# Why does an irreducible polynomial split into irreducible factors of equal degree over a Galois extension?

I've been struggling to prove this fact over the past day or so.

Suppose $f(x)\in F[X]$ is irreducible over a field $F$ with $\deg(f)=n$, and let $L$ be the splitting field of $f(x)$ over $F$ with $\alpha$ a root of $f(x)$ in $L$. Let $K/F$ be a Galois extension, with $K\subset L$. Then why does $f(x)$ split into a product of $m$ irreducible polynomials of degree $d$ over $K$, with $m=[F(\alpha)\cap K:F]$ and $d=[K(\alpha):K]$?

If $G$ is the Galois group of $L/F$, then I let $H$ be the corresponding subgroup of $K$, which is normal since $K/F$ is Galois. I know $H$ permutes the roots of $f(x)$, specifically fixing the roots of $f(x)$ in $K$. I know some basic equalities like $[F(\alpha):F]=n$, and $[K:F]=|G:H|$, but I'm not seeing how to start a good argument.

What's a good way to prove this, or at least proceed? Thank you for any insight.

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Are you assuming that $f(x)$ is separable? – Arturo Magidin Nov 29 '11 at 17:47
I am a bit rusty, but I think the argument should proceed something like this: treating $f$ as an element of $K[x]$, we see that $f(\alpha)=0$ implies that $p$ divides $f$, where $p$ is the minimal polynomial for $\alpha$ over $K$. Note that $\deg(p)=\deg_K(\alpha)=[K(\alpha):K]=d$. Thus, in $K[x]$, we have $$f=p\cdot\text{stuff}=\underbrace{\prod_{i=1}^d(x-\sigma_i(\alpha))}_{p}\quad \underbrace{\prod_{i={d+1}}^{n}(x-\sigma_i(\alpha))}_{\text{stuff}}$$ where $\sigma_i$ are the elements of $\operatorname{Gal}(L/F)$ (I've assumed WLOG $f$ is monic). – Zev Chonoles Nov 29 '11 at 18:06
I think we then ought to be able to show that we get one of the Galois groups floating around acts on the irreducible factors of $f$, in a nice way, and use standard results about group actions and Galois theory to get that $m=[F(\alpha)\cap K:F]$, but this is the part that I'm having trouble filling in. I'm also not seeing where separability comes into play, though since Arturo has mentioned it I am sure that it does. – Zev Chonoles Nov 29 '11 at 18:07
@Zev: I may have jumped the gun. It's just that weird things can happen when you have nonperfect fields and you try to extend them by an irreducible inseparable polynomial: splitting field = Galois extension usually requires separability; but you may talk about Galois groups regardless by refering to the Galois closure. I don't think we are using that $|\mathrm{Gal}(L/F)|=|[L:F]|$, so it may not matter. – Arturo Magidin Nov 29 '11 at 20:04
Thanks for thinking about it too, @Zev. – yunone Nov 30 '11 at 9:58

Factor $f(x)$ into irreducibles in $K[x]$, $$f(x) = q_1(x)\cdots q_m(x).$$

If $\alpha$ is a root of $q_1(x)$ and $\beta$ is a root of $q_i(x)$, then $q_1(x)$ is the irreducible polynomial of $\alpha$ over $K$, and $q_i(x)$ is the irreducible polynomial of $\beta$ over $K$.

Since the action of $G$ is transitive on the roots of $f$ there exists $\sigma\in G$ such that $\sigma(\alpha)=\beta$. Since $K$ is Galois, $\sigma(K)=K$, so $\sigma(q_1(x))\in K[x]$ is a polynomial that has $\sigma(\alpha)=\beta$ as a root. Therefore, $q_i(x)|\sigma(q_1(x))$. Since both are irreducible, it follows that $\deg(q_i) = \deg(\sigma(q_1))=\deg(q_1)$. So all irreducible factors of $f(x)$ in $K$ have the same degree.

The degree is equal to the degree of the extension $K(\alpha)/K$ (which is the degree of $q_1(x)$).

A for $m$, the number of factors, we have: $n=[F(\alpha):F] = [F(\alpha):F(\alpha)\cap K][F(\alpha)\cap K:F]$.

I claim that $[F(\alpha):F(\alpha)\cap K] = [K(\alpha):K]$.

Indeed, since $K$ is Galois over $F$, then $K(\alpha)$ is Galois over $F(\alpha)$ ($K$ is the splitting field of some polynomial over $F$, and this same polynomial works for $K(\alpha)$), and $K$ is Galois over $K\cap F(\alpha)$. If $\sigma\in\mathrm{Gal}(K(\alpha)/F(\alpha))$, then restricting $\sigma$ to $K$ gives a homomorphism $\mathrm{Gal}(K(\alpha)/F(\alpha))\to \mathrm{Gal}(K/K\cap F(\alpha))$. If $\sigma$ restricts to the identity on $K$, then it must be the identity on $K(\alpha)$ (it fixes $\alpha$ since it fixes $F(\alpha)$ pointwise), so the map $\mathrm{Gal}(K(\alpha)/F(\alpha))\to \mathrm{Gal}(K/K\cap F(\alpha))$ is one-to-one.

