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In abstract algebra, the splitting field of a polynomial P(X) over a given field K is a field extension L of K over which P factorizes ("splits", hence the name of a splitting field) into linear factors
- X − ai,
and such that the ai generate L over K. The extension L is then an extension of minimal 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).
For an example if K is the rational number field Q and
- P(X) = X3 − 2,
then a splitting field L will contain a primitive cube root of unity, as well as a cube root of 2. Thus
![{L=\mathbb{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\mathbb{Q} \}}](http://wpcontent.answers.com/math/0/7/3/073e6eeba95bf78889a99027f7ecf7df.png)
where
,
, and
are the cubic roots of unity.
An extension L which is the splitting field for multiple polynomials P(X) over K is called a normal extension.
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.
Therefore, for example, for K given as a subfield of the complex numbers, the existence is automatic. 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 is therefore proved directly to avoid 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 polynomials over K of elements a of K′.
Examples
- The splitting field of x2 + 1 over R, the real numbers, is C, the complex numbers.
- The splitting field of x2 + 1 over GF7 is GF49; as 7 is not equivalent to 1 (mod 4), −1 has no square root over GF7[1].
- The splitting field of x2 − 1 over GF7 is GF7 since x2 − 1 = (x + 1)(x − 1) already factors into linear factors.
See also
References
- ^ See also the diagonal of the multiplication table for GF7, computed by Wolfram Alpha.
- Dummit, David S., and Foote, Richard M. (1999). Abstract Algebra (2nd ed.). New York: John Wiley & Sons, Inc. ISBN 0-471-36857-1.
- Weisstein, Eric W., "Splitting field" from MathWorld.
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