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I am trying to clarify my understanding of Proposition 11 of Dummit and Foote Ch13 Field Theory concerning the degree of \alpha over F.
Proposition 11 reads as follows:
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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
-------------------------------------------------------------------------------------------------------------------------------------------------------------- However Corollary 7 (Dummit and Foote page 518) states the following: -----------------------------------------------------------------------------------------------------------------------------------------------------------------
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
[This has also been posted on MHF]
Proposition 11 reads as follows:
---------------------------------------------------------------------------------------------------------------------------------------------------------
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
-------------------------------------------------------------------------------------------------------------------------------------------------------------- However Corollary 7 (Dummit and Foote page 518) states the following: -----------------------------------------------------------------------------------------------------------------------------------------------------------------
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
-------------------------------------------------------------------------------------------------------------------------------------------------------------------
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
[This has also been posted on MHF]