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, 2018 at 22:42
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    $\begingroup$ $L$ 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. $\endgroup$ Mar 19, 2018 at 22:44
  • $\begingroup$ @SphericalTriangle: wait, what minimal polynomial now? $\endgroup$
    – kjo
    Mar 19, 2018 at 22:46
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    $\begingroup$ @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, 2018 at 22:48
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    $\begingroup$ @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. $\endgroup$ Mar 19, 2018 at 22:48

1 Answer 1


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.

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    $\begingroup$ 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 $0=\phi(0)=\phi(m_i(\zeta_{i,q}))=\sum_j k_j\phi(\zeta_{i,1})^j=m_i(\phi(\zeta_{i,1}))$. $\endgroup$ Mar 20, 2018 at 14:57

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