Dtyjlui

Wrapping up Charles C. Pinter's "Abstract Algebra", having been introduced to the very basics of Galois theory in the previous chapter (fundamental theorem and a few other results), I'm finding myself very confused by the argument presented in the final chapter that's dedicated to the radical solutions to polynomials.

The author starts off by defining a radical extension over a field $F$ as an extension $F(c_1,...,c_n)$ such that for each $i$, there exists a (non-zero) integer, k, such that $c_i^kin F(c_1,...,c_{i-1})$.

He then establishes that every radical extension over a field that contains the needed roots of unity is an abelian extension.

The key part I'm failing to get is what seems to be a proof that every radical extension over a field, $F$, (that contains its roots of unity) is a splitting field (referred to in the book as a root field) over $F$?

This is what the author says (it is assumed $F$ already has all the relevant roots of unity):

Thus if $K=F(c_1,...,c_n)$ is a radical extension of $F$:

$qquadunderline Fsubsetequnderline {F(c_1)}subsetequnderline {F(c_1,c_2)}subseteq ...subsetequnderline {F(c_1,...,c_n)} qquadqquad$ (2)
$qquad I_0qquad I_1 qquadqquad I_2 qquadqquad;;;; I_m=K$

is a sequence of simple abelian extensions. (The extensions are all abelian by the comments in the preceding three paragraphs.)
$quad$Still, this is not quite enough for our purposes: In order to use the machinery that was set up in the previous chapter, we must be able to say that each field in (2) is a root field of $F$.

At this point, it seems to me like the author is going to provide a proof that every radical extension over $F$ is also splitting field over $F$.

The above interpretation of what Pinter has said may be where my misconception in fact lies, because I can't see how the argument provided ahead is strong enough to go all the way to showing this (that every radical extension is a splitting field, I mean).

(my misunderstanding in what follows-the argument presented (seems to be) essentially induction over the fields $I_q$ to show that they're all root fields over $F$. But the way he wraps things up at the end makes it seem like we weren't even supposed to assume that $I_{q+1}$ was an extension field to start with)

Here's the argument:

This may be accomplished as follows. Suppose we have already constructed the extensions $I_0subseteq I_1subseteq...subseteq I_q$ in (2) so that $I_q$ is a root field over $F$. We must extend $I_q$ to $I_{q+1}$ so that $I_{q+1}$ is a root field over $F$. Also, $I_{q+1}$ must include the element $c_{q+1}$ which is the $n$th root of some element $ain I_q$.
$quad$Let $H={h_1,...,h_r}$ be the group of all the $F$-fixing automorphisms of $I_q$, and consider the polynomial:

$qquad b(x)=[x^n-h_1(a)][x^n-h_2(a)]...[x^n-h_r(a)]$

In what follows, the author basically uses the fact that any isomorphism $f:K_1rightarrow K_2$ can be extended to an isomorphism $bar f:K_1[X]rightarrow K_2[X]$ where

$bar f(a_0+a_1x+...a_sx^s)=f(a_0)+f(a_1)x+...+f(a_s)x^s$.

So (for any i):

$bar h_i(b(x))=bar h_i[x^n-h_1(a)]bar h_i[x^n-h_2(a)]...bar h_i[x^n-h_r(a)]\=[x-h_i(h_1(a)][x-h_i(h_2(a)]...[x-h_i(h_r(a)]$

But, as $h_icirc :Hrightarrow H$ (for any i) is a bijection, the terms of the form $[x-h_k(a)]$ are all just permuted, so $bar h_i(b(x))=b(x)$. So (given $bar h_i$'s definition) each $h_i$ fixes each coefficient of $b(x)$- every coefficient of $b(x)$ is then in the fixfield of $H$, i.e. $F$ (by the fundamental theorem) so $b(x)in F[X]$.

Then, the author finishes:

We now define $I_{q+1}$ to be the root field of $b(x)$ over $F$. Since all the roots of $b(x)$ are $n$th roots of elements in $I_q$, it follows that $I_{q+1}$ is a radical extension of $I_q$. The roots may be adjoined one by one, yielding a succession of abelian extensions, as discussed previously. To conclude, we may assume in (2) that $K$ is a root field over $F$.

Yeah, so essentially, how do we know that $I_{q+1}$ is actually the $q+1$th radical extension in (2), is my question? It's a root field for sure and it contains $I_q$ and $c_{k+1}$, but what it seems like we've shown is that a splitting field containing a splitting field (that happens to be a radical extension) is also a radical extension rather than what I assumed we were trying to prove (that each radical extension is a splitting field). Or is that not what's happening? Am I missing something obvious or am I just completely looking at this the wrong way?

I know this is a lot for a question, but I'd really appreciate any and all help.

(If there's anything unclear here, please leave a comment.)

