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
- The Theorem
- The Proof
First, let’s have some preliminaries to the Primitive Element Theorem.
We say that the extension of a field 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 ; and
- each is separable, i.e. if is the minimal polynomial of over , then completely factors in .
We say that the extension of a field is simple if it is expressible as for some , where means adjoining to the field , i.e
where $n$ is the degree of the minimal polynomial of $\alpha$.
We say that the extension over is finite if
The Primitive Element Theorem is stated as follows:
If is finite and separable, then is simple.
A short statement, but important nonetheless. For example, it implies that finite extensions of perfect fields We say that a field 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 that came out of almost nowhere. You will also be stumped by how a is suddenly defined in terms of . And why oh why did we think about choosing this extension 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.
Planning for the proof
If we want for some in some extension, one naive choice is to go with for some denotes the set of units of and hope that this will force . The argument is similar for either $\alpha$ or $\beta$ (by switching variables), so let’s think about only one of them. Let and be the minimal polynomials of and , respectively. Now is not necessarily a minimal polynomial of $\beta$ over $F(\gamma)$, so let’s make use of that.
If , then we must have . So let’s consider the minimal polynomial of in , which would divide . Of course, ideally, we want . Then let’s suppose that has some root other than .
Then in the splitting field of , where must then also split, since is separable, we have that must therefore be able to split into linear terms, where each linear term has a root of as its constant value. In other words, all roots of are roots of .
Then we notice another possible polynomial that such an can divide: we know that , and is a root of . Then if we let , we have
So . Now since all the roots of are roots of , these roots must also be roots of . Let these other roots of be labelled ’s. Then picking , we have
We already know what the roots of are so let’s label those as . Then
Note that since the roots are unique. Following that,
We notice that there are only finitely many such ’s in $F^\times$ since there are only as many as the roots ’s and ’s can allow. However, is infinite by our assumption, i.e. there are always units of that cannot be expressed as in .
So by picking a that is not determined by , we rule out the possibility that has these other ’s as roots, and hence forcing $h(x)$ to be what we want: that is . Thus our job is done for showing that !
We can then apply the same argument to showing that , by letting 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 .
Then to put the two together, we could have then started working with an extension where both and splits, and the splitting field of is exactly where we should be working in.
The Clean Proof
If is finite, then itself is finite, and in particular it is generated by some , i.e. . So . This is the easy case.
Suppose is infinite, which then implies that is infinite. Since the extension is finite, we may assume that
where the ’s are algebraic over . Note that it is sufficient for us to show that for some , since we may then repeatedly apply the same argument for each of the adjoined elements. For simplicity, let’s write and .
Now let be the splitting field of over . Let the roots of be
and the roots of be
By separability, and for all . Now let Note that had we wanted to start with , we would have declared $S$ with elements like .
Since is finite while is infinite, such that . Let .
We now claim that . Clearly, since . Let be the minimal polynomial of over . Since , we have that , and consequently if , then for some .
Now let . Notice that
Thus . Notice that for , we have
Thus we know that for these ’s, since we chose .
It follows that for , and so , implying that . Using the same argument, we can show that . 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.