Factorization of irreducible polynomial over Galois extension

Question

Paraphrased from problem 14.4.4 in Dummit and Footes "Abstract Algebra".

Let $K/F$ be a Galois-extension, and let $f(x) \in F[x]$ be an irreducible polynomial. Then we want to show that $f(x)$, seen as a polynomial in $K[x]$, factors as a product of $m$ irreducible polynomials, each of degree $d$ over $K$, where $m = [F(\alpha) \cap K:F]$ and $d = [K(\alpha):K]$, where $\alpha$ is a root of $f(x)$. Here $\alpha$ is a root of $f(x)$ in a Galois closure $L/F$ of $f(x)$ (by Galois closure of $f(x)$ we mean the smallest field extension that is Galois and contains the splitting field of $f(x)$).

Setup: By the hint on the problem, we first want to show that the factorization of $f(x)$ over $K$ is the same as the factorization of $f(x)$ over $L \cap K$. Then we want to show that the factors of $f(x)$ over $L \cap K$ correspond to the orbits of $H = \text{Gal}(L/L \cap K)$ on the roots of $f(x)$, and finally use exercise 9 of section 4.1. (in D & F). Exercise 4.1.9 gives

Assume that $G$ acts transitively on the finite set $A$ and let $H$ be a normal subgroup of $G$. Let $\mathcal{O}_1,\ldots,\mathcal{O}_r$ be the distinct orbits of $H$ on $A$. Then

1. $G$ permutes the sets $\mathcal{O}_1,\ldots,\mathcal{O}_r$ in the sense that for each $g \in G$ and each $i = 1,\ldots,r$ there is a $j$ such that $g\mathcal{O}_i = \mathcal{O}_j$, where $g\mathcal{O} = \{g \cdot a \;|\; a \in \mathcal{O}\}$, and $G$ is transitive on $\{\mathcal{O}_1,\ldots,\mathcal{O}_r\}$, and furthermore all orbits of $H$ on $A$ have the same cardinality.
2. If $a \in \mathcal{O}_1$ then $|\mathcal{O}_1| = |H : H \cap G_a|$ and $r = |G:HG_a|$, where we can note that $H \cap G_a = H_a$, and recall that $G_a = \{g \in G \;|\; g \cdot a = a\}$, i.e. the isotropy group/stabilizer group induced by the action of $G$ on $A$.

My attempt: Assume $f(x) = p_1(x) \cdots p_r(x)$ where $p_i(x) \in K[x]$ are irreducible. Note that $f(x)$ splits completely in $L[x]$, which means that any restrictions imposed on the decomposition of $f(x)$ into linear factors comes from $K$, in $L \cap K$. Since $L \cap K \subseteq K$, it is clear that any decomposition in $L \cap K$ of $f(x)$ is permitted over $K$ as well. On the other hand, since $L$ is the splitting field for $f(x)$, it contains the coefficients of all possible decompositions of $f(x)$ (I believe), hence it follows that the decomposition $p_1(x) \cdots p_r(x)$ is permitted in $(L \cap K)[x]$ as well (since the coefficients for the $p_i$ is in $L \cap K$), therefore I believe it follows that they have the same decomposition.

