1
$\begingroup$

Suppose that $k$ is a field with characteristic equal to zero, that $P \in k[X]$ is an irreducible polynomial and that $\alpha$ is a root of $P$ in an algebraic closure $\overline{k}$.

Suppose also that $\alpha$ belongs to a radical extension $R$ of $k$. How can I prove that $P$ is solvable by radical, i.e. that there exists an extension of $P$ splitting field which is a radical extension?

What happens if the characteristic of $k$ is not supposed to be equal to zero?

$\endgroup$
2
$\begingroup$

If I understood the question correctly we are given that $\alpha$ belongs to an extension field $k_n$, $n$ a positive integer, gotten as a root extension tower. That is, we have a sequence of intermediate fields $$k\subset k_1\subset k_2\subset\cdots\subset k_n$$ such that for all $i$ we have $k_{i+1}=k_i(x_i)$ for an element $x_i$ with the property that $x_i^{m_i}\in k_i$ for some integer $m_i>1$.

Let $\beta\in \overline{k}$ be another zero of $P$. Then we have the usual $k$-isomorphism $\sigma: k[\alpha]\to k[\beta]$ such that $\sigma(\alpha)=\beta$. A standard application of Zorn's lemma shows that $\sigma$ can be extended to an automorphism of $\overline{k}$ (this is, of couse, a bit overkill, but I don't want to specify a sufficiently large finite extension of $k$, so I sweep this detail under the rug by extending all the way up to $\overline{k}$).

Using such an extended automorphism (call it $\sigma$, too) it is then clear that $\beta$ belongs to the field $\sigma(k_n)$ that is a root tower extension $$ k\subset \sigma(k_1)\subset \sigma(k_2)\subset \cdots \subset \sigma(k_n),$$ where $\sigma(k_{i+1})=\sigma(k_i)[\sigma(x_i)]$ and $\sigma(x_i)^{m_i}=\sigma(x_i^{m_i})\in\sigma(k_i)$ for all $i$.

Continuing in this way we see that any zero of $P$ belongs to a radical extension - each to their private extension. The compositum of those (finitely many) extensions is then obviously also a root tower extension. Because that compositum contains all the zeros of $P$ in $\overline{k}$ it contains the splitting field also.


Nothing above really depended on the characteristic. AFAICT $P$ need not even be separable. What fails in characteristic $p$ is the connection between radical extensions and solvability of the Galois group (in the cases where we avoid separability problems). The key step in that part of Galois theory is the following.

Lemma. Let $K$ be a field that contains a primitive root of unity of order $m$. Let $L$ be such an extension of $K$ that $L/K$ is Galois with $Gal(L/K)\cong C_m$. Then there exists an element $z\in L$ such that $L=K(z)$ and $z^m\in K$. Conversely, if $a\in K$ is such an element that the polynomial $x^m-a$ is irreducible in $K[x]$, then the extension $M$ gotten by adjoining $z=a^{1/m}$, $M=K(z)$ is Galois with $Gal(M/K)\cong C_m$.

In characteristic zero the connection between radical extensions and solvability of the Galois group more or less comes from this Lemma. Going one way we apply it to each step in the composition series of the Galois group, and going the other way we build a solvable Galois group from a tower of such extensions. A little bit of extra comes from the need to make sure that bringing in the necessary roots of unity won't change the scene.

That brings us to one source of problems in characteristic $p$. Namely, primitive roots of order $p$ do not exist. This is because $$ x^p-1=(x-1)^p $$ only has the multiple root $1$. Therefore the Lemma is never available, when $m=p$. This in itself is not fatal, because we do understand what characteristic $p$ cyclic extensions of degree $p$ look like. They are so called Artin-Schreier extensions. If $L/K$ is such a cyclic extension, we can find an alemen $z\in L$ such that $L=K(z)$, $y=z^p-z\in K$ and the minimal polynomial of $z$ is $$ x^p-x-y. $$ See this answer for an on-site proof. If we allowed A-S steps in addition to radical extensions, we could get a satisfactory description of solvable characteristic $p$ extensions.

What breaks down in characteristic $p$ is inseparability - not all polynomials have a splitting field that is a Galois extension of the base field. Therefore the Galois correspondence is not available as a tool for studying the intermediate fields.

$\endgroup$

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.