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.

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!

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.

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.

Thanks!

Ooh! Ooh! I think I got it. Could anyone please check if the reasoning is flawless?

I had to use some properties of the reals, so I came up with these.

Well ordering property: suppose x>0, then x=y^2 for some y, so f(x)=f(y)^2>0. So if x>0 then f(x)>0.

If b>a then b-a>= so f(b-a)=f(b)-f(a)>0, so f preserves order.

Completeness. Given a partition of Q into two disjoint subsets A,B with every element in A smaller than any element of B, there exists a unique real number 'inbetween', and vice versa. So if x is irrational, take A to be the set of rational smaller than x and B the set of rationals larger than x. Then f is the identity on A and on B and because of order preservation sup(A)=inf(B)=x is also mapped to itself. Hence f is the identity.

Thanks Matt,

Well, the kernel is an ideal and the only ideals in a field F are the trivial ones (0) and F. Unless the map F-> maps into the trivial ring {0} (that isn't a field right?) the kernel can't be F, since 1 has to go to 1. So the kernel is {0} which means the map is injective.

So if the map is between fields it's always injective. I know from memory that {0} isn't a field, but I can't remember why, since the definition for a field is a commutative ring with R*=R\{0} which holds for {0}.

But anyway, that doesn't make F invertible right? For example, the inclusion f:Q->R is not surjective.

And I don't see how it helps, since it is given that f is invertible. I must show it is the identity.

Hi everyone!

This seemingly easy question is proving difficult.

I have to show that a field automorphism of R to R (real numbers)

is the identity.

What I've done:

Let f:R -> R

be an field automorphism.

Since f(1)=1 is a generator of Z, f is the identity on Z. Because multiplicative inverses are sent to their inverses: f(1/x)=1/f(x), it follows that f is also the identity on Q.

I`m not even sure if I`m on the right track. I obviously have to use some property of the reals now. A previous exercise which I very likely need to use asks to show that if x>0 then f(x)>0.

Any help is appreciated.