The Stacks project

Lemma 9.19.1 (Primitive element). Let $E/F$ be a finite extension of fields. The following are equivalent

  1. there exists a primitive element for $E$ over $F$, and

  2. there are finitely many subextensions $E/K/F$.

Moreover, (1) and (2) hold if $E/F$ is separable.

Proof. Let $\alpha \in E$ be a primitive element. Let $P$ be the minimal polynomial of $\alpha $ over $F$. Let $E \subset M$ be a splitting field for $P$ over $E$, so that $P(x) = (x - \alpha )(x - \alpha _2) \ldots (x - \alpha _ n)$ over $M$. For ease of notation we set $\alpha _1 = \alpha $. Next, let $E/K/F$ be a subextension. Let $Q$ be the minimal polynomial of $\alpha $ over $K$. Observe that $\deg (Q) = [E : K]$. Writing $Q = x^ d + \sum _{i < d} a_ i x^ i$ we claim that $K$ is equal to $L = F(a_0, \ldots , a_{d - 1})$. Indeed $\alpha $ has degree $d$ over $L$ and $L \subset K$. Hence $[E : L] = [E : K]$ and it follows that $[K : L] = 1$, i.e., $K = L$. Thus it suffices to show there are at most finitely many possibilities for the polynomial $Q$. This is clear because we have a factorization $P = QR$ in $K[x]$ in particular in $E[x]$. Since we have unique factorization in $E[x]$ there are at most finitely many monic factors of $P$ in $E[x]$.

If $F$ is a finite field (equivalently $E$ is a finite field), then $E/F$ has a primitive element by the discussion in Section 9.18. Next, assume $F$ is infinite and there are at most finitely many proper subfields $E/K/F$. List them, say $K_1, \ldots , K_ N$. Then each $K_ i \subset E$ is a proper sub $F$-vector space. As $F$ is infinite we can find a vector $\alpha \in E$ with $\alpha \not\in K_ i$ for all $i$ (a vector space can never be equal to a finite union of proper subvector spaces; details omitted). Then $\alpha $ is a primitive element for $E$ over $F$.

Having established the equivalence of (1) and (2) we now turn to the final statement of the lemma. Choose an algebraic closure $\overline{F}$ of $F$. Enumerate the elements $\sigma _1, \ldots , \sigma _ n \in \mathop{Mor}\nolimits _ F(E, \overline{F})$. Since $E/F$ is separable we have $n = [E : F]$ by Lemma 9.12.11. Note that if $i \not= j$, then

\[ V_{ij} = \mathop{\mathrm{Ker}}(\sigma _ i - \sigma _ j : E \longrightarrow \overline{F}) \]

is not equal to $E$. Hence arguing as in the preceding paragraph we can find $\alpha \in E$ with $\alpha \not\in V_{ij}$ for all $i \not= j$. It follows that $|\mathop{Mor}\nolimits _ F(F(\alpha ), \overline{F})| \geq n$. On the other hand $[F(\alpha ) : F] \leq [E : F]$. Hence equality by Lemma 9.12.11 and we conclude that $E = F(\alpha )$. $\square$


Comments (6)

Comment #4595 by David Tweedle on

In the proof of the lemma, the statement (in the second paragraph of the proof) "...(a finite union of proper subvector spaces is never a subvector space; details omitted)" is false. A counterexample is any proper subvector space of a vector space. It is sufficient for the lemma to know that a vector space can never be equal to a finite union of proper subspaces if is infinite.

Comment #4596 by on

@#4595 Yes, and just two sentences earlier we reduced to the case where is infinite. I think most people would read the parenthetical remark as applying to that case.

Comment #4597 by David Tweedle on

I'm sorry, either I am missing something else, or I did not express myself clearly. Let us assume is infinite (as in the proof of the lemma, I agree with you that most people would apply this to the parenthetical comment). This is false: a finite union of proper subvector spaces is never a subvector space. Counterexample: If are proper subvector spaces of , then is a subvector space as long as or . This is true (and I think this is what was intended): a vector space cannot be written as a finite union of proper subvector spaces.

Comment #4598 by on

Sorry, my bad. You are of course completely correct and you said it correctly the first time too. I will fix this when I next go through all the comments. Thanks!

Comment #4599 by David Tweedle on

No problem!


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