Rupture field

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In abstract algebra, a rupture field of a polynomial $P(X)$ over a given field $K$ such that $P(X)\in K[X]$ is the field extension of $K$ generated by a root $a$ of $P(X)$ .^[1]

For instance, if $K={\mathbb Q}$ and $P(X)=X^{3}-2$ then ${\mathbb Q}[{\sqrt[ {3}]2}]$ is a rupture field for $P(X)$ .

The notion is interesting mainly if $P(X)$ is irreducible over $K$ . In that case, all rupture fields of $P(X)$ over $K$ are isomorphic, non canonically, to $K_{P}=K[X]/(P(X))$ : if $L=K[a]$ where $a$ is a root of $P(X)$ , then the ring homomorphism $f$ defined by $f(k)=k$ for all $k\in K$ and $f(X\mod P)=a$ is an isomorphism. Also, in this case the degree of the extension equals the degree of $P$ .

The rupture field of a polynomial does not necessarily contain all the roots of that polynomial: in the above example the field ${\mathbb Q}[{\sqrt[ {3}]2}]$ does not contain the other two (complex) roots of $P(X)$ (namely $\omega {\sqrt[ {3}]2}$ and $\omega ^{2}{\sqrt[ {3}]2}$ where $\omega$ is a primitive third root of unity). For a field containing all the roots of a polynomial, see the splitting field.

Examples

The rupture field of $X^{2}+1$ over ${\mathbb R}$ is ${\mathbb C}$ . It is also its splitting field.

The rupture field of $X^{2}+1$ over ${\mathbb F}_{3}$ is ${\mathbb F}_{9}$ since there is no element of ${\mathbb F}_{3}$ with square equal to $-1$ (and all quadratic extensions of ${\mathbb F}_{3}$ are isomorphic to ${\mathbb F}_{9}$ ).

References

↑ Escofier, Jean-Paul (2001). Galois Theory. Springer. p. 62. ISBN 0-387-98765-7.

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Rupture field

Examples

See also

References