Thread: Galois Theory: when does [E : F] \= |Gal(E/F)| ?

I'm trying to find a field F and a polynomial with splitting field E such that . I know that f(x) must be inseparable for this to work, and so F must be infinite with characteristic p > 0. But apart from that, I'm stumped. The fields (the field of rational functions with elements of as coefficients) for some prime p are the only such fields I can think of, and I can't think of an example in that field that makes it easy for me to compute or . Any help or hints would be much appreciated.

Originally Posted by mtdim

I'm trying to find a field F and a polynomial with splitting field E such that . I know that f(x) must be inseparable for this to work, and so F must be infinite with characteristic p > 0. But apart from that, I'm stumped. The fields (the field of rational functions with elements of as coefficients) for some prime p are the only such fields I can think of, and I can't think of an example in that field that makes it easy for me to compute or . Any help or hints would be much appreciated.

If you are looking for which is inseparable I cannot help you but if you are looking for such that I have a solution.

If you take , where then you have an extension with and . This happens because the extension is not a normal extension.

Hi zoek,

This is confusing to me because I have a theorem in my textbook (Rotman, "Galois Theory") which says:

If is a separable polynomial and if is its splitting field, then .

I also have a theorem which says every field of characteristic 0 is perfect i.e. every polynomial is separable. How does this make sense in light of your example? How did you get that and ? It seems to me that a splitting field for your polynomial would contain both the real cube root of 2 and the two imaginary cube roots of 2, and so .

Thanks for the help.

in this example and this field is not a splitting field for the polynomial (because the other two roots does not belong to )

if then for all and , so we have , so finally for all .

therefore .

Originally Posted by mtdim

It seems to me that a splitting field for your polynomial would contain both the real cube root of 2 and the two imaginary cube roots of 2

this is absolutely right!

one thing that occurs to me is to set K = Z3(u), rational functions in u over Z3, and set f(x) = x^3 - u in K[x]. i believe f is irreducible over K, but if α is a cube root of u, then

f(x) = (x - α)^3. so E = K(α) is a splitting field of f over K, and [E:K] = 3. since any automorphism of E has to send α-->α, Gal(E/K) = {1}.

Originally Posted by zoek

in this example and this field is not a splitting field for the polynomial (because the other two roots does not belong to )

Ah, I see. That makes sense now; I thought you meant E to be the splitting field for that polynomial.

Your example isn't quite what I'm looking for; I want the field extension to specifically be the splitting field of the polynomial in question.

Originally Posted by Deveno

one thing that occurs to me is to set K = Z3(u), rational functions in u over Z3, and set f(x) = x^3 - u in K[x]. i believe f is irreducible over K, but if α is a cube root of u, then

f(x) = (x - α)^3. so E = K(α) is a splitting field of f over K, and [E:K] = 3. since any automorphism of E has to send α-->α, Gal(E/K) = {1}.

That's it! I was trying to use the field of rational functions over Z_p, but I couldn't come up with the polynomial that works. I'm quite certain you are correct about f being irreducible over K since there is no cube root of u in K.

Thanks.