12 Mar 2019  0

We look into the proof of the Primitive Element Theorem, reverse-engineer a logical reason of how to push forward with the proof.

Table of Contents


First, let’s have some preliminaries to the Primitive Element Theorem.

Separable Extension

We say that the extension KK of a field FF We usually write $K/F$ for $K$ being an extension of $F$. is separable if it is

  • algebraic: i.e. each of its elements is a root of some polynomial of F[x]F[x]; and
  • each αK\alpha \in K is separable, i.e. if p(x)p(x) is the minimal polynomial of α\alpha over FF, then p(x)p(x) completely factors in K[x]K[x].

Simple Extension

We say that the extension KK of a field FF is simple if it is expressible as K=F(α)K = F(\alpha) for some αK\alpha \in K, where F(α)F(\alpha) means adjoining α\alpha to the field FF, i.e

F(α)=spanF{1,α,,αn1},F(\alpha) = \text{span}_{F} \{ 1, \alpha, \ldots, \alpha^{n - 1} \},

where $n$ is the degree of the minimal polynomial of $\alpha$.

Finite Extension

We say that the extension KK over FF is finite if

K=F(α1,α2,,αn)=F(α1)(α2)...(αn).K = F(\alpha_1, \alpha_2, \ldots, \alpha_n) = F(\alpha_1)(\alpha_2)...(\alpha_n).

The Theorem

The Primitive Element Theorem is stated as follows:

If K/FK / F is finite and separable, then K/FK / F is simple.

A short statement, but important nonetheless. For example, it implies that finite extensions of perfect fields We say that a field FF is perfect if every one of its algebraic extension is separable. are simple, which helps us realize that,

  • fields of characteristic 0 A field has characteristic 0 is a field where no matter how many times we add 1 to 1, we will never get 0. always have simple extensions, and
  • any extension of a finite field is always just one algebraic number away.

If you skip ahead and try to read the clean proof, you will probably be stumped by the weirdly defined SS that came out of almost nowhere. You will also be stumped by how a k(x)k(x) is suddenly defined in terms of p(x)p(x). And why oh why did we think about choosing this extension LL in the first place? At first glance, these seem like a struck of genius, but these are choices selected well within careful reasoning that requires some actual hands-on into what we are given with.

The Proof

Planning for the proof

If we want F(α,β)=F(γ)F(\alpha, \beta) = F(\gamma) for some γ\gamma in some extension, one naive choice is to go with γ=α+uβ\gamma = \alpha + u \beta for some uF×u \in F^\times F×F^\times denotes the set of units of FF and hope that this will force α,βF(γ)\alpha, \beta \in F(\gamma). The argument is similar for either $\alpha$ or $\beta$ (by switching variables), so let’s think about only one of them. Let p(x)p(x) and q(x)q(x) be the minimal polynomials of α\alpha and β\beta, respectively. Now q(x)q(x) is not necessarily a minimal polynomial of $\beta$ over $F(\gamma)$, so let’s make use of that.

If βF(γ)\beta \in F(\gamma), then we must have xβq(x)x - \beta \mid q(x). So let’s consider the minimal polynomial h(x)h(x) of β\beta in F(γ)F(\gamma), which would divide q(x)q(x). Of course, ideally, we want h(x)=xβh(x) = x - \beta. Then let’s suppose that h(x)h(x) has some root other than β\beta.

Then in the splitting field of q(x)q(x), where h(x)h(x) must then also split, since q(x)q(x) is separable, we have that h(x)h(x) must therefore be able to split into linear terms, where each linear term has a root of q(x)q(x) as its constant value. In other words, all roots of h(x)h(x) are roots of q(x)q(x).

Then we notice another possible polynomial that such an h(x)h(x) can divide: we know that α=γuβ\alpha = \gamma - u \beta, and α\alpha is a root of p(x)p(x). Then if we let k(x)=p(γux)k(x) = p(\gamma - u x), we have

k(β)=p(γuβ)=p(α)=0.k(\beta) = p(\gamma - u \beta) = p(\alpha) = 0.

So h(x)k(x)h(x) \mid k(x). Now since all the roots of h(x)h(x) are roots of q(x)q(x), these roots must also be roots of k(x)k(x). Let these other roots of q(x)q(x) be labelled βj\beta_j’s. Then picking βjβ\beta_j \neq \beta, we have

k(βj)=0    p(γuβj)=0.k(\beta_j) = 0 \iff p(\gamma - u \beta_j) = 0.

We already know what the roots of p(x)p(x) are so let’s label those as αi\alpha_i. Then

γuβj=αi.\gamma - u \beta_j = \alpha_i.

Note that αiα\alpha_i \neq \alpha since the roots are unique. Following that,

α+uβuβj=αi,\alpha + u \beta - u \beta_j = \alpha_i,

which then

u=αiαββj.()u = \frac{\alpha_i - \alpha}{\beta - \beta_j}. \qquad (*)

We notice that there are only finitely many such uu’s in $F^\times$ since there are only as many as the roots αi\alpha_i’s and βj\beta_j’s can allow. However, F×F^\times is infinite by our assumption, i.e. there are always units of FF that cannot be expressed as in ()(*).

