§ Primitive element theorem
- Let be a finite extension. We will characterize when a primitive element exists, and show that this will always happen for separable extensions.
§ Try 0: Naive attempt
- We try to find some element such that the multiplicative subgroup generated by , generate .
- However, to find such an element is our mandate!
- What we can do, however, is to take arbitrary elements of the form and so on, and try to write one of them in terms of the other. If we can show that and where are polynomials, then we have reduced everything to powers of .
- Now, clearly, if we can do this for two elements, ie, given , find a polynomial such that , we have won. So the question is in finding this .
§ Part 1: Primitive element iff number of intermediate subfields is finite
§ Forward: Finitely many intermediate subfields implies primitive
- If is a finite field, then is a finite extension and is a cyclic group. The generator of is the primitive element.
- So suppose is an infinite field. Let have many intermediate fields.
- Pick non-zero such that . We will generalize to arbitrarily many generators via recursion.
- As varies in , the extension varies amongst the extensions of .
- Since only has finitely many extensions while is infinite, pigeonhole tells us that there are two in such that .
- Define . We claim that , which shows that is a primitive element for .
- Since , this implies that .
- Thus, we find that and . Thus, . Since , we have , and thus , which implies .
- Thus .
- Done. Prove for more generators by recursion.
§ Backward: primitive implies finitely many intermediate subfields
- Let be a simple field (field generated by a primitive element). We need to show that only has finitely many subfields.
- Let be the minimal polynomial for in . By definition, is irreducible.
- For any intermediate field , define to be the minimal polynomial of in .
- Since is also a member of and share a common root and is irreducible in , this means that divides .
- Proof sketch that irreducible polynomial divides any polynomial it shares a root with (Also written in another blog post): The GCD must be non constant since share a root). But the irreducible polynomial cannot have a smaller polynomial ( ) as divisor. Thus the GCD itself is the irreducible polynomial . This implies that divides since GCD must divide .
- Since is a polynomial, it only has finitely many divisors (upto rescaling, which does not give us new intermediate fields).
- Thus, there are only finitely many intermediate fields if a field is primitive.
§ Interlude: finite extension with infinitely many subfields
- Let where are independent variables. This is an infinite field extension of , since each of the are independent.
- Recall that the equation does not help here, as that only tells us that upon evaluation, the polynomials agree at all points. However, they are still two different polynomials / rational functions.
- Consider the subfield . This too is an infinite field extension of .
- Now consider the tower , ie, . This is of finite degree, as we can only have of power upto . lies in the base field . Exactly the same with .
- So the extension has basis , and so has degree .
- We will first show that cannot be generated by a single element .
- Suppose we have . Then we must have that frobenius fixes the base field. Thus, frobenius sends to .
- Thus, if we now had that , this could not be, since by the previous argument, the extension has degree . But we previously saw that has degree at least . Thus, this field extension does not have a primitive element .
- We will now show that has infinitely many intermediate subfields.
- Pick elements of the form . We claim that are all different fields for different .
- Suppose for contradiction that for ( ). [ for contradictory extension ]
- This means that , or that , which implies that , since we know that , as must contain both and .
- Since , , and , we must have .
- This is a contradiction, since thus now means that , where , which makes a primitive element, somthing that we saw is impossible.
- The key idea is that the extension is generated by a minimal polynomial , which factorizes as . We lose the connection between minimality and irreducibility, which makes the extension inseparable, since the minimal polynomial now has repeated roots.
§ Part 2: If is finite and separable then it has a primitive element
- Let be separable for . Then we will show that there exists a primitive element such that .
- By repeated application, this shows that for any number of generators , we can find a primitive element.
- If is a finite field, then the generator of the cyclic group is a primitive element.
- So from now on, suppose is infinite, and for .
- Let be the minimal polynomial for , and the minimal polynomial for . Since the field is separable, have unique roots.
- Let the unique roots of be such that , and similarly let the unique roots of be such that .
- Now consider the equations for and .
- Rearranging, we get . Since and , this shows that there is a unique that solves the above equation.
- Since the extension is infinite, we can pick a which avoids the finite number of .
- Thus, once we choose such an , let . Such a can never be equal to for any , since the only choices of that make true are the , and was chosen to be different from these!
- Now let . Since , is a subfield of .
- See that .
- We will prove that .
- Let denote the minimal polynomial for over . Since , if is trivial, the .
- By definition, is a root of . Since is an irreducible over , we have that divides [proof sketch: irreducible polynomial shares a root with . Thus, must be linear or higher. Since divides , we must have as is irreducible and cannot have divisors. Thus, , being the GCD, also divides ].
- Thus, the roots of must be a subset of the roots of .
- Consider the polynomial . is also a root of the polynomial , since , which is equal to . [since is a root of ].
- Thus, we must have divides .
- We will show that is not a root of for . implies , which implies since the roots of are . But then we would have , a contradiction as was chosen precisely to avoid this case!
- Thus, every root of must come from . Also, the roots of must come from the roots of . But only shares the root with the set of roots . Also, does not have multiple roots since it is separable. Thus, is linear, and the degree of the field extension is 1. Therefore, .