MHB Normal Extensions and Normal Subgroups

[Kiwi1]

View attachment 5929

I have answered Q1-Q3. I am unsure of my answer for Q4 and am stuck on Q5.

Notation: \(I^*=Gal(K:I)\). That is the subgroup of G=Gal(K:F) that fixes I.

Q4.
I2 is conjugate to I1.
iff \(\exists i \in G:I_2=i(I_1)\)
iff \( I^*_2=[i(I_1)]^*=iI_1^*i^{-1}\) by Q3.
iff \( I^*_2=iI_1^*i^{-1}\)
iff \( I^*_2=\phi_i(I_1^*)\) where \(\phi_i\) is the function conjugate by i.
iff \(I_1^*\) is conjugate to \(I_2^*\)

Q5. I have lots of ideas but keep going in circles.
Suppose \(I_2\) is a normal extension of \(I_1\) then

\(I_2\) is the root field of some a(x) over \(I_1\)

and we have the isomorphism \(\phi \) defined by:

\(I_2 \cong \frac{I_1[x]}{<a(x)>}\)

Now I gloss over the fact that phi is an isomorphism when I need an automorphism. I take phi to mean that I_1 and I_2 are conjugate. Therefore using Q4 \(I^*_2\) and the fixer of the RHS of my isomorphism are conjugate. That is there exists an automorphism pi such that:

\(I_2^* \cong \left(\frac{I_1[x]}{<a(x)>}\right)^*\)

again I am glossing over the difference between isomorphism and automorphism.

Now since these two fixers are conjugate I can write

there exists \(x \in I_1\) such that \(I^*_2 \cong x \left(\frac{I_1[x]}{<a(x)>}\right)^* x^{-1}\)

Now x and \(x^{-1}\) both fix the coefficients of a(x) and \(I_1^*[x]\) so:

there exists \(x \in I_1\) such that \(I^*_2 \cong \left(\frac{I_1[x]}{<a(x)>}\right)^* \)

So \(I^*_2\) is a normal subgroup of \(I^*_1\)

I know I have 'cheated' at various points but am I on the right track to finding a solution? I need to reverse the order of my argument so that this argument complements Q1.

 

[Kiwi1]

Solved it!

Suppose \(I_2\) is a normal extension of \(I_1\). Then there exists a(x) such that:

\(I_2 \cong \frac{I_1[x]}{<a(x)>} \supseteq I_1\). Now this isomorphism fixes \(I_1\) so by Theorem (i) the isomorphism is in fact an automorphism.

So by problem 4:
\(I_2^* \cong \left(\frac{I_1[x]}{<a(x)>}\right)^* \subseteq I_1^*\) is an automorphism so:

\(I_2^* \subseteq I_1^*\) and \(I_2^* = \left(\frac{I_1[x]}{<a(x)>}\right)^*\)

but is \(I_2^*\) a NORMAL subgroup?

Let \(\pi\) be any \(\pi \in I_1^*\) then

\(\pi I_2^* \pi^{-1} = \pi \left(\frac{I_1[x]}{<a(x)>}\right)^* \pi^{-1}= \left( \pi \left(\frac{I_1[x]}{<a(x)>}\right)\right)^* \) (using problem 3)

but \(\pi\) fixes all \(I_1\) and therefore fixes the coefficients of \(I_1[x]\) and a(x), therefore:

\(\pi I_2^* \pi^{-1} = \left(\frac{I_1[x]}{<a(x)>}\right)^* = I_2^*\)

so indeed \(I_2^*\) a NORMAL subgroup of \(I_1^*\). The reverse direction follows directly from Question 1.