If $g,h \in L[X]$ prime divisors of f irreducible, than it exists $\phi \in Aut(L/K)$ with $\phi(g) = h$.

Question

Let f $\in$ K[X] be irreducible and L/K a normal field extension. If g,h $\in$ L[X] normalized prime divisor of f in L[X], so it exist a automorphism $\tau \in$ Aut(L/K) with $\tau$(g)=h.

I got so far:

Set f = gt = hs with t,s $\in$ L[X]. Than

• g divides hs
• h divides gt

and because of f irreducible follows with definition that

• g or t $\in L[X]^x=L^x$
• h or s $\in L[X]^x=L^x$

Since g,h prime, so irreducible, so $\notin L[X]^x=L^x$, it must be t,s $\in L^x$ and so we can set g = $h \cdot \frac{s}{t} := h \cdot b$ with $b \in L^x$.

Any comments or ideas how to go on?

Answer 1

You can write $f=\Pi_if^{n_i}_i$ where $f_i$ is prime in $L[X] $. The polynomial $\Pi_{h\in Aut(L/K), {f_1}^h\neq f_1} {f^{n_1}_1}^h$ is in $K[X] $ and divides $f$ so it is $f$. This implies the result.