abstract-algebrafield-theorygalois-theory
I am self studying Fields and Galois Theory from Algebra by Hungerford and I am unable to think about how to prove (iii) of lemma 2.6 given on Page 246 of the book.
Adding it's image->
Defination that could be useful -> H' is fixed field.
I tried firstly by simply using the definition and then assuming some x belongs to L and then trying to prove x belongs to L" .
But I am not able to successfully use these.
A Sylow $p$-subgroup of a group $G$ has order $p^{n_p}$, where $p^{n_p}$ is the largest power of $p$ dividing $|G|$. In your setting, $$ |\mathrm{Aut}_{\Bbb{R}}\, F| = 2^n \cdot \text{[product of powers of odd primes]} \text{,} $$ so the index is a (possibly empty) product of powers of odd primes, so is odd.
As noted, the index of $E$ over $\Bbb{R}$ is the same as this index, so is odd.
This makes the minimal polynomial have odd degree. Hypothesis (B) says this polynomial has a root in $\Bbb{R}$, so no extension (by $u$) is needed and the root is real, so its minimal polynomial (over $\Bbb{R}$) is linear. (Every element of the base field is a root of a linear polynomial. Let $u$ be such an element; $x - u$ is the polynomial.)
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The previous paragraph ends with the assertion that $\mathrm{Aut}_{\Bbb{C}}\, F$ has order $2^m$, $0 \leq m$. This paragraph starts by assuming $m > 0$. There is a $2$-subgroup of order $2^{m-1}$ properly contained in a Sylow $2$-subgroup (of order $2^m$). The Fundamental theorem show a correspondence between the degrees of the fields (also called the dimensions of the extensions) and the indices of the Galois groups. So if the index of $J$ is $2$, the degree over the fixed field is also $2$.
The most fundamental thing that one can do in a field of characteristic zero, that one cannot do in a field of characteristic $p$, is divide by $p$. To briefly explain why, the operation of division assumes that the result of division exists : namely that if $a$ is an element of a field of characteristic zero, then there is another element of that field $b$, which exists and satisfies $bp = a$. Therefore, we write $\frac ap = b$ and can work with $\frac ap$ in arguments restricted to that field.
Why is the above true? Let $\mathbb F$ be a field of characteristic zero, and consider , for a prime $p$, the map $M_p : \mathbb F \to \mathbb F$ given by $M_p(x) = px$ , which is the element of $\mathbb F$ obtained by adding $x$ to itself $p$ times. Consider the image of $M_p$, the set $$S = \{x \in \mathbb F : \exists y \in \mathbb F, M_p(y) = x\}$$ We claim that $S$ is an ideal. To see this, note that $S$ is closed under addition (because $px+px' = p(x+x')$) and if $x= M_p(y) \in S, x' \in \mathbb F$ then $$xx' = M_p(y)x' = pyx' = p(yx') = M_p(yx')$$ hence $xx' \in S$. Therefore, $S$ is an ideal of $\mathbb F$, and it's non-zero , because $\mathbb F$ not being of characteristic $p$ implies that there is an element $a$ with $M_p(a) \neq 0$. But $\mathbb F$ has only two ideals, itself and $\{0\}$. It follows that $S = \mathbb F$ i.e. that $M_p$ is surjective. Thus, for all $a \in \mathbb F$ there is a $b \in \mathbb F$ such that $pb = a$.
Suppose we are in a field of characteristic $p$, and are trying to bring in an argument from characteristic zero, into characteristic $p$. It will go through, except at points where division by $p$ occurs, because in a field of characteristic $p$, I can't talk about $\frac ap$ for any $a \neq 0$ because multiplication by $p$ always yields zero.
The reason why characteristic $2$ is avoided in the theory of discriminants, is that in the most fundamental arguments involving discriminants, division by $2$ is a critically unavoidable step, which would go through in all characteristics except for it being $2$. That's also the reason why characteristic $2,3$ is avoided in the theory of cubic discriminants.
That's not to say that the discriminant itself wouldn't actually exist in characteristics $2$ : it would (because the existence of it is just based on the polynomial having all its roots in some extension and that's always the case), but it wouldn't give any useful information at all, compared to the other characteristics.
For example, in characteristic $2$ we know that $\sigma(\Delta) = +\Delta$ for even OR odd permutations $\sigma$ because $\Delta = -\Delta$ in characteristic $2$, so Proposition 4.5 (ii) is useless because the discriminant can't say anything about a permutation now.
Similarly, consider a quadratic equation of the form $ax^2+bx+c$. The roots of this are often specified using the formula $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Now, the division by $2$ cannot be performed in characteristic $2$ : what this means is that quadratic formulas need separate treatment in characteristic $2$, in particular some of the properties that a polynomial may have had in other fields, would not carry over to characteristic $2$ (e.g. separability : a lack of separability would make the discriminant zero and hence totally useless e.g. $X^2-d$ is not necessarily separable in characteristic $2$). Much more on characteristic $2$ can be found here.
In proposition 4.8, there is a statement in the proof that goes like :
... $u \in F$ is a root of $f$ if and only if $u + \frac b3$ is a root of $g = f(x - \frac b3)$ ...
Clearly this particular step involves division by $3$, and would not go through in characteristic $3$. Another way to see the importance of $2$ and $3$ is to observe the final discriminant formula which is $-4p^3 - 27q^2$ (let's imagine that we started out with $x^3+px+q$ instead of creating it from the standard form), and the important thing to observe here is that either in characteristic $2$ or $3$, some of the terms in the discriminant vanish, and this is how I instantly know that in these characteristics, something goes wrong.
The truth, in fact is that there's different ways to Cardano's formula to work out roots of a cubic equation in characteristics 2,3. That is detailed here and it's a fairly soft read.
The author doesn't talk about these, because these are such special cases that they need treatment that differs from fields of other characteristics. They will be avoided in a general treatment and considered only when the situation is relevant.
If the characteristic is $2$ or $3$, then the propositions don't hold as is : the roots will be characterized differently and the theory will be different.
The last one is fairly easy : if a polynomial has repeated roots, it is not separable, and therefore the Galois group corresponding to its splitting field is not necessarily given by a transitive subgroup of $S_n$. Furthermore, the Galois group isn't even of the same order as the degree of the polynomial anymore!
The transitive subgroups of $S_n$ are those that , for each $i,j \in \{1,2,...,n\}$ are such that for each $i,j$ there is an element of the subgroup taking $i$ to $j$. Transitive subgroups, for $n$ small, are extremely small in number : even for $n=4,5$ there are only five transitive subgroups in all. This means that Galois theory is vastly simplified for separable polynomials. Not insisting on separability would remove a huge constraint from the Galois group. The degree to size constraint being removed only continues to make it a bigger task to resolve. The best thing to do is to assume these things and go along the theory.
Best Answer
I think you may have not understood who $L''$ and $H''$ are. Let's look at $L''$ for example:
$L$ is an intermediate field, so $L'=\{\sigma \in Aut_KF: \ \sigma(v)=v \ \forall v \in L\}$ is now a subgroup of $Aut_{K}F$, and therefore $L''=(L')'$ is now of the form of $H'$ in part $(i)$ of Theorem 2.3 (and not of the form of $E'$ in part $(ii)$). Therefore:
$$L''=\{v \in F: \ \sigma(v)=v \ \forall \sigma \in L'\}$$
Now let $l\in L$. By definition, $\sigma(l)=l \ \forall \sigma \in L'$, so we conclude that $l\in L''$.