From prop. 14.4.19 we see that $\text{Gal}(LK/K) \cong \text{Gal}(L/L \cap K)$ and this is a subgroup of $\text{Gal}(L/F)$. We know that $L$ contains the splitting field for $f(x)$ and $L/F$ is Galois, and so $\text{Gal}(L/F)$ acts transitively on the roots of $f(x)$ (since $f(x)$ is irreducible; this is generally not true, i.e. if $f(x)$ is not irreducible). Since $F \subseteq K \cap L \subseteq K$ and $K/F$ is Galois, we have that $H = \text{Gal}(L/K \cap L)$ is Galois by the fundamental theorem of Galois theory $\iff$ $H$ is normal in $\text{Gal}(L/F)$. If we take $A = \{\alpha_1,\ldots,\alpha_n\}$ as the roots of $f(x)$, and we take $\mathcal{O}_1,\ldots,\mathcal{O}_r$ as the orbits of the action $H \curvearrowright A$, then we conclude from 1) above that $\text{Gal}(L/F)$ permutes the orbits $\mathcal{O}_1,\ldots,\mathcal{O}_r$ and that all orbits have the same cardinality. Now I believe that each $\mathcal{O}_i$ should correspond to one irreducible factor of $f(x)$ (the orbit $\mathcal{O}_i$ will be roots $\alpha_{i_1},\ldots,\alpha_{i_d}$ and then the corresponding irreducible factor I presume would be $(x-\alpha_{i_1}) \cdots (x-\alpha_{i_d})$) in $L \cap K$, and the size of the orbit should then correspond to the degree of the irreducible factor; by the exercise it follows that all irreducible factors then have the same degree $d$ (since all orbits of $H$:s action on $A$ have the same degree). We want to show that $d = [K(\alpha):K]$; ~~here I am stuck~~. Well, by the exercise, all orbits are of equal size, so take $\alpha \in \mathcal{O}$ for some orbit $\mathcal{O}$ of the action $H \curvearrowright A$, and then we have that $d = |H:H_\alpha|$ for some root $\alpha$ of $f(x)$. By the orbit-stabilizer theorem we have that $d = |H \cdot \alpha|$. We note that $H = \text{Gal}(L/L \cap K) \leq \text{Gal}(L/F)$ and that $H$ fixes everything in the intersection $L \cap K$. This means that if $\alpha \in L \cap K$ then $|H \cdot \alpha| = 1$. But in this case $f(x)$ splits completely over $K$ and so the statement of the theorem holds. So assume instead that $K$ contains no roots of $f(x) \ldots$ INSERT ARGUMENT

On the other hand, the number of orbits, $r$, would naturally correspond to the number of irreducible factors. From exercise 9.2), section 4.1 we had that \begin{align*} r &= |G:HG_a|\\ &= \Bigg|\frac{\text{Gal}(L/F)}{\text{Gal}(L/L \cap K) \text{Gal}(L/F)_a}\Bigg| \end{align*} where $a$ is some root of $f(x)$. Here I am also sort of stuck and don't know how to proceed. Any hints on how to proceed would be well-appreciated.

Answer 1

By inspiration from Why does an irreducible polynomial split into irreducible factors of equal degree I'll try to provide atleast a partial answer.

Notice that since $K/F$ is Galois, it follows from the fundamental theorem of Galois theory that for any $\sigma \in G = \text{Gal}(L/F)$, its restriction to $K$, $\sigma|_K$ is an embedding of $K$ into $L$. Since $K/F$ is Galois, it follows that $\sigma(K) = K$ (i.e. all embeddings of $K/F$ are actually automorphisms of $K/F$). Let $\alpha,\beta$ be roots of $f(x) = q_1(x) \cdots q_r(x)$. If $\alpha$ is a root of $q_1(x)$ and $\beta$ is a root of $q_i(x)$ then since $G$ acts transitively on the roots of $f(x)$ (since $f(x)$ is irreducible) it follows that there is some automorphism $\sigma \in G$ such that $\sigma(\alpha) = \beta$. By the natural extension $\sigma'$ of $\sigma$ to polynomial rings (linear extension), we see that $\sigma'(q_1(x))$ must have $\beta$ as a root, and one checks that $\sigma'(q_1(x))$ must be irreducible (and it is still a polynomial in $K[x]$ since it acts as $\sigma$ on the coefficients of $q_1(x)$). Therefore, $\sigma'(q_1(x)) = q_i(x) \in K[x]$. But $\sigma$ preserves the degree, so that $\deg(q_1(x)) = \deg(q_i(x))$ for arbitrary $i = 1,\ldots,r$. This shows that all irreducible factors $q_i(x)$ are of the same degree.

