# proof of primitive element theorem

###### Theorem.

Let $F$ and $K$ be arbitrary fields, and let $K$ be an extension^{} of $F$ of finite degree. Then there exists an element $\alpha \mathrm{\in}K$ such that $K\mathrm{=}F\mathit{}\mathrm{(}\alpha \mathrm{)}$ if and only if there are finitely many fields $L$ with $F\mathrm{\subseteq}L\mathrm{\subseteq}K$.

###### Proof.

Let $F$ and $K$ be fields, and let $[K:F]=n$ be finite.

Suppose first that $K=F(\alpha )$. Since $K/F$ is finite, $\alpha $ is algebraic over $F$. Let $m(x)$ be the minimal polynomial of $\alpha $ over $F$. Now, let $L$ be an intermediary field with $F\subseteq L\subseteq K$ and let ${m}^{\prime}(x)$ be the minimal polynomial of $\alpha $ over $L$. Also, let ${L}^{\prime}$ be the field generated by the coefficients of the polynomial ${m}^{\prime}(x)$. Thus, the minimal polynomial of $\alpha $ over ${L}^{\prime}$ is still ${m}^{\prime}(x)$ and ${L}^{\prime}\subseteq L$. By the properties of the minimal polynomial, and since $m(\alpha )=0$, we have a divisibility ${m}^{\prime}(x)|m(x)$, and so:

$$[K:L]=\mathrm{deg}({m}^{\prime}(x))=[K:{L}^{\prime}].$$ |

Since we know that ${L}^{\prime}\subseteq L$, this implies that ${L}^{\prime}=L$. Thus, this shows that each intermediary subfield^{} $F\subseteq L\subseteq K$ corresponds with the field of definition of a (monic) factor of $m(x)$. Since the polynomial $m(x)$ has only finitely many monic factors, we conclude that there can be only finitely many subfields of $K$ containing $F$.

Now suppose conversely that there are only finitely many such intermediary fields $L$. If $F$ is a finite field^{}, then so is $K$, and we have an explicit description of all such possibilities; all such extensions are generated by a single element. So assume $F$ (and therefore $K$) are infinite^{}. Let ${\alpha}_{1},{\alpha}_{2},\mathrm{\dots},{\alpha}_{n}$ be a basis for $K$ over $F$. Then $K=F({\alpha}_{1},\mathrm{\dots},{\alpha}_{n})$. So if we can show that any field extension generated by two elements is also generated by one element, we will be done: simply apply the result to the last two elements ${\alpha}_{j-1}$ and ${\alpha}_{j}$ repeatedly until only one is left.

So assume $K=F(\beta ,\gamma )$. Consider the set of elements $\{\beta +a\gamma \}$ for $a\in {F}^{\times}$. By assumption^{}, this set is infinite, but there are only finitely many fields intermediate between $K$ and $F$; so two values must generate the same extension $L$ of $F$, say $\beta +a\gamma $ and $\beta +b\gamma $. This field $L$ contains

$$\frac{(\beta +a\gamma )-(\beta +b\gamma )}{a-b}=\gamma $$ |

and

$$\frac{(\beta +a\gamma )/a-(\beta +b\gamma )/b}{1/a-1/b}=\beta $$ |

and so letting $\alpha =\beta +a\gamma $, we see that

$$F(\alpha )=L=F(\beta ,\gamma )=K.$$ |

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Title | proof of primitive element theorem |
---|---|

Canonical name | ProofOfPrimitiveElementTheorem |

Date of creation | 2013-03-22 14:16:27 |

Last modified on | 2013-03-22 14:16:27 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 8 |

Author | alozano (2414) |

Entry type | Proof |

Classification | msc 12F05 |