More Algebra

[Zareon]

Why is algebra so difficult? Other students in my class seem to have far less difficulty with it then I do and other math subjects are so much easier in comparison. I have trouble remembering all the theorems relating properties between definitions. I can't visualize anything, and I do practice alot. If I don't look at it for one week I already forget all kinds of stuff, while I can remember most things from other topics easily. I can follow the proofs in the book, but mostly by checking the logic, I can't see (and thus don't understand) the significance or the consequences of what has been proved.

Any tips on how to make my algebra-life easier is greatly appreciated.

 

I want to prove the following: If F is a field and the element 'a' transcendental over F, then the algebraic closure of F in F(a) is equal to F.

 

Not sure where to start. The algebraic closure of F in F(a) is {x in F(a)|x algebraic over F}. Every element of F is ofcourse an element of F(a).

So take an x in F(a) which is algebraic over F. I want to show it is an element of F. So I guess I could show x has an inverse? And I probably have to use that F(a) is isomorphic to F(X) (field of rational functions over F).

But I don't see how it all comes together.

 

Thanks for any help.

[matt grime]

Just consider the minimal polys. write out the definitions, if something has a poly in F[a] with certain properties then it has one in F too since 'a' is transcendental.

 

 

and it's hard exactly because you need to memorize lots of definitions; but this makes the theorems harder to state and thus easier to prove.

[Zareon]

if something has a poly in F[a] with certain properties then it has one in F too since 'a' is transcendental.

Well' date=' that's one of those things I don't see.

So do I look at the minimum polynomial of x over F? If F is equal to its alg. closure, then it means F is already alg. closed, right? So I can show that every nonconstant poly. in F[X']=F[a] has a root in F, or that every poly. in F[X] can be written as a product of linear factors.

[matt grime]

But if you've got something in the alg. closure of the small field satisfying an irred poly with coefficients somehow involving the a's, then you've gotten swapping roles over a poly that a satsifies over the small field, contradicting the fact that a is transcendental.

[Zareon]

I think I see. So what you're saying is that if x=sum c_i a^i in F(a) is algebraic over F, then there should a polynomial f=sum d_j X^j with x as a root. And plugging in f(x) would give a polynomial with 'a' as a root, contradicting its transcendentalness.

 

That makes sense, thanks a lot!

[matt grime]

Check, it please, it was just a guess on my part.

[Zareon]

Well, I worked out the details and it looks good. Unless x is in F you will have a polynomial with a as a root. In retrospect the problems are usually not that difficult, but you only find that out afterwards :/

Guess I just have to practice and play with these concepts more.