§ 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.
§ 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 . 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 has finite degree since it has a vector space basis .
- Let be roots of and . Define .
- Consider intermediate subfields for .
- Suppose for two elements in . We want to show that . This gives us infinitely many subfields as has infinitely many elements. TODO
§ 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, .