup>5</sup> ≡ ''x''}}, etc.; and so, for example {{nowrap|1=''p'' + ''qx'' + ''rx''<sup>2</sup> + ''sx''<sup>3</sup> ≡ ''p'' + ''qx'' + ''r''(−1) + ''s''(−''x'') = (''p'' − ''r'') + (''q'' − ''s'')''x''.}}

The addition and multiplication operations are given by firstly using ordinary polynomial addition and multiplication, but then reducing modulo {{nowrap|1=''x''<sup>2</sup> + 1}}, i.e. using the fact that {{nowrap|1=''x''<sup>2</sup> ≡ −1}}, {{nowrap|1=''x''<sup>3</sup> ≡ −''x''}}, {{nowrap|1=''x''<sup>4</sup> ≡ 1}}, {{nowrap|1=''x''<sup>5</sup> ≡ ''x''}}, etc. Thus:
:<math>(a_1 + b_1x) + (a_2 + b_2x) = (a_1 + a_2) + (b_1 + b_2)x, </math>
:<math>(a_1 + b_1x)(a_2 + b_2x) = a_1a_2 + (a_1b_2 + b_1a_2)x + (b_1b_2)x^2 \equiv (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)x \, . </math>
If we identify {{nowrap|1=''a'' + ''bx''}} with (''a'',''b'') then we see that addition and multiplication are given by
:<math>(a_1,b_1) + (a_2,b_2) = (a_1 + a_2,b_1 + b_2), </math>
:<math>(a_1,b_1)\cdot (a_2,b_2) = (a_1a_2 - b_1b_2,a_1b_2 + b_1a_2). </math>

We claim that, as a field, the quotient ring {{nowrap|1='''R'''[''x''] / (''x''<sup>2</sup> + 1)}} is [[isomorphic]] to the [[complex number]]s, '''C'''. A general complex number is of the form {{nowrap|1=''a'' + ''bi''}}, where ''a'' and ''b'' are real numbers and {{nowrap|1=''i''<sup>2</sup> = −1.}} Addition and multiplication are given by

:<math>(a_1 + b_1 i) + (a_2 + b_2 i) = (a_1 + a_2) + i(b_1 + b_2),</math>
:<math>(a_1 + b_1 i) \cdot (a_2 + b_2 i) = (a_1a_2 - b_1b_2) + i(a_1b_2 + a_2b_1).</math>

If we identify {{nowrap|1=''a'' + ''bi''}} with (''a'', ''b'') then we see that addition and multiplication are given by

:<math>(a_1,b_1) + (a_2,b_2) = (a_1 + a_2,b_1 + b_2),</math>
:<math>(a_1,b_1)\cdot (a_2,b_2) = (a_1a_2 - b_1b_2,a_1b_2 + b_1a_2).</math>

The previous calculations show that addition and multiplication behave the same way in {{nowrap|1='''R'''[''x'']&thinsp;/&thinsp;(''x''<sup>2</sup> + 1)}} and '''C'''. In fact, we see that the map between {{nowrap|1='''R'''[''x'']&thinsp;/&thinsp;(''x''<sup>2</sup> + 1)}} and '''C''' given by {{nowrap|1=''a'' + ''bx'' → ''a'' + ''bi''}} is a [[homomorphism]] with respect to addition ''and'' multiplication. It is also obvious that the map {{nowrap|1=''a'' + ''bx'' → ''a'' + ''bi''}} is both [[injective]] and [[surjective]]; meaning that {{nowrap|1=''a'' + ''bx'' → ''a'' + ''bi''}} is a [[bijective]] homomorphism, i.e., an [[ring isomorphism|isomorphism]]. It follows that, as claimed: {{nowrap|1='''R'''[''x'']&thinsp;/&thinsp;(''x''<sup>2</sup> + 1) ≅ '''C'''.}}

In 1847, [[Augustin-Louis Cauchy|Cauchy]] used this approach to ''define'' the complex numbers.<ref>{{Citation|last = Cauchy|first = Augustin-Louis|author-link = Augustin-Louis Cauchy|title = Mémoire sur la théorie des équivalences algébriques, substituée à la théorie des imaginaires|journal = [[Comptes Rendus Hebdomadaires des Séances de l'Académie des Sciences]]|volume = 24|year = 1847|language = fr|pages = 1120–1130}}</ref>

=== Cubic example ===
Let {{mvar|K}} be the [[rational number field]] {{math|'''Q'''}} and {{math|''p''(''x'') {{=}} ''x''<sup>3</sup> − 2}}. Each root of {{mvar|p}} equals {{math|{{radic|2|3}}}} times a [[cube root of unity]]. Therefore, if we denote the cube roots of unity by

:<math>\omega_1 = 1,\,</math> <!-- do not delete "\,": it improves the display of formula on certain browsers. -->
:<math>\omega_2 = -\frac{1}{2} + \frac{\sqrt{3}}{2} i,</math>
:<math>\omega_3 = -\frac{1}{2} - \frac{\sqrt{3}}{2} i.</math>

any field containing two distinct roots of {{mvar|p}} will contain the quotient between two distinct cube roots of unity. Such a quotient is a [[primitive root of unity|primitive]] cube root of unity—either <math>\omega_2</math> or <math>\omega_3=1/\omega_2</math>. It follows that a splitting field {{mvar|L}} of {{mvar|p}} will contain ''ω''<sub>2</sub>, as well as the real [[cube root]] of 2; [[converse (logic)|conversely]], any extension of {{math|'''Q'''}} containing these elements contains all the roots of {{mvar|p}}. Thus

