$N|K$ normal $\Rightarrow F \in K[X]$ factors into $F=Q_1\cdots Q_r$ with $\text{deg}(Q_1)=\dots=\text{deg}(Q_r)$

Question

Let be $N|K$ a normal extension, $F \in K[X]$ monic, irreducible polynomial and $F=Q_1\cdots Q_r$ its factorization into monic, irreducible polynomials $Q_1, \dots, Q_r\in N[X]$. Prove that $\deg(Q_1)=\dots=\deg(Q_r)$.

Hint: Think of the set $\{Q_1^{\sigma}\mid \sigma\in \text{Aut}(N|K)\}$

My idea is to prove that $\{Q_1^{\sigma}\mid \sigma\in \text{Aut}(N|K)\}=\{Q_1, \dots, Q_r\}$, which proves that $\text{deg}(Q_k)=\text{deg}(Q_1)$ for every $k$, and then we are done.

I now how to prove $\subset$:

Take any $\sigma\in \text{Aut}(N|K)$. By definition it fixes $K$ and since $F\in K[X]$, we have $F=F^{\sigma}=(Q_1\cdots Q_r)^{\sigma}=Q_1^{\sigma}\cdots Q_r^{\sigma}$. By the unique factorization in $N[X]$, we must have $Q_1^{\sigma}=Q_k$ for some $k$, therefore $\{Q_1^{\sigma}\mid \sigma\in \text{Aut}(N|K)\}\subset\{Q_1, \dots, Q_r\}$.

To prove $\supset$, I went this far:

Suppose there is a $Q_k$ such that $Q_1^{\sigma}\neq Q_k$ for every $\sigma\in \text{Aut}(N|K)$ or, equivalently, $Q_k^{\sigma}\neq Q_1 \,\,\,\forall\sigma\in \text{Aut}(N|K)$.

I was hoping to conclude something like this: each $\sigma$ fixes $Q_k$ and therefore, $Q_k\in K[X]$, which is absurd, since $F$ is irreducible in $K[X]$

I don't even know if this argument can work, but that's the only thing I could come up with. Any ideas? Thanks!

Answer 1

What you need to show is that the Galois group action on the irreducible factors of $F(X)$ in $N[X]$ is transitive.

The quotient ring $N[X]/(F(X))$ injects into the direct product of rings $\prod_j N[X]/(Q_j(X))$. The ideals $Q_j$'s are distinct prime ideals hence maximal (we are in a PID $N[X]$), and so they are relatively prime. So the injective ring homomorphism is in fact surjective (Chinese Remainder theorem).

Answer 2

Another way to proceed. Let $k \in \{2, \ldots, r\}$. We'll show there exist an $\sigma \in \text{Gal}(N|K)$ such that $Q_1^\sigma = Q_r$. Let $\Omega$ be an algebraically closed field containing $N$. Let $\alpha, \beta \in \Omega$ be roots of $Q_1$ and $Q_k$. It is straightforward to note that $F(\alpha) = F(\beta) = 0$. Therefore, they are $K$-conjugates (since $F$ is irreducible over $K$) and there exists an isomorphism $\xi: K(\alpha) \to K(\beta)$ extending the identity isomorphism of $K$ and mapping $\alpha$ to $\beta$.

On the other hand, $N(\alpha)|K(\alpha)$ is an algebraic extension. So, it follows that we are able to extend $\xi$ to $\tilde{\xi}:N(\alpha) \to \Omega$. Since $N|K$ is a normal extension, we have that $\tilde\xi(N) \subseteq N$. Thus, the restriction $\left.\tilde{\xi}\right|_N:N \to N$ is an isomorphism.

Afterwards, it is not difficult to check that $\tilde\xi(N(\alpha)) = N(\beta)$. In fact, let $w \in N(\beta)$, there exists a polynomial $g(X) \in N[X] $ such that $w = g(\beta)$, because the extension $N(\beta)|N$ is algebraic. Since $\left.\tilde{\xi}\right|_N:N \to N$ is an isomorphism, we are able to find polynomial $h\in N[X]$ such that $h^\xi = g$. Now, just consider $z = h(\alpha) \in N(\alpha)$ and by construction $\xi(z)=w$. There other inclusion is easier.

Finally, we have proven that $\sigma :=\left.\xi\right|_{N(\alpha)}:N(\alpha) \to N(\beta) $ is an isomorphism. This fact, is equivalent to $P_{\alpha, N}^\sigma = P_{\beta, N}$. Since the polynomials $Q_1$ and $Q_k$ are monic irreducible polynomials, such that $Q_1(\alpha) = Q_k(\beta) = 0$. It follows $Q_1 = P_{\alpha, N}$ and $Q_k = P_{\beta, N}$. This allow us to conclude what it was asked.