In the argument, the extensions $I_q$ are being defined recursively $-$ he defines $I_{q+1}$ using the definition of $I_q$ and then proves that $I_{q+1}$ is indeed a root field using what he (inductively) knows about $I_q$. In the end, we want $I_q = F(c_1,dots,c_q)$, but he is not assuming that in the proof; it is part of what he proves about $I_{q+1}$ in the inductive step. I don't really like this proof, so let me modify it a little, defining the $I_q$ from the start.

It seems like you are unsure about the inductive step, or at least the part where we show that $I_{q+1}$ is a root field over $F$. The author also recursively defines the $I_q$, which is not necessary. Let me try to rephrase the proof, remove that bit, and add more details.

Claim. Let $K$ be a field extension of $F$ such that there exist $c_i in K$, $1le ile n$, with $K = F(c_1,dots,c_n)$, and such that for each $1le i le n$, $c_i^{k_i} in F(c_1,dots,c_{i-1})$ for some integer $k_i ge 1$. Suppose also that $mbox{char}(F) = 0$ and that, for each $1le i le n$, the $k_i^{tinymbox{th}}$ roots of unity lie in $F$. Then $K$ is a root field over $F$ and $K/F$ is Galois.

Proof. For $0le qle n$, define $I_q = F(c_1,dots,c_q)$. Trivially, $I_0$ is a root field over $F$ and $I_0/F$ is Galois; say $I_q$ is the splitting field of $h(x)$ over $F$. Now suppose $0le qle n-1$ and we have shown that $I_q$ is a root field over $F$ and that $I_q/F$ is Galois. We will show that $I_{q+1}$ is a root field over $F$ and that $I_{q+1}/F$ is Galois. Let
$f_{q+1}(x) = x^{k_{q+1}} - c_{q+1}^{k_{q+1}}$. Since $c_{q+1}^{k_{q+1}} in I_q$, this polynomial has coefficients in $I_q$. Let $zeta_1,dots,zeta_{k_{q+1}}$ be the $k_{q+1}^{tinymbox{th}}$ roots of unity in $F$ (since $mbox{char}(F) = 0$, these are distinct). Notice that in $I_{q+1}$, $f_{q+1}(x)$ factors completely as
$$f_{q+1}(x) = prod_{i=1}^{k_{q+1}}(x-zeta_i c_{q+1})$$
with distinct roots, so that $I_{q+1}$ is the splitting field of $f_{q+1}$ over $I_q$, so that $I_{q+1}/I_q$ is Galois. Let $G_q$ be the Galois group of $I_q/F$, i.e., the group of $F$-fixing automorphisms of $I_q$.
Now consider the polynomial
$$g_{q+1}(x) = prod_{sigma in G_q} sigma(f_{q+1}(x)) = prod_{sigma in G_q} (x^{k_{q+1}} - sigma(c_{q+1}^{k_{q+1}})),$$
which has coefficients in $F$ since the subfield of $I_q$ consisting of elements fixed by $G_q$ is equal to $F$. Notice that for each $sigmain G_q$, since $I_{q+1}/I_q$ is Galois, we may extend $sigma$ to an automorphism of $I_{q+1}$. Thus $g_{q+1}(x)$ has all its roots in $I_{q+1}$.

We claim that $I_{q+1}$ is the splitting field of $g_{q+1}(x)h(x)$ over $F$. Indeed, adding the roots of $h(x)$ to $F$ gives us $I_q$, then all the roots of $g_{q+1}(x)$ lie in $I_{q+1}$, and finally $g_{q+1}$ has $c_{q+1}$ as a root. So $I_{q+1}$ is a root field over $F$ and is Galois over $F$.

By induction, we conclude that $K = I_n$ is a root field over $F$ and is Galois over $F$. $hspace{1cm}Box$

Edit to answer comment:

Pinter's proof is essentially the same; the difference is that he defines $I_{q+1}$ as the splitting field of $b(x)$ within the proof, where I define $I_{q+1}$ from the start and instead show that's it's the splitting field of some polynomial over $I_q$, and then also over $F$. His $b(x)$ is exactly the same as my $g_{q+1}(x)$. In fact, the way you represent his proof makes it seem like he has a small error in his proof when he defines $I_{q+1}$ to be the splitting field of $b(x)$ over $F$ (in my proof, I used $h(x)g_{q+1}(x)$, not just $g_{q+1}(x)$). For example, what if $F = mathbb{Q}$ and $K = F(sqrt{2},sqrt{3})$? Then $I_1 = mathbb{Q}(sqrt{2})$, which has Galois group over $mathbb{Q}$ consisting of the identity map and the map which swaps $sqrt{2}$ and $-sqrt{2}$. Now we have $sqrt{3}^2 in I_1$, and we would have $f_{2}(x) = x^2 - 3$, and then $g_2(x) = (x^2-3)^2$. If we just define $I_2$ to be the splitting field of $g_2(x)$ over $mathbb{Q}$, then $I_2 = mathbb{Q}(sqrt{3})$, which does not even contain $I_1$.

To your question at the end of your comment: that's right.