Rupture field

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 $P(X)$.

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 $kin\; K$ and $f(Xmod\; P)=a$ is an isomorphism.

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 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 $jsqrt\; [3]\; 2$ and $j^2sqrt\; [3]\; 2$ where $j$ 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 not element of $mathbb\; F\_3$ with square equal to $-1$ (and all quadratic extensions of $mathbb\; F\_3$ are isomorphic to $mathbb\; F\_9$).

ee also

* Splitting field