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 $K=\mathbb{\mathbb{Q}}$ and $P(X)=X^3-2$ then $\mathbb{\mathbb{Q}}[\sqrt[3]{2}]$ 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 $f(XmodP)=a$ 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 $\mathbb{\mathbb{Q}}[\sqrt[3]{2}]$ 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 ${\mathbb{𝔽}}_9$).

See also

• Splitting field