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

Proposition 11 reads as follows:

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

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

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

i.e. the degree of $\displaystyle \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

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

Can someone please clarify this situation for me

Peter