Let $H'$ be the image of this map. Then $H'$ fixes $K\cap F(\alpha)$ pointwise, and if $k\in K$ is fixed by all elements of $H'$, then $k$ must be fixed by all elements of $\mathrm{Gal}(K(\alpha)/F(\alpha))$, hence $k\in F(\alpha)\cap K$. So $F(\alpha)\cap K$ is the fixed field of $H'$, hence $H'=\mathrm{Gal}(K/K\cap F(\alpha))$.

Thus, $[K(\alpha):F(\alpha)] = [K:K\cap F(\alpha)]$.

Now, \begin{align*} [K(\alpha):K\cap F(\alpha)] &= [K(\alpha):K][K:K\cap F(\alpha)]\\ \text{ and }[K(\alpha):K\cap F(\alpha)] &= [K(\alpha):F(\alpha)][F(\alpha):K\cap F(\alpha)].\end{align*} Therefore, $[K(\alpha):K] = [F(\alpha):K\cap F(\alpha)]$, as claimed.

So we have that \begin{align*} n &= [F(\alpha):F]= [F(\alpha):F(\alpha)\cap K][F(\alpha)\cap K:F]\\ &= [K(\alpha):K][F(\alpha)\cap K:F] \\ &= d[F(\alpha)\cap K:F]. \end{align*} Since $n=dm$, then $$m = [F(\alpha)\cap K:F],$$ as claimed.

To see the action of $G$ is transitive, note that $F(\alpha)$ is isomorphic to $F(\beta)$ over $F$, since $\alpha$ and $\beta$ have the same irreducible over $F$; hence there is an isomorphism $\sigma\colon F(\alpha)\to F(\beta)$ that restricts to the identity on $F$ and sends $\alpha$ to $\beta$. Since $L$ is Galois over $F$, and $F(\alpha),F(\beta)\subseteq L$, $\sigma$ extends to an element of $G$.

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Thank you Arturo. Perhaps it's obvious, but why does $[F(\alpha):F(\alpha)\cap K] = [F(\alpha)K:(F(\alpha)\cap K)K]$? – yunone Nov 29 '11 at 20:03
@yunone: Ah, yes; I was afraid you were going to ask that. (-; I was on my way to a class, and one inequality is easy but the other one didn't seem so clear. Let me work it through and add it. – Arturo Magidin Nov 29 '11 at 20:07
Ok, thanks for that! I'm in no rush, so please take your time. – yunone Nov 29 '11 at 20:17
@yunone: I'm pretty sure there is a simpler way of doing it (just lifting the extension and making an appropriate observation), but the above works. – Arturo Magidin Nov 29 '11 at 21:46
You could simplify the discussion of the claim by appealing to a theorem from Galois theory: when $K/F$ is a (finite) Galois extension and $E/F$ is any extension, then $[KE:E] = [K:K\cap E]$. Taking $E = F(\alpha)$, we get $[K(\alpha):F(\alpha)] = [K:K \cap F(\alpha)]$. Therefore $[K(\alpha):K]=[F(\alpha):K\cap F(\alpha)]$. You are essentially reproving this standard theorem in the claim, and writing out those details makes things look more involved than they are. – KCd Apr 12 '12 at 12:07

This may also be viewed in the context of factorization of prime ideals in integral extensions of Dedekind domains (possibly a sledgehammer for a small task).

The prime (ideal) $(h(x)) \subset F[x]$ splits into products of powers of prime ideals $\prod_{i=1}^g (h_i(x))^{e_i}$ in $K[x]$. Then a standard theorem states $\sum_{i=1}^g e_if_i= n=[K:F]$ where $e_i$ are ramification indices, $f_i$ residual degrees, and $g$ the number of primes lying above $f(x)$. In the special case $K$ is Galois over $F$, all $e_i$'s are equal and all $f_i$'s are equal.

And so we have isomorphisms of residual fields by the transitive action of $Gal\,(K/F)$ on the set of $h_j$'s: $K[x]/(h_i(x))\cong K[x]/(h_j(x))$. So degree of $h_i(x) =$ degree of $h_j$. (As transitive action of Galois group needs just Chinese Remainder Theorem, this answer, I hope, is not a circular argument.)

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Your "standard theorem" should really have $[K(x):F(x)]$ on the right side, which then simplifies to $[K:F]$, and likewise for ${\rm Gal}(K(x)/F(x))$ rather than ${\rm Gal}(K/F)$ (they are isomorphic naturally). – KCd Apr 12 '12 at 11:55
@KCd: Thanks. This point is needed. These natural isomorphisms follow from linear disjointness of the two extensions K and F(x) over F – P Vanchinathan Apr 16 '12 at 4:33