So by picking a uF×u \in F^\times that is not determined by ()(*), we rule out the possibility that k(x)k(x) has these other βj\beta_j’s as roots, and hence forcing $h(x)$ to be what we want: that is h(x)=xβh(x) = x - \beta. Thus our job is done for showing that βF(γ)\beta \in F(\gamma)!

We can then apply the same argument to showing that αF(γ)\alpha \in F(\gamma), by letting γ=β+uα\gamma = \beta + u' \alpha by choosing $u’$ in a similar fashion as above. In this case, we would have to extend our working field to the splitting field of p(x)p(x).

Then to put the two together, we could have then started working with an extension where both p(x)p(x) and q(x)q(x) splits, and the splitting field of p(x)q(x)p(x) q(x) is exactly where we should be working in.

The Clean Proof

If FF is finite, then KK itself is finite, and in particular it is generated by some αK\alpha \in K, i.e. K=αK = \langle \alpha \rangle. So K=F(α)K = F(\alpha). This is the easy case.

Suppose FF is infinite, which then implies that KK is infinite. Since the extension is finite, we may assume that

K=F(η1,η2,,ηn),K = F(\eta_1, \eta_2, \ldots, \eta_n),

where the ηi\eta_i’s are algebraic over FF. Note that it is sufficient for us to show that K=F(η1,η2)=F(γ)K = F(\eta_1, \eta_2) = F(\gamma) for some γK\gamma \in K, since we may then repeatedly apply the same argument for each of the adjoined elements. For simplicity, let’s write η1=α\eta_1 = \alpha and η2=β\eta_2 = \beta.

Now let LL be the splitting field of p(x)q(x)p(x) q(x) over KK. Let the roots of p(x)p(x) be

α=α1,α2,,αn,\alpha = \alpha_1, \, \alpha_2, \, \ldots, \, \alpha_n,

and the roots of q(x)q(x) be

β=β1,β2,,βm.\beta = \beta_1, \, \beta_2, \, \ldots, \, \beta_m.

By separability, αiαj\alpha_i \neq \alpha_j and βiβj\beta_i \neq \beta_j for all iji \neq j. Now let Note that had we wanted to start with γ=β+uα\gamma = \beta + u \alpha, we would have declared $S$ with elements like βjβ1α1αi\frac{\beta_j - \beta_1}{\alpha_1 - \alpha_i}.

S:={αiα1β1βj:1<in,1<jm}.S := \left\{ \frac{\alpha_i - \alpha_1}{\beta_1 - \beta_j} : 1 < i \leq n, \, 1 < j \leq m \right\}.

Since SS is finite while FF is infinite, uF×\exists u \in F^\times such that uSu \notin S. Let γ=α+uβ\gamma = \alpha + u \beta.

We now claim that F(α,β)=F(γ)F(\alpha, \beta) = F(\gamma). Clearly, F(γ)F(α,β)F(\gamma) \subseteq F(\alpha, \beta) since γF(α,β)\gamma \in F(\alpha, \beta). Let h(x)h(x) be the minimal polynomial of β\beta over F(γ)F(\gamma). Since q(β)=0q(\beta) = 0, we have that h(x)q(x)h(x) \mid q(x), and consequently if h(η)=0h(\eta) = 0, then η=βj\eta = \beta_j for some j{1,,m}j \in \{ 1, \ldots, m \}.

Now let k(x)=p(γux)F(γ)[x]k(x) = p(\gamma - ux) \in F(\gamma)[x]. Notice that

k(β)=p(γuβ)=p(α)=0.k(\beta) = p(\gamma - u \beta) = p(\alpha) = 0.

Thus h(x)k(x)h(x) \mid k(x). Notice that for j>1j > 1, we have

k(βj)=0    p(γuβj)=0    γuβj=αi for some i    α1+uβ1uβj=αi    u=αiα1β1βjS.\begin{aligned} k(\beta_j) = 0 &\iff p(\gamma - u \beta_j) = 0 \\ &\iff \gamma - u \beta_j = \alpha_i \text{ for some } i \\ &\iff \alpha_1 + u \beta_1 - u \beta_j = \alpha_i \\ &\iff u = \frac{\alpha_i - \alpha_1}{\beta_1 - \beta_j} \in S. \end{aligned}

Thus we know that for these jj’s, k(βj)0k(\beta_j) \neq 0 since we chose uSu \notin S.

It follows that h(βj)0h(\beta_j) \neq 0 for j>1j > 1, and so h(x)=xβF(γ)[x]h(x) = x - \beta \in F(\gamma)[x], implying that βF(γ)\beta \in F(\gamma). Using the same argument, we can show that αF(γ)\alpha \in F(\gamma). This completes the proof,



This is indeed a very profound result, making use of relatively simple notions such as minimal polynomials and splitting fields, and then proving for us a theorem that helps us narrow down the choice of the algebraic number to a single number that extends the base field to the extension.


I am usually not someone who is satisfied with just being told that a given solution works because it does. There are many times where the solution is right before us, just buried in some soil or covered with some thick layer of dust that it is not immediately clear why or how someone knew what was hidden behind it. Calling someone smart or a genius in these cases is somewhat off-putting, in that it glosses over their hard work and, perhaps, just having slightly more patience than we do.

- Japorized -


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