§ Separable Polynomials and extensions
§ Separable polynomial
- I didn't really study galois theory over char. all that well the first time I studied it, so let's review.
- Let be an inclusion of fields, so is a field extension of .
- An irreducible polynomial is separable iff it has distinct roots in the algebraic closure of , .
- Said differently, the polynomial has no repeated roots in any extension of .
- Said differently, the polynomial has distinct roots in its splitting field over . The roots as separable since we can separate all the roots from each other --- they are all distinct.
- Said differently, the polynomial derivative of is not the zero polynomial.
§ Proof that is not separable iff share a root
§ Forward: is not separable implies do not share a root.
- Let have a repeated root . Thus in .
- Computing by product rule, we see that it is which can be written as .
- This shows that has as a root.
§ Backward: share a root implies is not separable
- Let be such that .
- Write for roots .
- Let [WLOG ].
- We know by product rule of calculus that .
- Computing , only the first term survives, which is [all other terms have an term which vanishes.
- For this to vanish, we must have some such that .
- This implies that has a repeated root and is thus not separable.
§ Proof that is separable iff
§ Forward: is separable implies
- Let .
- Suppose that it is not a unit, and not a constant (ie, a real polynomial).
- Then has a root (by previous)
- Since divides , this means that in , since and .
- This implies that is a repeated root of since it vanishes at both and its derivative!
- This contradicts is separable.
- Thus must be a unit, and , are relatively prime.
§ Backward: implies is separable
- Since , there are polynomials such that .
- Suppose is not separable. Thus it has a repeated root . This means that and vanish at . Thus divides '.
- Thus means that divides , by the eqn . This is absurd, and thus is separable.
§ Separable extension
- For all elements where , there is a polynomial such that is separable .
- Thus, all elements have separable polynomials that they are roots of.
§ Separable extension is transitive
- Claim: If and are separable field extensions, then is separable.
§ All Polynomials over character 0 is separable
- Let and . Then we claim that is separable.
- Let be an irreducible polynomial in . (the minimal polynomial of some element in )
- Recall that a polynomial is irreducible over a field iff it cannot be written as the product of two non-constant polynomials.
- We wish to show that is separable (has no repeated roots).
- For contradiction, suppose that has a repeated root in the algebraic closure. so for , .
- Thus, and share a common factor in .
- But the GCD algorithm works in , thus and share a common factor in [SID: I find this dubious! ]
- Hence, this means that is not a constant polynomial.
- If , then can be factored, which contradicts its irreducibility.
- Thus, [to prevent contradiction ].
- However, has smaller degree than . Thus the only way it can be divided by its GCD ( ) which has larger degree than it is if .
- This means (in characteristic zero) that is linear, and thus cannot have repeated roots!
- That this means that is zero can be seen by computing te derivative. Suppose with . Then . Since , , and thus the derivative of an th degree polynomial is degree.
§ All Polynomials over character 0 is separable, alternative proof.
- Let be an irreducible polynomial in (the minimal polynomial of some element in ). We claim that is separable.
- The key lemma is to show that if is ANY polynomial which shares a root with , then .
- Idea: since , this means that divides . Thus, is non-constant.
- Further, since the gcd divides both its factors.
- But since divides while is irreducible, we must have equals .
- Since divides , we have .
- Now, going back to our claim, let be some irreducible in . Suppose for contradiction that is not separable. Then share a common root. By the above lemma, this implies that divides . But this is absurd, since .
- Hence, no irreducible polynomial in can share a root with its derivative, which implies is always separable.
- This breaks down for character since can simply "die out".
§ All finite field extensions over character 0 is separable
- Can write any finite field extension as . This is the same as .
- Since separability of field extensions is transitive, and at each step, we add an element with separable minimal polynomial (all polynomials over char. 0 are separable), the full extension is separable.
§ All field extensions over character is separable
- Consider .
- build the "Fermat's little theorem" polynomial .
- All elements of satisfy this, thus has roots, which means all of its roots are distinct.
- Alternatively, see that so and don't share a commmon root.
§ Purely inseparable extensions
- of char. p. Then the field is purely inseparable iff any is a root of for some , with .
- Over the algebraic closure, this can be written as which over finite fields factorizes as by freshman's dream .
- Thus, over the algebraic closure, this has many copies of one root, which is "as far as possible" from being separable.
§ Breaking down extension into separable + purely inseparable
- Given any algebraic, can break it down into .
- The extension contains all elements of which have separable polynomials. Can show that this is a field.
- We can show that will be purely inseparable.
§ Example of inseparable extension
- Let , which are rational functions in over , and which are rational functions in .
- Clearly, , and the extension is of degree , because is a root of .
- See that is irreducible over .
- Over , factorizes as by freshman's dream , with all roots the same.
- Thus, we have an element whose minimal polynomial is not separable. Here, the function has derivative zero, and is thus inseparable.
- In some sense, the failure is because of freshman's dream, where .
- Reference: Borcherds, separable extension
§ Primitive element theorem / Theorem of the primitive element
- Let be a field extesion. We say is primitive for the extension (the extension) if .
- If is a finite separable extension, then for some , we have .
- Recall that .
§ Tensor product of field extensions
- Let be a finite separable extension of and be an arbitrary extension of . (usually, is the p-adics, is , is a number field).
- Then, is a product of finite separable extensions of . So , where each is a finite extension of .
- If is a primitive element for the extension (so ), then the image of in is a primitive element for over .
- If is the minimal poly. for over and is the minimal polynomial for over then .
- Proof : Start with a primitive element for . Then for the minimal polynomial of over . So witnesses this isomorphism.
- Consider . This is isomorphic to . Call the map that sends the LHS to the RHS as
- We claim that the ring is isomorphic to . Intuition: tensoring by doesn't do anything useful, and we can re-interpret as living in . The isomorphism is .
- Now suppose factors as over . Since is separable over and is an extension of , all the are distinct (otherwise it contradicts separability). Thus the family of ideals is pairwise coprime.