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 $$\mathbb F_9$$).