:<math>L = \mathbf{Q}(\sqrt[3]{2}, \omega_2) = \{ a + b\sqrt[3]{2} + c{\sqrt[3]{2}}^2 + d\omega_2 + e\sqrt[3]{2}\omega_2 + f{\sqrt[3]{2}}^2 \omega_2 \mid a,b,c,d,e,f \in \mathbf{Q} \}</math>

Note that applying the construction process outlined in the previous section to this example, one begins with <math>K_0 = \mathbf{Q}</math> and constructs the field <math>K_1 = \mathbf{Q}[X] / (X^3 - 2)</math>. This field is not the splitting field, but contains one (any) root.  However, the polynomial <math>Y^3 - 2</math> is not [[irreducible polynomial|irreducible]] over <math>K_1</math>  and in fact:

:<math>Y^3 -2 = (Y - X)(Y^2 + XY + X^2).</math>

Note that <math>X</math> is not an [[indeterminate (variable)|indeterminate]], and is in fact an element of <math>K_1</math>.  Now, continuing the process, we obtain <math>K_2 = K_1[Y] / (Y^2 + XY + X^2)</math>, which is indeed the splitting field and is spanned by the <math>\mathbf{Q}</math>-basis <math>\{1, X, X^2, Y, XY, X^2 Y\}</math>. Notice that if we compare this with <math>L</math> from above we can identify <math>X = \sqrt[3]{2}</math> and <math>Y = \omega_2</math>.

===Other examples===
* The splitting field of ''x<sup>q</sup>'' − ''x'' over '''F'''<sub>''p''</sub> is the unique [[finite field]] '''F'''<sub>''q''</sub> for ''q'' = ''p<sup>n</sup>''.<ref>{{Cite book|title=A Course in Arithmetic|first = Jean-Pierre|last=Serre|authorlink = Jean-Pierre Serre}}</ref> Sometimes this field is denoted by GF(''q'').

* The splitting field of ''x''<sup>2</sup> + 1 over '''F'''<sub>7</sub> is '''F'''<sub>49</sub>; the polynomial has no roots in '''F'''<sub>7</sub>, i.e., −1 is not a [[square (algebra)|square]] there, because 7 is not [[modular arithmetic|congruent]] to 1 modulo 4.<ref>Instead of applying this characterization of [[parity (mathematics)|odd]] [[prime number|prime]] moduli for which −1 is a square, one could just check that the set of squares in '''F'''<sub>7</sub> is the set of classes of 0, 1, 4, and 2, which does not include the class of −1&thinsp;≡&nbsp;6.</ref>

* The splitting field of ''x''<sup>2</sup> − 1 over '''F'''<sub>7</sub> is '''F'''<sub>7</sub> since ''x''<sup>2</sup> − 1 = (''x'' + 1)(''x'' − 1) already splits into linear factors.

* We calculate the splitting field of ''f''(''x'') = ''x''<sup>3</sup> + ''x'' + 1 over '''F'''<sub>2</sub>. It is easy to verify that ''f''(''x'') has no roots in '''F'''<sub>2</sub>; hence ''f''(''x'') is irreducible in '''F'''<sub>2</sub>[''x'']. Put ''r'' = ''x'' + (''f''(''x'')) in '''F'''<sub>2</sub>[''x'']/(''f''(''x'')) so '''F'''<sub>2</sub>(''r'') is a field and ''x''<sup>3</sup> + ''x'' + 1 = (''x'' + ''r'')(''x''<sup>2</sup> + ''ax'' + ''b'') in '''F'''<sub>2</sub>(''r'')[''x'']. Note that we can write + for − since the [[characteristic (algebra)|characteristic]] is two. Comparing coefficients shows that ''a'' = ''r'' and ''b'' = 1 + ''r''<sup>&thinsp;2</sup>. The elements of '''F'''<sub>2</sub>(''r'') can be listed as ''c'' + ''dr'' + ''er''<sup>&thinsp;2</sup>, where ''c'', ''d'', ''e'' are in '''F'''<sub>2</sub>. There are eight elements: 0, 1, ''r'', 1 + ''r'', ''r''<sup>&thinsp;2</sup>, 1 + ''r''<sup>&thinsp;2</sup>, ''r'' + ''r''<sup>&thinsp;2</sup> and 1 + ''r'' + ''r''<sup>&thinsp;2</sup>. Substituting these in ''x''<sup>2</sup> + ''rx'' + 1 + ''r''<sup>&thinsp;2</sup> we reach (''r''<sup>&thinsp;2</sup>)<sup>2</sup> + ''r''(''r''<sup>&thinsp;2</sup>) + 1 + ''r''<sup>&thinsp;2</sup> = ''r''<sup>&thinsp;4</sup> + ''r''<sup>&thinsp;3</sup> + 1 + ''r''<sup>&thinsp;2</sup> = 0, therefore ''x''<sup>3</sup> + ''x'' + 1 = (''x'' + ''r'')(''x'' + ''r''<sup>&thinsp;2</sup>)(''x'' + (''r'' + ''r''<sup>&thinsp;2</sup>)) for ''r'' in '''F'''<sub>2</sub>[''x'']/(''f''(''x'')); ''E'' = '''F'''<sub>2</sub>(''r'') is a splitting field of ''x''<sup>3</sup> + ''x'' + 1 over '''F'''<sub>2</sub>.
<!-- 
==See also==
* [deg 4 example]
-->

==Notes==
<references />

==References==
* Dummit, David S., and Foote, Richard M. (1999). ''Abstract Algebra'' (2nd ed.). New York: John Wiley & Sons, Inc. {{isbn|0-471-36857-1}}.
* {{springer|title=Splitting field of a polynomial|id=p/s086860}}
* {{MathWorld |title=Splitting field |urlname=SplittingField}}

{{DEFAULTSORT:Splitting Field}}
[[Category:Field (mathematics)]]