# Is Simple Extension Minimal?

• May 21st 2006, 09:29 AM
ThePerfectHacker
Is Simple Extension Minimal?
If you have fields $\displaystyle F\leq E$ with $\displaystyle \alpha \in E$ algebraic over $\displaystyle F$, show that,
$\displaystyle F(\alpha)$ is the minimal field containing $\displaystyle F$ and $\displaystyle \alpha$.

(Note: $\displaystyle F(\alpha)=\phi_{\alpha}[F[x]]$-a simple extension).
• May 21st 2006, 10:59 AM
rgep
The map $\displaystyle \phi_\alpha : F[X] \rightarrow E$ given by evaluation at $\displaystyle \alpha$ is a ring homomorphism and hence its image, which you denote $\displaystyle F(\alpha)$, is a subring of E. All you need to show is that it is closed under taking multiplicative inverses. Note for use in a moment that a subring of a field is an integral domain. The fact that $\displaystyle \alpha$ is algebraic is equivalent to saying that $\displaystyle F(\alpha)$ is a finite-dimensional F-vector space. Let $\displaystyle \beta$ be a non-zero element of $\displaystyle F(\alpha)$. Multiplication by $\displaystyle \beta$ is an F-linear endomorphism and has trivial kernel (integral domain), hence is surjective by finiteness of the dimension and the rank--nullity formula. So 1 is in the image, hence a multiple of $\displaystyle \beta$, which is thus invertible.
• May 21st 2006, 01:29 PM
ThePerfectHacker
I was thinking about this today, maybe you can do this?

You have $\displaystyle F\leq E$ with $\displaystyle \alpha$ algebraic over $\displaystyle F$. Let there exist a field $\displaystyle K$ with $\displaystyle \alpha \in K$ such as,
$\displaystyle F\leq K\leq F(\alpha)\leq E$,
Any element $\displaystyle \beta \in F(\alpha)$ can be expressed as,
$\displaystyle \beta =a_0+a_1\alpha +...+ a_{n-1}\alpha ^{n-1}$ with $\displaystyle a_i \in K$ because every finite dimensional vector basis has a finite basis. Therefore $\displaystyle \beta \in K$ thus, $\displaystyle K\leq F$ and $\displaystyle F\leq K$ thus, $\displaystyle K=F$.

Am I missing something?
• May 21st 2006, 01:34 PM
rgep
I realise that I didn't actually address the issue of being the minimal field, just that it actually was a field. But any field containing $\displaystyle \alpha$ clearly contains any polynomial in $\displaystyle \alpha$. Hence the minimality.
• May 21st 2006, 01:36 PM
ThePerfectHacker
Quote:

Originally Posted by rgep
$\displaystyle \alpha$ clearly contains any polynomial in $\displaystyle \alpha$. Hence the minimality.

What do you mean by that?
• May 21st 2006, 09:56 PM
rgep
Any field that contains F and contains $\displaystyle \alpha$ must contain any polynomial in $\displaystyle \alpha$ with coefficients in F. The previous part of my argument showed that the ring comprised of polynomials in $\displaystyle \alpha$ with coefficients in F is actually a field. Hence it is the minimal subfield of E containing both F and $\displaystyle \alpha$.
• May 22nd 2006, 02:12 PM
ThePerfectHacker
Quote:

Originally Posted by rgep
Any field that contains F and contains $\displaystyle \alpha$ must contain any polynomial in $\displaystyle \alpha$ with coefficients in F. The previous part of my argument showed that the ring comprised of polynomials in $\displaystyle \alpha$ with coefficients in F is actually a field. Hence it is the minimal subfield of E containing both F and $\displaystyle \alpha$.

You mean a polynomial F[x] evaluated at $\displaystyle \alpha$.
• May 23rd 2006, 02:47 AM
DMT
Isn't this all circular? I thought $\displaystyle F(\alpha)$ was defined to be the smallest the field containing both F and $\displaystyle \alpha$.
• May 23rd 2006, 01:17 PM
ThePerfectHacker
Quote:

Originally Posted by DMT
Isn't this all circular? I thought $\displaystyle F(\alpha)$ was defined to be the smallest the field containing both F and $\displaystyle \alpha$.

Maybe in some texts. The version I seen is that F(a) is defined as above.