# Field of Fractions and Extensions

• Oct 8th 2010, 03:10 PM
bleys
Field of Fractions and Extensions
Hello, I'd like some help regarding field extensions.
Let L:K be a field extension. $\displaystyle \alpha \in L$ is algebraic if and only if $\displaystyle K[\alpha] = K(\alpha) \simeq F[X]/(f)$ where f is the minimal polynomial of alpha in K.
Now before I even get to the $\displaystyle F[X]/(f)$ I'd like some clarification about the equality between the polynomials in alpha and the rational functions in alpha.
I know one inclusion is trivial; the other inclusion (showing that all non-zero elements in $\displaystyle K[\alpha]$ are invertible) is fine except the proof I've seen states 'it is enough to show that the inverse of $\displaystyle \alpha$ is in the ring'. I don't understand why this is enough? Generally $\displaystyle K[\alpha]$ is not of the form $\displaystyle k+t\alpha$ with k,t in K (right?) so how is showing $\displaystyle \alpha ^{-1}$ is in the ring enough to show all the other inverses are in the ring?
For the isomorphism I'm not sure what map to use..
• Oct 8th 2010, 06:36 PM
tonio
Quote:

Originally Posted by bleys
Hello, I'd like some help regarding field extensions.
Let L:K be a field extension. $\displaystyle \alpha \in L$ is algebraic if and only if $\displaystyle K[\alpha] = K(\alpha) \simeq F[X]/(f)$ where f is the minimal polynomial of alpha in K.
Now before I even get to the $\displaystyle F[X]/(f)$ I'd like some clarification about the equality between the polynomials in alpha and the rational functions in alpha.
I know one inclusion is trivial; the other inclusion (showing that all non-zero elements in $\displaystyle K[\alpha]$ are invertible) is fine except the proof I've seen states 'it is enough to show that the inverse of $\displaystyle \alpha$ is in the ring'. I don't understand why this is enough? Generally $\displaystyle K[\alpha]$ is not of the form $\displaystyle k+t\alpha$ with k,t in K (right?) so how is showing $\displaystyle \alpha ^{-1}$ is in the ring enough to show all the other inverses are in the ring?
For the isomorphism I'm not sure what map to use..

It is enough because clearly $\displaystyle K[\alpha]$ is an integral domain...(Wink)

Tonio
• Oct 9th 2010, 11:39 AM
bleys
Call me stupid, but I still don't understand :(
• Oct 9th 2010, 02:43 PM
tonio
Quote:

Originally Posted by bleys
Call me stupid, but I still don't understand :(

Not stupid, but perhaps al little careless: what's missing an integral domain to be a field? Well then, what must be proved in order to show some int. domain is actually a field?

Tonio
• Oct 10th 2010, 07:15 AM
bleys
Well, for any non-zero $\displaystyle a$ in the integral domain, you must show $\displaystyle a^{-1}$ is also in the integral domain. We've shown it for $\displaystyle \alpha$, what about the rest? How is knowing $\displaystyle \alpha ^{-1}$ is in $\displaystyle K[\alpha]$ imply any polynomial in $\displaystyle \alpha$ is invertible?
I'm not really sure where you're going; the best I could do is use induction on the degree of polynomials in $\displaystyle K[\alpha]$, but I don't think that's the kind of proof you're (or the book is) thinking about, even though it is quite simple and short.
• Oct 10th 2010, 08:14 AM
tonio
Quote:

Originally Posted by bleys
Well, for any non-zero $\displaystyle a$ in the integral domain, you must show $\displaystyle a^{-1}$ is also in the integral domain. We've shown it for $\displaystyle \alpha$, what about the rest? How is knowing $\displaystyle \alpha ^{-1}$ is in $\displaystyle K[\alpha]$ imply any polynomial in $\displaystyle \alpha$ is invertible?
I'm not really sure where you're going; the best I could do is use induction on the degree of polynomials in $\displaystyle K[\alpha]$, but I don't think that's the kind of proof you're (or the book is) thinking about, even though it is quite simple and short.

I think it's simpler than that, but deeper from an abstract algebra point of view: look at the ring homom. $\displaystyle \phi: K[x]\rightarrow K[\alpha]$

defined by $\displaystyle \phi(f(x)):=f(\alpha)$ . Since $\displaystyle K[\alpha]$ in an int. domain, $\displaystyle <f(x)>:=\ker\phi$ is a prime ideal, but since

$\displaystyle K[x]$ is an Euclidean domain this prime ideal is in fact maximal (and thus $\displaystyle f(x)$ is irreducible: the irr.

pol. of $\displaystyle \alpha$ , and thus $\displaystyle K[\alpha]$ is a field, and since $\displaystyle K[\alpha]\subset K(\alpha)$ , then...

Tonio
• Oct 10th 2010, 09:06 AM
bleys
Alright, I see. It didn't occur to me to use the substitution homomorphism...
Anyway, thanks for the help, Tonio!