Do All Factors of a Polynomial in a Normal Extension Have Equal Degree?

[chocok]

Question:
If g(x)\in K[x] and 1< deg(g)=n.
Given that G/K is a normal field ext., if g(x)=g1(x)*...*gk(x)\in G[x],
then deg(g1)=...=deg(gk)

My attempt:
I let G = K adjoins the coefficients of gi's.
Let \alpha be a root of g.
Notice that K \subseteq G \subseteq K( \alpha) = G( \alpha)

We can express g(x) = irr( \alpha, G) *p(x) for some p(x) \in K[x].
if i let g1 be irr( \alpha, G),
then deg(g1) = deg( irr( \alpha, G)) = [G(\alpha):G]=[K(\alpha):G]

Next, we do the same thing again with another root, say \beta.
g(x) = irr(\beta, G) *q(x) for some q(x) \in K[x]
if i let g2 be irr( \beta, G),
then deg(g2) = deg( irr( \beta, G)) = [G(\beta):G]=[K(\beta):G]

if we proceed in the same way, deg(gi) can be found to be equal to [K(\theta):G] for some root \theta of g.

Next, we observe that since K(\alpha) and K(\beta) are normal extnesion of K, they split into linear factors for g(x). so K(\alpha) = K(\beta) (and this in fact implies to K adjoining other roots of g)

so deg(g1)=...=deg(gk)

please tell me if there's anything wrong with it.. this question is a bit too advanced for me

 

[morphism]

I suppose we're assuming that g(x) is irreducible over K. I skimmed through your solution, and I think you have the right ideas, but your write-up is very hard to read - and some things don't make sense, e.g. what do you mean when you say "K(a) splits into linear factors for g(x)"? If I'm not mistaken, this problem is from Chapter V of Hungerford's Algebra, correct? It was one of my favorite problems!

Here are some tips that will help make your post more readable: When you want to use inline TeX (i.e. during sentences), use instead . This will make everything align nicely. Also, try to state clearly what you&amp;#039;re trying to show at each step.