Is Simple Extension Minimal?

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May 21st 2006, 09:29 AM
ThePerfectHacker

Is Simple Extension Minimal?

If you have fields $F\leq E$ with $\alpha \in E$ algebraic over $F$ , show that,
$F(\alpha)$ is the minimal field containing $F$ and $\alpha$ .

(Note: $F(\alpha)=\phi_{\alpha}[F[x]]$ -a simple extension).
May 21st 2006, 10:59 AM
rgep

The map $\phi_\alpha : F[X] \rightarrow E$ given by evaluation at $\alpha$ is a ring homomorphism and hence its image, which you denote $F(\alpha)$ , 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 $\alpha$ is algebraic is equivalent to saying that $F(\alpha)$ is a finite-dimensional F-vector space. Let $\beta$ be a non-zero element of $F(\alpha)$ . Multiplication by $\beta$ 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 $\beta$ , which is thus invertible.
May 21st 2006, 01:29 PM
ThePerfectHacker

I was thinking about this today, maybe you can do this?

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

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 $\alpha$ clearly contains any polynomial in $\alpha$ . Hence the minimality.
May 21st 2006, 01:36 PM
ThePerfectHacker

Quote:

Originally Posted by rgep

$\alpha$ clearly contains any polynomial in $\alpha$ . Hence the minimality.

What do you mean by that?
May 21st 2006, 09:56 PM
rgep

Any field that contains F and contains $\alpha$ must contain any polynomial in $\alpha$ with coefficients in F. The previous part of my argument showed that the ring comprised of polynomials in $\alpha$ with coefficients in F is actually a field. Hence it is the minimal subfield of E containing both F and $\alpha$ .
May 22nd 2006, 02:12 PM
ThePerfectHacker

Quote:

Originally Posted by rgep

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

You mean a polynomial F[x] evaluated at $\alpha$ .
May 23rd 2006, 02:47 AM
DMT

Isn't this all circular? I thought $F(\alpha)$ was defined to be the smallest the field containing both F and $\alpha$ .
May 23rd 2006, 01:17 PM
ThePerfectHacker

Quote:

Originally Posted by DMT

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

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