# Basic Proof

1. Apr 6, 2005

### Palindrom

Suppose k is separable over E. Prove that E(k)/E is separable.

A hint, if I may ask.

2. Apr 6, 2005

### Euclid

What are E, k, and E(k)/E? Can you define these please? I might be able to help.

3. Apr 6, 2005

### mathwonk

well what do you know about separability? it is related to the degree of the extension, and its connection with automorphisms.

if you want more than a hint, here is an extract from my book: since many symbols do, not translate well it is also something like a hint.

Now we recall the concept of "separable" field extension. (The terminology is due to Van der Waerden.)
Definition: A "separable" k-polynomial is one whose roots are all distinct, in any splitting field.

Remark: Since all splitting fields are k-isomorphic it does not matter which one we use, and since äk contains a splitting field, we can simply say f is separable iff its roots are distinct in äk.

Definition: An element of an extension of k is separable over k iff its minimal k-polynomial is separable. A field extension kfiL is separable iff every element of L is separable over k.

Definition: If L=k(å1,....,ån) is a finite extension of k, the "separable degree" of L over k, [L:k]s, is the number of k-homomorphisms ƒ:L¨äk, from L into an algebraic closure of k.

Although the terminology is new, the next result is the same as one proved last quarter, but it is worth reviewing the ideas.
Lemma: Assume L = k(å1,....,ån) is a finite extension of k. Then the degree of L over k equals the separable degree iff all the generators {å1,....,ån} are separable over k. If any åi is not separable over k, the separable degree is less than the degree.
proof: Let us compute the separable degree, by computing the number of k-homomorphisms ƒ:L¨äk. We can obtain every such homomorphism ƒ in stages, by starting with the identity map k¨kfiäk, extending it to a map k(å1)¨äk, then extending it further to a map k(å1,å2)¨äk, etc.....until we get a map k(å1,....,ån) = L¨äk. Moreover, if there are n1 ways to extend the identity map k¨kfiäk. to a map of k(å1)¨äk., and n2 ways to extend further to a map k(å1,å2)¨äk, then for each choice of an image of å1 there are n2 choices for the image of å2. Hence there are altogether n1n2 ways to define ƒ on å1,å2, hence n1n2 extensions of k¨k to k(å1,å2)¨äk. Reasoning in this way, the number of k-homomorphisms of L¨äk equals the product of the number of possible extensions at each stage. From our usual extension theory, we know the number of possible extensions to k(å1) is just the number of distinct roots in äk of the minimal polynomial g of å1 over k. This number equals the degree of g over k if g is separable, and otherwise is less than that degree. Moreover the degree of g over k equals the degree of the field extension [k(å1):k]. Hence the number of extensions to k(å1) is at most the degree [k(å1):k], and equals that degree iff å1 is separable over k. Similarly, the number of further extensions to k(å1,å2) is at most the degree of the field extension [k(å1,å2):k(å1)], and equals that degree iff å2 is separable over k(å1). In particular, since the minimal polynomial of å2 over k(å1) is a factor of the minimal polynomial over k, if å2 is separable over k, it is also separable over k(å1). Consequently the number of extensions from k=kfiL to k(å1,å2)¨L is at most the product of these degrees, [k(å1):k]\[k(å1,å2):k(å1)] = [k(å1,å2):k], (by multiplicativity of degrees of field extensions), and equals that product provided both å1,å2 are separable over k. If å1 is not separable over k, the number of such homomorphisms is less than the degree [k(å1,å2):k]. Continuing, we get that the number of k-homomorphisms from L¨äk, equals the degree [L:k] if every åi is separable over k, but is less if å1 is not separable overk. Since we can reorder the generators, any åi can be taken as å1. Thus the separable degree equals the degree iff all the åi are separable over k. QED.

Cor: A finite field extension kfiL=k(å1,....,ån) is separable over k iff every generator åi is separable over k.
proof: Since all åi belong to L, if L is separable then all åi are separable. Coversely, if all åi are separable over k, then the separable degree of L over k equals the degree. If some element ? of L is not separable over k, then we could use ? as a generator, i.e. L = k(?,å1,....ån), and the result just proved then would show that the seprable degree of L is less than the degree. This contradiction shows that there is no such element ?. QED.

Last edited: Apr 6, 2005
4. Apr 7, 2005

### Palindrom

Thanks a lot!