As part of a proof, my textbook1 essentially asserts (without proof, AFAICT) that if $L:K$ is a finite extension, and $\mathrm{Gal}(L:K)$ is the Galois group of $L$ over $K$, then $\left| \mathrm{Gal}(L:K) \right|$ (i.e. the cardinality of this Galois group) is finite.

I don't see how this follows from the results derived up to this point in the book. All the theorems that would say something about the cardinality of $\left| \mathrm{Gal}(L:K) \right|$ make assumptions that do not hold above. (For example, corollary 7.29, on p. 117, states that $\left| \mathrm{Gal}(L:K) \right| = [L:K]$, provided that $L$ is both normal and separable.)

EDIT: To avoid misunderstandings, here are the definitions my book uses (copied verbatim, from p. 94):

Let $L$ be an extension of a field $K$. An automorphism $\alpha$ of $L$ is called a $K$-automorphism if $\alpha(x) = x$ for every $x$ in $K$. The set of all $K$-automorphisms of $L$ is denoted by $\mathrm{Gal}(L:K)$.

EDIT 2: An additional source of confusion for me is the statement of Theorem 7.12 (on p. 100, well before the assertion cited above):

Theorem 7.12

Let $L$ be a finite extension of a field $K$ and let $G$ be a finite subgroup of $\mathrm{Gal}(L:K)$. Then $[L:\Phi(G)] = \left| G \right|$.

...where the map $\Phi$ is defined (on p. 95) as $$ \Phi(G) = \{x \in L | \alpha(x) = x , \; \forall \alpha \in G \} $$

To be more specific, my confusion stems in part from the fact that the statement of the theorem bothers to specify that the subgroup $G$ is finite. This seems strange if $\mathrm{Gal}(L:K)$ (which contains $G$) is finite to begin with.

(On top of everything, right before proving Theorem 7.12, the author points out that "[t]he proof is longer than one might have expected -- or hoped". This makes me wonder if these seemingly obvious cardinality facts are actually pretty hard to prove rigorously.)

1 John M. Howie, Fields and Galois Theory, p. 117.

  • $\begingroup$ @TorstenSchoeneberg: beyond the trivial assertion $\mathrm{Gal}(L:K) \subseteq \mathrm{Aut}(L)$, I don't know what else to say about the general case you cite. $\endgroup$ – kjo Mar 19 '18 at 22:42
  • 2
    $\begingroup$ $L$ is a finite dimensional $K$-vector space. An automorphism of $L$ that fixes $K$ is in particular a $K$-linear transformation. A linear transformation is determined once you fix its values at a basis. But in addition to being linear an element of the basis can only be sent to one of the finitely many roots of its minimal polynomial over $K$. So, finitely many choices. $\endgroup$ – SphericalTriangle Mar 19 '18 at 22:44
  • $\begingroup$ @SphericalTriangle: wait, what minimal polynomial now? $\endgroup$ – kjo Mar 19 '18 at 22:46
  • 1
    $\begingroup$ @TorstenSchoeneberg: not in this book. Since this could be a source of confusion, I'll edit my post to include the definitions I'm using. Give me a few minutes. $\endgroup$ – kjo Mar 19 '18 at 22:48
  • 2
    $\begingroup$ @kjo An element $e\in L$, if $L:K$ is a finite extension, must satisfy a polynomial with coefficients in $K$. This is because otherwise $1,e,e^2,...$ would be linearly independent over $K$ contradicting that $L$ is a finite-dimensional vector space over $K$. Among all polynomials with coefficients in $K$ that $e$ satisfies, there is a monic one with minimal degree. That is the (it is unique by Euclid's algorithm) minimal polynomial. Each element of your basis of $L$ over $K$ satisfies one such polynomial. $\endgroup$ – SphericalTriangle Mar 19 '18 at 22:48

This is my best attempt to fill in the details of the answer that SphericalTriangle sketched in his/her comments.

Since $L:K$ is a finite extension, there is a finite basis $\{1,z_1,z_2,\dots,z_n\}$ of $L$ over $K$, with all the $z_i$ in $L\backslash K$. Each $z_i$ is therefore the root of a corresponding minimal polynomial $m_i \in K[X]$.

(Otherwise, if some $z_k$ was not the root of any polynomial in $K[X]$, the infinite set $\{1, z_k, z_k^2, \dots \}\subseteq L$ of non-negative powers of $z_k$ would be linearly independent, thereby contradicting the finiteness of $[L:K]$.)

For each $i \in \{1,\dots,n\}$, let $d_i$ be the degree of $m_i$, let $\{\zeta_{i,1},\zeta_{i,2},\dots,\zeta_{i,d_i}\}$ be its roots, and, without loss of generality, assume that $\zeta_{i,1} = z_i$.

For any $K$-automorphism $\varphi$ of $L$, specifying the value of $\varphi$ at the basis elements $\{1,z_1,z_2,\dots,z_n\} = \{1, \zeta_{1,1}, \zeta_{2,1}, \dots, \zeta_{n,1}\}$ fully determines $\varphi$.

By virtue of being a $K$-automorphism, $\varphi$ must (a) fix $1$, and (b) map each root $\zeta_{i,1} = z_i$ of $m_i$ to some other root $\zeta_{i,j}$ of $m_i$. (In more detail: $\forall a \in L, \varphi(m_i(a)) = m_i(\varphi(a))$, because $\varphi$ is an automorphism of $L$ that fixes the coefficients of $m_i \in K[X]$. Hence, $m_i(\varphi(\zeta_{i,1})) = \varphi(m_i(\zeta_{i,1})) = \varphi(0) = 0,$ which means that $\varphi(\zeta_{i,1})$ is a root of $m_i$, so it must equal $\zeta_{i,j}$, for some $j$.)

Since there are at most $d_i$ distinct values that each $z_i$ can be mapped to by a $K$-automorphism of $L$ (i.e. by an element of $\mathrm{Gal}(L:K)$), there can be at most $\prod_1^n d_i < \infty$ distinct $K$-automorphisms of $L$. Hence $\mathrm{Gal}(L:K)$ is finite.

  • 1
    $\begingroup$ That's all. The reason why it must send $\zeta_{i,1}$ to some root of $m_i$, which you probably know, is because if $m_i(x)=\sum_j k_jx^j$ for $k_j\in K$. Then $0=\phi(0)=\phi(m_i(\zeta_{i,q}))=\sum_j k_j\phi(\zeta_{i,1})^j=m_i(\phi(\zeta_{i,1}))$. $\endgroup$ – SphericalTriangle Mar 20 '18 at 14:57

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.