# Question on normal field ext.

1. Dec 12, 2007

### 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

2. Dec 12, 2007

### 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 [itex] instead [tex]. This will make everything align nicely. Also, try to state clearly what you're trying to show at each step.