# a condition of algebraic extension

Theorem. A field extension $L/K$ is algebraic (http://planetmath.org/AlgebraicExtension) if and only if any subring of the extension field $L$ containing the base field^{} $K$ is a field.

Proof. Assume first that $L/K$ is algebraic. Let $R$ be a subring of $L$ containing $K$. For any non-zero element $r$ of $R$, naturally $K[r]\subseteq R$, and since $r$ is an algebraic element over $K$, the ring $K[r]$ coincides with the field $K(r)$. Therefore we have ${r}^{-1}\in K[r]\subseteq R$, and $R$ must be a field.

Assume then that each subring of $L$ which contains $K$ is a field. Let $a$ be any non-zero element of $L$. Accordingly, the subring $K[a]$ of $L$ contains $K$ and is a field. So we have ${a}^{-1}\in K[a]$. This means that there is a polynomial^{} $f(x)$ in the polynomial ring $K[x]$ such that ${a}^{-1}=f(a)$. Because $af(a)-1=0$, the element $a$ is a zero of the polynomial $xf(x)-1$ of $K[x]$, i.e. is algebraic over $K$. Thus every element of $L$ is algebraic over $K$.

## References

- 1 David M. Burton: A first course in rings and ideals. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).

Title | a condition of algebraic extension |
---|---|

Canonical name | AConditionOfAlgebraicExtension |

Date of creation | 2013-03-22 17:53:34 |

Last modified on | 2013-03-22 17:53:34 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 7 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 12F05 |

Related topic | RingAdjunction |

Related topic | FieldAdjunction |

Related topic | Overring |

Related topic | AConditionOfSimpleExtension |

Related topic | SteinitzTheoremOnFiniteFieldExtension |