Field of Fractions and Extensions

**bleys** · October 8th 2010, 03:10 PM

Hello, I'd like some help regarding field extensions.
Let L:K be a field extension. $\alpha \in L$ is algebraic if and only if $K[\alpha] = K(\alpha) \simeq F[X]/(f)$ where f is the minimal polynomial of alpha in K.
Now before I even get to the $F[X]/(f)$ 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 $K[\alpha]$ are invertible) is fine except the proof I've seen states 'it is enough to show that the inverse of $\alpha$ is in the ring'. I don't understand why this is enough? Generally $K[\alpha]$ is not of the form $k+t\alpha$ with k,t in K (right?) so how is showing $\alpha ^{-1}$ 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..

**tonio** · October 8th 2010, 06:36 PM

Originally Posted by bleys

Hello, I'd like some help regarding field extensions.
Let L:K be a field extension. $\alpha \in L$ is algebraic if and only if $K[\alpha] = K(\alpha) \simeq F[X]/(f)$ where f is the minimal polynomial of alpha in K.
Now before I even get to the $F[X]/(f)$ 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 $K[\alpha]$ are invertible) is fine except the proof I've seen states 'it is enough to show that the inverse of $\alpha$ is in the ring'. I don't understand why this is enough? Generally $K[\alpha]$ is not of the form $k+t\alpha$ with k,t in K (right?) so how is showing $\alpha ^{-1}$ 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 $K[\alpha]$ is an integral domain...

Tonio

**bleys** · October 9th 2010, 11:39 AM

Call me stupid, but I still don't understand

**tonio** · October 9th 2010, 02:43 PM

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

**bleys** · October 10th 2010, 07:15 AM

Well, for any non-zero $a$ in the integral domain, you must show $a^{-1}$ is also in the integral domain. We've shown it for $\alpha$ , what about the rest? How is knowing $\alpha ^{-1}$ is in $K[\alpha]$ imply any polynomial in $\alpha$ 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 $K[\alpha]$ , 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.

**tonio** · October 10th 2010, 08:14 AM

Originally Posted by bleys

Well, for any non-zero $a$ in the integral domain, you must show $a^{-1}$ is also in the integral domain. We've shown it for $\alpha$ , what about the rest? How is knowing $\alpha ^{-1}$ is in $K[\alpha]$ imply any polynomial in $\alpha$ 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 $K[\alpha]$ , 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. $\phi: K[x]\rightarrow K[\alpha]$

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

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

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

Tonio

**bleys** · October 10th 2010, 09:06 AM

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

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