# a finite extension of fields is an algebraic extension

###### Theorem 1.

Let $L\mathrm{/}K$ be a finite field extension. Then $L\mathrm{/}K$ is an algebraic
extension^{}.

###### Proof.

In order to prove that $L/K$ is an algebraic extension, we need to show that any element
$\alpha \in L$ is algebraic, i.e., there exists a non-zero
polynomial^{} $p(x)\in K[x]$ such that $p(\alpha )=0$.

Recall that $L/K$ is a finite extension of fields, by definition,
it means that $L$ is a finite dimensional vector space^{} over $K$.
Let the dimension^{} be

$$[L:K]=n$$ |

for some $n\in \mathbb{N}$.

Consider the following set of “vectors” in $L$:

$$\mathcal{S}=\{1,\alpha ,{\alpha}^{2},{\alpha}^{3},\mathrm{\dots},{\alpha}^{n}\}$$ |

Note that the cardinality of $S$ is $n+1$, one more than the dimension of the vector space. Therefore, the elements of $S$ must be linearly dependent over $K$, otherwise the dimension of $S$ would be greater than $n$. Hence, there exist ${k}_{i}\in K,\mathrm{\hspace{0.25em}0}\le i\le n$, not all zero, such that

$${k}_{0}+{k}_{1}\alpha +{k}_{2}{\alpha}^{2}+{k}_{3}{\alpha}^{3}+\mathrm{\dots}+{k}_{n}{\alpha}^{n}=0$$ |

Thus, if we define

$$p(X)={k}_{0}+{k}_{1}X+{k}_{2}{X}^{2}+{k}_{3}{X}^{3}+\mathrm{\dots}+{k}_{n}{X}^{n}$$ |

then $p(X)\in K[X]$ and $p(\alpha )=0$, as desired.

∎

NOTE: The converse is not true. See the entry “algebraic extension” for details.

Title | a finite extension of fields is an algebraic extension |
---|---|

Canonical name | AFiniteExtensionOfFieldsIsAnAlgebraicExtension |

Date of creation | 2013-03-22 13:57:30 |

Last modified on | 2013-03-22 13:57:30 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 6 |

Author | alozano (2414) |

Entry type | Theorem |

Classification | msc 12F05 |

Related topic | Algebraic |

Related topic | AlgebraicExtension |

Related topic | ProofOfTranscendentalRootTheorem |