To see that $d = [K(\alpha):K]$ we just note that $[K(\alpha):K] = \deg(q_1(x))$ since $q_1(x)$ was irreducible, so is the minimal polynomial for $\alpha$ over $K$. Since all $q_i(x)$ for $i = 1,\ldots,r$ were of the same degree by the above reasoning, it follows that $[K(\alpha):K] = \deg(q_i(x))$ for arbitrary $q_i(x)$.

It remains to find an expression for $m$, the # of irreducible factors of $f(x)$ over $K$.

Here we note (as in the question posed) that $K$ and $K \cap L$ have the same decomposition of $f(x) = q_1(x) \cdots q_r(x)$ (I'd like to make this argument more precise). We want to show that the factors $q_i(x)$ of $f(x)$ correspond to orbits $\mathcal{O}_i$ of the action of $\text{Gal}(L/L \cap K)$ on the roots of $f(x)$.

Proposition. Let $L/F$ be a Galois-extension, and let $F'/F$ be an arbitrary extension. Then $LF'/F'$ is Galois, with Galois group $$\text{Gal}(LF'/F') \cong \text{Gal}(L/L \cap F')$$ isomorphic to a subgroup of $\text{Gal}(L/F)$.

If we take $L = L$ as the Galois-closure of $f(x)$, and $F' = K$, we see that \begin{align*} \text{Gal}(LK/K) &\cong \text{Gal}(L/L \cap K)\\ &\leq \text{Gal}(L/F). \end{align*}

Since $\text{Gal}(L/F)$ is Galois, $\text{Gal}(L/K \cap L)$ is Galois by the fundamental theorem. That $q_i(x)$ is irreducible over $K \implies$ it is irreducible over $L \cap K \iff \text{Gal}(L/L \cap K)$ acts transitively on the roots of $q_i(x)$. As before, note that $G = \text{Gal}(L/F)$ acts transitively on $f(x)$ and that since $L/(L \cap F)$ is Galois $\iff$ $\text{Gal}(L/L \cap F) \triangleleft \text{Gal}(L/F)$. The orbits are by the exercise then all of equal size, and by the orbit-stabilizer theorem this is exactly $|H \cdot \alpha|$ for $\alpha$ a root of $q_i(x)$. From the fact that the action is transitive, so that $|H \cdot \alpha| \supseteq \{\text{roots of} \ q_i(x)\}$ and since $h \in H$ must take a root of $q_i(x)$ to a root of $q_i(x)$ (otherwise $q_i = q_j$ for $i \neq j$, contradiction), it follows that $|H \cdot \alpha| = \{\text{roots of} \ q_i(x)\}$.

We need to show that the number of orbits are $[F(\alpha) \cap K: F]$. By the exercise we have that \begin{align*} r &= |\text{Gal}(L/F):\text{Gal}(L/L \cap K) \text{Gal}(L/F)_{\alpha}|. \end{align*}

It should not be hard to check that $\text{Gal}(L/F)_{\alpha} = \text{Gal}(L/F(\alpha))$ (use the fundamental correspondence of Galois theory between fixed fields and subgroups). Then we have that \begin{align*} r &= \frac{|\text{Gal}(L/F)|}{|\text{Gal}(L/L \cap K) \text{Gal}(L/F(\alpha))|} \\ &= \frac{[L:F]}{\frac{[L:L \cap K][L:F(\alpha)]}{|\text{Gal}(L/L \cap F) \cap \text{Gal}(L/F(\alpha))|}}\\ &= \frac{[L:F]}{\frac{[L:L \cap K][L:F(\alpha)]}{|\text{Gal}(L/L \cap F(\alpha)|}}\\ &= \frac{[L:F][L:L \cap F(\alpha)]}{[L:L \cap K][L:F(\alpha)]}\\ &= \frac{[L:F]}{[L:L \cap K]}\\ &= [L \cap K:F] \end{align*}

Hm, in the end, I do not manage to derive $r = [F(\alpha) \cap K:F]$. I'll post this for now, and if I can figure it out I plan on updating the answer.