Thread: 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..

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...

Tonio

Call me stupid, but I still don't understand

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

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.

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

Alright, I see. It didn't occur to me to use the substitution homomorphism...
Anyway, thanks for the help, Tonio!