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- #1

- Jun 22, 2012

- 2,918

Proposition 11 reads as follows:

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**Proposition 11.**Let [TEX] \alpha [/TEX] be algebraic over the field F and let [TEX] F(\alpha) [/TEX] be the field generated by [TEX] \alpha [/TEX] over F.

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

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

i.e. the degree of [TEX] \alpha [/TEX] 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

[TEX] 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 [/TEX]

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

Can someone please clarify this situation for me

Peter

[This has also been posted on MHF]