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Homeproof of primitive element theorem

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# 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 $\alpha\in K$ such that $K=F(\alpha)$ if and only if there are finitely many fields $L$ with $F\subseteq L\subseteq K$.

###### Proof.

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:

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

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 $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ be a basis for $K$ over $F$. Then $K=F(\alpha_{1},\ldots,\alpha_{n})$. 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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