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en>Richard Arthur Norton (1958- ) |
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| In [[abstract algebra]], a '''splitting field''' of a [[polynomial]] with coefficients in a [[field (mathematics)|field]] is a smallest [[field extension]] of that field over which the polynomial ''splits'' or decomposes into [[linear factors]].
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| ==Definition==
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| 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
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| :<math>p(X) = \prod_{i=1}^{\deg(p)} (X - a_i) \in L[X]</math>
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| and such that the coefficients ''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 to be the [[Galois group]] of ''p'' (if we assume it is [[separable polynomial|separable]]).
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| ==Facts==
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| An extension ''L'' which is a splitting field for a set of polynomials ''p''(''X'') over ''K'' is called a [[Field_extension#Normal.2C_separable_and_Galois_extensions|normal extension]] of ''K''.
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| Given an [[algebraically closed field]] ''A'' containing ''K'', there is a unique splitting field ''L'' of ''p'' between ''K'' and ''A'', generated by the [[root of a function|root]]s of ''p''. If ''K'' is a subfield of the [[complex number]]s, the existence is immediate. On the other hand, the existence of algebraic closures in general is usually proved by 'passing to the limit' from the splitting field result, which therefore requires an independent proof to avoid [[Circular definition|circular reasoning]].
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| 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 polynomials over ''K'' of elements ''a'' of ''K''′.
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| ==Constructing splitting fields==
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| ===Motivation===
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| Finding [[Root of a function|roots]] of polynomials has been an important problem since the time of the ancient Greeks. Some polynomials, however, have no roots such as ''X''<sup>2</sup>+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.
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| ===The construction===
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| 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 <math>F=K_0, K_1, \ldots K_{r-1}, 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:
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| *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>.
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| *Choose any nonlinear irreducible factor ''f''(''X'') = ''f<sub>i</sub>''(''X'').
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| *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'')
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| *Repeat the process for ''K''<sub>''i''+1</sub> until ''p''(''X'') completely factors.
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| The irreducible factor ''f<sub>i</sub>'' 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.
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| Since ''f''(''X'') is irreducible, (''f''(''X'')) is a [[maximal ideal]] and hence ''K''<sub>''i''</sub>[''X'']/(''f''(''X'')) is, in fact, a field. Moreover, if we let <math>\pi : K_i[X] \to K_i[X]/(f(X))</math> be the natural projection of the ring onto its quotient then
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| :<math>f(\pi(X)) = \pi(f(X)) = f(X)\ \bmod\ f(X) = 0</math>
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| so π(''X'') is a root of ''f''(''X'') and of ''p''(''X'').
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| 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''!.
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| === The field ''K''<sub>''i''</sub>[''X'']/(''f''(''X'')) ===
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| 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
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| :<math>c_{n-1}\alpha^{n-1} + c_{n-2}\alpha^{n-2} + \cdots + c_1\alpha + c_0</math>
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| 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>''j''</sup> for {{nowrap|0 ≤ ''j'' ≤ ''n''−1}} form a basis.)
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| 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> the product ''g''(α)''h''(α) = ''r''(α) where ''r''(''X'') is the remainder of ''g''(''X'')''h''(''X'') divided by ''f''(''X'') in ''K<sub>i</sub>''[''X''].
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| 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''(α) = ''g''(α)''h''(α) directly. First let
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| :<math>f(X) = X^n + b_{n-1} X^{n-1} + \cdots + b_1 X + b_0.</math>
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| The polynomial is over a field so one can take ''f''(''X'') to be [[monic]] without loss of generality. Now α is a root of ''f''(''X''), so
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| :<math>\alpha^n = -(b_{n-1} \alpha^{n-1} + \cdots + b_1 \alpha + b_0).</math>
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| If the product ''g''(α)''h''(α) has a term α<sup>''m''</sup> with {{nowrap|''m'' ≥ ''n''}} it can be reduced as follows:
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| :<math>\alpha^n\alpha^{m-n} = -\left( b_{n-1} \alpha^{n-1} + \cdots + b_1 \alpha + b_0 \right) \alpha^{m-n} = -\left( b_{n-1} \alpha^{m-1} + \cdots + b_1 \alpha^{m-n+1} + b_0 \alpha^{m-n} \right) </math>.
