Re: Galois Theory Problem

In article <1118262411.489447.144130@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<jamesdickerson00@xxxxxxxxxxx> wrote:
>Arturo Magidin wrote:
>> In article <1118165025.779384.174470@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
>> <jamesdickerson00@xxxxxxxxxxx> wrote:
>> >Let E be a finite degree normal extension of F and let K and L be
>> >fields between E and F. Suppose that E is separable over both K and L.
>> >Prove that E is separable over (K intersect L).
>>
>> No need to post it twice. Have some patience.
>>
>> What have you done so far, and what are you having trouble with?
>
>Sorry about posting it twice. I'm just not sure whether or not it's
>right and even if correct how prove this without using the next
>chapter. Here is what I have:
>
>First 2 theorems:
>
>1. Let F be a subfield of K and f belong to F[x]. Then f has distinct
>roots as a member of F[x] if and only if it has distinct roots as a
>member of K[x].

I'm not sure what this means. Certainly "f has distinct roots as a
member of F[x]" does ->not<- mean "distinct roots in F", because we
can take a purely inseparable polynomial with no roots in F and let K
be its splitting field.

So what does it mean? Does it mean "distinct roots in some algebraic
closure of F"?

>*2. Let F be a subfield of L and L be a subfield of E where E is
>separable over L and L is separable over F. THen E is separable over
>F. (* this is from the next chapter).
>
>Proof: By assumption for all f in E, f is separable over K. Also f is
>separable over L. By Theorem 1, for any f in (K intersect L)[x] f has
>distinct roots as a member of K[x]

This makes no sense. f was an element of E, now you are talking about
f as if it were a polynomial. Perhaps you should not use "f" to denote
elements.

>implies that f has distint roots as
>a member of (K intersect L)[x].

This is nonsense.

> Also by Theorem 1, f has distinct
>roots as a member of L[x] implies that f has distinct roots as a
>member of (K intersect L)[x].

This is also nonsense. In general, the minimal polynomial for your
element over K will DIVIDE the minimal polynomial over (K intersect
L), but only in K[x]. The minimal over K need not lie in M[x], where
M=(K intersect L).

You need to start over.

--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
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Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.