Rupture field
In abstract algebra, a rupture field of a polynomial [math]\displaystyle{ P(X) }[/math] over a given field [math]\displaystyle{ K }[/math] is a field extension of [math]\displaystyle{ K }[/math] generated by a root [math]\displaystyle{ a }[/math] of [math]\displaystyle{ P(X) }[/math].[1]
For instance, if [math]\displaystyle{ K=\mathbb Q }[/math] and [math]\displaystyle{ P(X)=X^3-2 }[/math] then [math]\displaystyle{ \mathbb Q[\sqrt[3]2] }[/math] is a rupture field for [math]\displaystyle{ P(X) }[/math].
The notion is interesting mainly if [math]\displaystyle{ P(X) }[/math] is irreducible over [math]\displaystyle{ K }[/math]. In that case, all rupture fields of [math]\displaystyle{ P(X) }[/math] over [math]\displaystyle{ K }[/math] are isomorphic, non-canonically, to [math]\displaystyle{ K_P=K[X]/(P(X)) }[/math]: if [math]\displaystyle{ L=K[a] }[/math] where [math]\displaystyle{ a }[/math] is a root of [math]\displaystyle{ P(X) }[/math], then the ring homomorphism [math]\displaystyle{ f }[/math] defined by [math]\displaystyle{ f(k)=k }[/math] for all [math]\displaystyle{ k\in K }[/math] and [math]\displaystyle{ f(X\mod P)=a }[/math] is an isomorphism. Also, in this case the degree of the extension equals the degree of [math]\displaystyle{ P }[/math].
A rupture field of a polynomial does not necessarily contain all the roots of that polynomial: in the above example the field [math]\displaystyle{ \mathbb Q[\sqrt[3]2] }[/math] does not contain the other two (complex) roots of [math]\displaystyle{ P(X) }[/math] (namely [math]\displaystyle{ \omega\sqrt[3]2 }[/math] and [math]\displaystyle{ \omega^2\sqrt[3]2 }[/math] where [math]\displaystyle{ \omega }[/math] is a primitive cube root of unity). For a field containing all the roots of a polynomial, see Splitting field.
Examples
A rupture field of [math]\displaystyle{ X^2+1 }[/math] over [math]\displaystyle{ \mathbb R }[/math] is [math]\displaystyle{ \mathbb C }[/math]. It is also a splitting field.
The rupture field of [math]\displaystyle{ X^2+1 }[/math] over [math]\displaystyle{ \mathbb F_3 }[/math] is [math]\displaystyle{ \mathbb F_9 }[/math] since there is no element of [math]\displaystyle{ \mathbb F_3 }[/math] which squares to [math]\displaystyle{ -1 }[/math] (and all quadratic extensions of [math]\displaystyle{ \mathbb F_3 }[/math] are isomorphic to [math]\displaystyle{ \mathbb F_9 }[/math]).
See also
References
- ↑ Escofier, Jean-Paul (2001). Galois Theory. Springer. pp. 62. ISBN 0-387-98765-7. https://archive.org/details/galoistheorygrad00esco.
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