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| As an example of the reduction rule, take ''K<sub>i</sub>'' = '''Q'''[''X''], the ring of polynomials with rational coefficients, and take ''f''(''X'') = ''X''<sup>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>7</sup> − 2). The reduction rule given by ''f''(''X'') is α<sup>7</sup> = 2 so
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| :<math>g(\alpha) h(\alpha) = \left(\alpha^5 + \alpha^2\right) \left(\alpha^3 + 1\right) = \alpha^8 + 2 \alpha^5 + \alpha^2 = \left(\alpha^7\right) \alpha + 2\alpha^5 + \alpha^2 = 2 \alpha^5 + \alpha^2 + 2\alpha.</math>
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| ==Examples==
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| === The complex numbers ===
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| Consider the [[polynomial ring]] '''R'''[''x''], and the [[irreducible polynomial]] {{nowrap|1=''x''<sup>2</sup> + 1.}} The [[quotient ring]] {{nowrap|1='''R'''[''x''] / (''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''] / (''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''.}}
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| 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:
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| :<math> (a_1 + b_1x) + (a_2 + b_2x) = (a_1 + a_2) + (b_1 + b_2)x, </math>
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| :<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>
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| If we identify {{nowrap|1=''a'' + ''bx''}} with (''a'',''b'') then we see that addition and multiplication are given by
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| :<math> (a_1,b_1) + (a_2,b_2) = (a_1 + a_2,b_1 + b_2), </math>
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| :<math> (a_1,b_1)\cdot (a_2,b_2) = (a_1a_2 - b_1b_2,a_1b_2 + b_1a_2). </math>
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| We claim that, as a field, the quotient {{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'' + i''b''}}, where ''a'' and ''b'' are real numbers and {{nowrap|1=i<sup>2</sup> = −1.}} Addition and multiplication are given by
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| :<math> (a_1 + ib_1) + (a_2 + ib_2) = (a_1 + a_2) + i(b_1 + b_2), </math>
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| :<math> (a_1 + ib_1) \cdot (a_2 + ib_2) = (a_1a_2 - b_1b_2) + i(a_1b_2 + a_2b_1). </math>
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| If we identify {{nowrap|1=''a'' + i''b''}} with (''a'',''b'') then we see that addition and multiplication are given by
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| :<math> (a_1,b_1) + (a_2,b_2) = (a_1 + a_2,b_1 + b_2), </math>
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| :<math> (a_1,b_1)\cdot (a_2,b_2) = (a_1a_2 - b_1b_2,a_1b_2 + b_1a_2) \, . </math>
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| The previous calculations show that addition and multiplication behave the same way in {{nowrap|1='''R'''[''x''] / (''x''<sup>2</sup> + 1)}} and '''C'''. In fact, we see that the map between {{nowrap|1='''R'''[''x'']/(''x''<sup>2</sup> + 1)}} and '''C''' given by {{nowrap|1=''a'' + ''bx'' → ''a'' + i''b''}} is a [[homomorphism]] with respect to addition ''and'' multiplication. It is also obvious that the map {{nowrap|1=''a'' + ''bx'' → ''a'' + i''b''}} is both [[injective]] and [[surjective]]; meaning that {{nowrap|1=''a'' + ''bx'' → ''a'' + i''b''}} is a [[bijective]] homomorphism, i.e. an isomorphism. It follows that, as claimed: {{nowrap|1='''R'''[''x''] / (''x''<sup>2</sup> + 1) ≅ '''C'''.}}
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| === Cubic example ===
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| Let ''K'' be the [[rational number field]] '''Q''' and
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| :''p''(''X'') = ''X''<sup>3</sup> − 2.
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| Each root of ''p'' equals <math>\sqrt[3]{2}</math> times a [[cube root of unity]]. Therefore, if we denote the cube roots of unity by
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| :<math>\omega_1 = 1 \,</math>, <!-- do not delete "\,": it improves the display of formula on certain browsers. -->
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| :<math>\omega_2 = - \frac{1} {2} + \frac {\sqrt{3}} {2} i, </math>
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| :<math>\omega_3 = - \frac{1} {2} - \frac {\sqrt{3}} {2} i. </math>
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| any field containing two distinct roots of ''p'' will contain the quotient between two distinct cube roots of unity. Such a quotient is a primitive cube root of unity—either ω<sub>2</sub> or <math>\omega_3=1/\omega_2</math>). It follows that a splitting field ''L'' of ''p'' will contain ω<sub>2</sub>, as well as the real [[cube root]] of 2; conversely, any extension of '''Q''' containing these elements contains all the roots of ''p''. Thus
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| :<math>{L=\mathbf{Q}(\sqrt[3]{2},\omega_2)=\{a+b \omega_2+c\sqrt[3]{2} +d \sqrt[3]{2} \omega_2+ e \sqrt[3]{2^2} + f \sqrt[3]{2^2} \omega_2 \,|\,a,b,c,d,e,f\in\mathbf{Q} \}}</math>
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| ===Other examples===
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| *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 there, because 7 is not equivalent to 1 (mod 4).<ref>Instead of applying this characterization of odd 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≡6.</ref>
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| *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 factors into linear factors.
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| *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 is two. Comparison of coefficients shows that ''a'' = ''r'' and ''b'' = 1 + ''r''<sup>2</sup>. The elements of '''F'''<sub>2</sub>(''r'') can be listed as ''c'' + ''dr'' + ''er''<sup>2</sup>, where ''c'', ''d'', ''e'' are in '''F'''<sub>2</sub>. There are eight elements: 0, 1, ''r'', 1 + ''r'', ''r''<sup>2</sup>, 1 + ''r''<sup>2</sup>, ''r'' + ''r''<sup>2</sup> and 1 + ''r'' + ''r''<sup>2</sup>. Substituting these in ''x''<sup>2</sup> + ''rx'' + 1 + ''r''<sup>2</sup> we reach (''r''<sup>2</sup>)<sup>2</sup> + ''r''(''r''<sup>2</sup>) + 1 + ''r''<sup>2</sup> = ''r''<sup>4</sup> + ''r''<sup>3</sup> + 1 + ''r''<sup>2</sup> = 0, therefore ''x''<sup>3</sup> + ''x'' + 1 = (''x'' + r)(''x'' + ''r''<sup>2</sup>)(''x'' + (''r'' + ''r''<sup>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==
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| * [[Rupture field]]
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| ==References==
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| <references/>
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| * Dummit, David S., and Foote, Richard M. (1999). ''Abstract Algebra'' (2nd ed.). New York: John Wiley & Sons, Inc. ISBN 0-471-36857-1.
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| * {{springer|title=Splitting field of a polynomial|id=p/s086860}}
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| * {{MathWorld |title=Splitting field |urlname=SplittingField}}
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| {{DEFAULTSORT:Splitting Field}}
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| [[Category:Field theory]]
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| [[de:Körpererweiterung#Zerfällungskörper]]
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