Rupture field

In abstract algebra, a rupture field of a polynomial $P(X)$ over a given field $K$ is a field extension of $K$ generated by a root $a$ of $P(X)$.

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 $P$.

A 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 cube root of unity). For a field containing all the roots of a polynomial, see Splitting field.

Examples

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

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