Editing Splitting field (section)

===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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