2
$\begingroup$

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 $c_i^k\in F(c_1,...,c_{i-1})$.

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 $F$:

$\qquad\underline F\subseteq\underline {F(c_1)}\subseteq\underline {F(c_1,c_2)}\subseteq ...\subseteq\underline {F(c_1,...,c_n)} \qquad\qquad$ (2)
$\qquad I_0\qquad I_1 \qquad\qquad I_2 \qquad\qquad\;\;\;\; I_m=K$

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 $I_0\subseteq I_1\subseteq...\subseteq I_q$ 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 $a\in I_q$.
$\quad$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 $f:K_1\rightarrow K_2$ 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 $h_i\circ :H\rightarrow H$ (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.)

$\endgroup$
  • $\begingroup$ Root field ? Let $F_{n+1} = F_n(c_n^{1/d_n}),c_n \in F_n, d_n \in \mathbb{Z}_{\ge 1}$ so $F_N/F_0$ is a radical extension. Let $E_n = F_n(\zeta_k),k = \prod_{n=1}^N d_n$ and $E_{n+1} = E_n(c_n^{1/d_n}), d_n \in F_n \subset E_n$. Then $E_{n+1}/E_n$ is an abelian extension. Proof : there is some $a_n$ such that $Gal(E_{n+1}/E_n) $ is generated by $c_n^{1/d_n} \mapsto \zeta_{d_n}^{a_n}c_n^{1/d_n}$. $\endgroup$ – reuns Jan 17 at 22:28
  • $\begingroup$ @reuns Hi. As you can probably see, I'm pretty new to Galois theory. Do you think you can provide an explanation more in terms of the concepts mentioned in my question? I don't know a whole lot outside of the relative basics $\endgroup$ – Cardioid_Ass_22 Jan 17 at 22:32
1
+50
$\begingroup$

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

Proof. For $0\le q\le n$, 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 $0\le q\le n-1$ 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 $k_{q+1}^{\tiny\mbox{th}}$ 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 $\sigma\in G_q$, 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.

$\endgroup$
  • $\begingroup$ This is great but (since it seems you do understand it) could you also add in an edit or something explaining how the proof Pinter presents relates to yours? I'd really appreciate it. Also, (just to be sure) when you say "Notice that for each $\sigma\in G_q$, 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 $I_{q+1}$.", you basically mean (using the version of $\sigma$ extended to $I_{q+1}$) that each root of $g_{q+1}(x)$ can be written as $\zeta_t \sigma(c_{q+1})$ for some $t$, right? $\endgroup$ – Cardioid_Ass_22 Jan 23 at 14:07
  • 2
    $\begingroup$ Added an edit at the bottom. Hope it helps! $\endgroup$ – csprun Jan 23 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.