Math Help - Field Theory - The Degree of alpha over F

1. Field Theory - The Degree of alpha over F

I am trying to clarify my understanding of Proposition 11 of Dummit and Foote Ch13 Field Theory concerning the degree of $\alpha$ over F.

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Proposition 11. Let $\alpha$ be algebraic over the field F and let $F(\alpha)$ be the field generated by $\alpha$ over F.

Then $F(\alpha) \cong F[x]/(m_{\alpha}(x))$

so that in particular $[F(\alpha) \ : \ F] = deg \ m_{\alpha}(x) = deg \ \alpha$

i.e. the degree of $\alpha$ over F is the degree of the extension it generates over F

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However Corollary 7 (Dummit and Foote page 518) states the following:

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Corollary 7. Suppose in Theorem 6 that p(x) is of degree n. Then

$F(\alpha) = \{ a_0 + a_1 {\alpha} + a_2 {\alpha}^2 + ... ... a_{n-1} {\alpha}^{n-1} \ | \ a_0, a_1, ... ... a_{n-1} \in F \} \subseteq K$

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Given that $F(\alpha)$ consists of polynomials of degree (n-1) should not the degree of $[F(\alpha) \ : \ F] = deg \ m_{\alpha}(x) - 1 = deg \ \alpha -1$ - that is the degree of $\alpha$ over F be one less than the degree of the minimal polynomial?

Can someone please clarify this situation for me

Peter

2. Re: Field Theory - The Degree of alpha over F

I have just been reflecting about my own question above.

It is possible that $F(\alpha)$ viewed as a vector space of deg (n-1) polynomials would have the following as a basis:

$1, {\alpha}, {\alpha}^2, {\alpha}^3, ... ... {\alpha}^{n-1}$

Then the degree of the space would be n which would equal the degree of the minimal polynomial involved.

If this thought is actually correct, could someone please confirm this as the case.

Peter

3. Re: Field Theory - The Degree of alpha over F

That is correct. Any higher degree polynomials would contain a linear dependency (some linear combination with nonzero coefficients would sum to zero).

4. Re: Field Theory - The Degree of alpha over F

Thanks SlipEternal

Sorry my thank you was so late - my day job intervened! :-(