Show [F(a):F] = r Suppose that K/F is Galois and a in K has precisely r distinct images under G(K/F). Show [F(a):F] = r
I'll really appreciate if someone could give some hints? Thank you!
 A: If $\left[K:F\right]$ is finite you can also do the following:
Let $G:=\mathrm{Gal}(K/F)$. Then $\left[F(a):F \right]=\frac{|G|}{|H|}$ where $H$ is the pointwise stabilizer of the elements of $F(a)$ in $G$ which is just the stabilizer of $a$ in $G$. Now $a$ has exactly $r$ distinct images under $G$ hence its stabilizer is of order $|H|=\frac{|G|}{r}$.
So we may conclude $\left[F(a):F \right]=\frac{|G|}{|H|}=r$.
A: I think it works like this: let $a_1 = a, a_2, \ldots, a_r$, be the $r$ distinct images of $a$ under the action of $G(K/F)$. Note $a = a_1$ is in this list, since it is the image of itself under the action of the identity map $e \in G(K/F)$. Consider the polynomial $p(x) \in K[x]$ defined by
$p(x) = \prod_1^r (x - a_j); \tag{1}$
I claim that $p(x)$, though defined in $K[x]$, is in fact a polynomial in $F[x]$; to see this, observe that the coefficients of $p(x)$, defined by (1) as it is, are each and every one fixed by all the elements of $G(K/F)$; and this may be seen by noting that any $g \in G(K/F)$ must act as a permutation on the set $\{ a_1, a_2, \ldots, a_r \}$; and this follows from the observation that $G(K/F)a_1 = \{ a_1, a_2, \ldots, a_r \}$ as sets, so that for $g \in G(K/F)$ we have $g\{ a_1, a_2, \ldots, a_r \} = gG(K/F)a_1 = G(K/F)a_1$ since $gG(K/F) = G(K/F)$. Since $g \in G(K/F)$ merely re-arrangles the factors $(x - a_j)$ of $p(x)$ amongst themselves, we must have
$p(x) = \prod_1^r(x - a_j) = \prod_1^r (x - g(a_j)) \tag{2}$
hence, as asserted, the coefficients of $p(x)$ must be fixed under action of $g \in G(K/F)$; alternatively, one can argue that the coefficients of $p(x)$ are in fact, up to a sign, the elementary symmetric functions of the $a_j$, $1 \le j \le r$; that is, writing
$p(x) = \sum_0^r p_j x^j = \prod_1^r(x - a_j) \tag{3}$
we have
$p_{r - 1} = -\sum_1^r a_j, \tag{4}$
$p_{r - 2} = \sum_{j = 1; j < k}^r a_j a_k, \tag{5}$
and so on, down to
$p_0 = (-1)^r \prod_1^r a_j; \tag{6}$
and again, since $g \in G(K/F)$ acts as a permutation on $\{ a_1 = a, a_2, \ldots, a_r \}$, each of these coefficients is clearly $g$-invariant. Now since $K/F$ is a Galois extension, the fixed field of $G(K/F)$ is precisely $F$, hence we must have $p_j \in F$ for $0 \le j \le r$, since as we have just seen each $p_j$ is $G(K/G)$ invariant. Thus $p(x) \in F[x]$, and since $\deg p(x) = r$ and $\sum_0^r p_j a^j = p(a) = p(a_1) = 0$, we can conclude in the usual manner that $[F(a):F] \le r$, that is, the dimension of $F(a)$ as a vector space over $F$, is at most $r$ by virtue of the fact that there exists a non-trivial linear relation having coefficients in $F$ between the elements of the set $\{ 1, a, a^2, \ldots, a^r \}$.  ($p(a)$ is non-trivial since $p(x)$ is monic, $p_r = 1$, and $p_0 =  (-1)^r \prod_1^r a_j \ne 0$.)  I now claim that furthermore $p(x)$ is irreducible over $F$, for if $p(x) = p_1(x)p_2(x)$ with $p_1(x), p_2(x) \in F[x]$ and $\deg p_1(x), \deg_2(x) \ge 1$, then since $0 = p(a) = p_1(a)p_2(a)$  we must have, say, $p_1(a) = 0$; but then, since the coefficients of $p_1(x)$ are all in $F$, for any $g \in G(K/F)$ we have
$p_1(g(a)) = g(p_1(a)) = g(0) = 0;  \tag{7}$
now as $g$ varies over $G(K/F)$, $g(a)$ takes on the $r$ distinct values $a_j$, $1 \le j \le r$, implying that $p_1(x)$ has at least $r$ distinct roots in $K$, further implying that $\deg p_1(x) \ge r$; but $r = \deg p(x) = \deg p_1(x) + \deg p_2(x)$ is then impossible since $\deg p_2(x) \ge 1$.  Hence $p(x)$ is irreducible over $F$ and hence the set $\{ 1, a, a^2, \ldots, a^{r - 1} \}$ is linearly independent over $F$; if not, then there would exist $0 \ne d(x) \in F[x]$ of minimal degree $\deg d(x) < r$ with $d(a) = 0$; but then applying the division algorithm to $p(x), d(x)$ yields $p(x) = d(x)q(x) + s(x)$ and if $s(x) \ne 0$, then $\deg s(x) < \deg d(x)$ which contradicts the minimality of $\deg d(x)$, since $s(a) = p(a) - d(a)q(a) = 0$.  And if $s(x) = 0$, the irreducibility of $p(x)$ is contradicted. Thus the set $\{ 1, a, a^2, \ldots, a^{r - 1} \}$ is linearly independent over $F$ and thus $[F(a):F] \ge r$, which in turn implies that in fact $[F(r):F] = r$ as required.   QED.
and
Hope this helps.  Cheerio,
and as always, Fiat Lux!!!
A: It's easier to argue it the other way round. $|F(a):F|=r$ implies that the minimal polynomial $f$ of $a$ over $F$ has degree $r$. Since $K:F$ is Galois, it contains all $r$ roots $a=a_1,a_2,\ldots,a_r$ of $f$, and they are all distinct, because Galois extensions are separable. Then for each $i$, the map $a \to a_i$ induces an isomorphism $F(a) \to F(a_i)$, which extends to an element of ${\rm Gal}(K,F)$, so $a$ has $r$ distinct images.
