Editing Splitting field (section)

==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>.
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==See also==
* [deg 4 example]
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