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

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

I'm trying to find a field F and a polynomial $f(x) \in F[x]$ with splitting field E such that $[E : F] \not = |Gal(E/F)|$. 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 $\mathbb{Z}_p(x)$ (the field of rational functions with elements of $\mathbb{Z}_p$ 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 $[E : F]$ or $|Gal(E/F)|$. Any help or hints would be much appreciated.

2. Originally Posted by mtdim
I'm trying to find a field F and a polynomial $f(x) \in F[x]$ with splitting field E such that $[E : F] \not = |Gal(E/F)|$. 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 $\mathbb{Z}_p(x)$ (the field of rational functions with elements of $\mathbb{Z}_p$ 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 $[E : F]$ or $|Gal(E/F)|$. Any help or hints would be much appreciated.
If you are looking for $E|F$ which is inseparable I cannot help you but if you are looking for $E|F$ such that $[E:F] \neq Gal(E|F)$ I have a solution.

If you take $\mathbb Q (\alpha )| \mathbb Q$, where $\alpha \in \mathbb R, \alpha ^3 = 2$ then you have an extension with $[E:F] = 3$ and $Gal(E|F) = 1$. This happens because the extension is not a normal extension.

3. Hi zoek,

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

If $f(x) \in F[x]$ is a separable polynomial and if $E/F$ is its splitting field, then $|Gal(E/F)|=[E:F]$.

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 $|Gal(E/F)|=1$ and $[E:F]=3$? 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 $|Gal(E/F)|>1$.

Thanks for the help.

4. in this example $E = \mathbb Q(\alpha)$ and this field is not a splitting field for the polynomial $f(x) = x^3 - 2$ (because the other two roots does not belong to $\mathbb Q(\alpha)$)

5. if $g \in Gal(\mathbb Q(\alpha)|\mathbb Q)$ then $g(x)=x$ for all $x\in \mathbb Q$ and $2= g(2)=g(\alpha^3) = g(\alpha)^3$, so we have $g(\alpha)=\alpha$, so finally $g(x)=x$ for all $x\in \mathbb Q(\alpha)$.

therefore $|Gal(\mathbb Q(\alpha)|\mathbb Q)|= 1$.

6. 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!

7. 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}.

8. Originally Posted by zoek
in this example $E = \mathbb Q(\alpha)$ and this field is not a splitting field for the polynomial $f(x) = x^3 - 2$ (because the other two roots does not belong to $\mathbb Q(\alpha)$)
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.

9. 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.