7
$\begingroup$

I'm working on the proof of Theorem 4.26 (pg. 215) in Rotman's "Advanced Modern Algebra". The theorem is stated as follows:

Let $f(x)\in k[x]$, where $k$ is a field, and let $E$ be a splitting field of $f(x)$ over $k$. If $f(x)$ is solvable by radicals, then its Galois group $\operatorname{Gal}(E/k)$ is a solvable group.

Here $E$ is the splitting field of $f(x)$ over $k$. The proof relied on Theorem 4.20 (pg.213)

Let $k$ be a field and let $f(x)\in k[x]$ be solvable by radicals, so there is a radical extension $k=K_0\subseteq K_1\subseteq \dots \subseteq K_t$ with $K_t$ containing a splitting field $E$ of $f(x)$. If each $K_i/K_{i-1}$ is a pure extension of prime type $p_i$, where $p_i\neq \operatorname{char}(k)$, and if $k$ contains all $p_ith$ roots of unity, then the Galois group $\operatorname{Gal}(E/k)$ is a quotient of a solvable group.

(Hence it is a solvable group.)

In the proof, he constructed an extension $E^*$ of $E$ and an extension $k^*$ of $k$ so that $k^*$ contains appropriate roots of unity, thus $\operatorname{Gal}(E^*/k^*)$ is solvable by Theorem 4.20. The problem is, I don't know why he can get rid of the extra hypothesis $p_i\neq \operatorname{char}(k)$ for all $i$. I read the chapter twice, but still no clue.

$\endgroup$
1
$\begingroup$

I have the second edition of Rotman's book. The material is now slightly different.

The theorem in question is now Theorem 3.27, pp. 189:

Theorem 3.27 (Galois). Let $f(x)\in k[x]$, where $k$ is a field, and let $E$ be a splitting field of $f(x)$ over $k$. If $f(x)$ is solvable by radicals, then its Galois group $\mathrm{Gal}(E/k)$ is a solvable group.

The proof relies on Lemma 3.21, which is the counterpart of what you quote. This one now reads:

Lemma 3.21. Let $k$ be a field, let $f(x)\in k[x]$ be solvable by radicals, and let $k=K_0\subseteq K_1\subseteq\cdots\subseteq K_t$ be a tower with $K_i/K_{i-1}$ a pure extension of prime type $p_i$ for all $i$. If $K_t$ contains a splitting field $E$ of $f(x)$ and $k$ contains all the $p_i$th roots of unity, then the Galois group $\mathrm{Gal}(E/k)$ is a quotient of a solvable group.

So, the assumption on the characteristic of $k$ has been dropped. This Lemma in turn relies on Theorems 3.17, Lemma 3.18, and Lemma 3.19. They are:

Theorem 3.17. Let $k\subseteq B\subseteq E$ be a tower of fields. If $B/k$ and $E/k$ are normal extensions, then $\sigma(B)=B$ for all $\sigma\in \mathrm{Gal}(E/k)$, $\mathrm{Gal}(E/B)\triangleleft \mathrm{Gal}(E/k)$, and $\mathrm{Gal}(E/k)/\mathrm{Gal}(E/B)\cong \mathrm{Gal}(B/k)$.

Lemma 3.18. (i) If $B=k(u_1,\ldots,u_t)/k$ is a finite extension field, then there is a normal extension $E/k$ containing $B$; that is, $E$ is a splitting field for some $f(x)\in k[x]$. If each $u_i$ is seprable over $k$, then $f(x)$ is a separable polynomial and, if $G=\mathrm{Gal}(E/k)$, then $E=k(\sigma(u_1),\ldots,\sigma(u_t)\colon \sigma\in G)$. (ii) If $B/k$ is a radical extension, then the normal extension $E/k$ is a radical extension.

Lemma 3.19. Let $k=K_0\subseteq K_1\subseteq K_2\subseteq \cdots\subseteq K_t$ be a tower with each $K_i/K_{i-1}$ a pure extension of prime type $p_i$. If $K_t/k$ is a normal extension and $k$ contains all the $p_i$th roots of unity, for $i=1,\ldots,t$, then there is a sequence of groups $\mathrm{Gal}(K_t/k)=G_0\supseteq G_1\supseteq G_2\supseteq\cdots\supseteq G_t=\{1\}$, with each $G_{i+1}\triangleleft G_i$ and $G_i/G_{i+1}$ cyclic of prime order $p_{i+1}$ or $\{1\}$.

So the assumption on the characteristic of $k$ has been dropped entirely in the Second edition.

$\endgroup$
  • 1
    $\begingroup$ Thanks in advance! His new treatment indeed solve the problem. Now I see it, he uses the assumption on the characteristic of $k$ to conclude that $K_i/K_{i-1}$ is a splitting field of a separable polynomial, then $G_{i-1}/G_i=\operatorname{Gal}(K_i/K_{i-1})$ is a cyclic group of prime order $p_i$. In the case where $\text{char}(K_0)=p_i$, the Galois group is $\{1\}$, as we see in the statement of Lemma 3.19. It does not effect the solvability of Galois group, so the assumption is not required. $\endgroup$ – withoutfeather Apr 30 at 6:12

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.