Algebraic Extensions - Dummit and Foote, Propositions 11 and 12 .... ....

[Math Amateur]

I am reading Dummit and Foote, Chapter 13 - Field Theory.

I am currently studying Section 13.2 : Algebraic Extensions

I need some help with an aspect of Propositions 11 and 12 ... ...

Propositions 11 and 12 read as follows:
https://www.physicsforums.com/attachments/6606https://www.physicsforums.com/attachments/6607
Now Proposition 11 states that the degree of \(\displaystyle F( \alpha )\) over \(\displaystyle F\) is equal to the degree of the minimum polynomial ... ... that is

\(\displaystyle [ F( \alpha ) \ : \ F ] = \text{ deg } m_\alpha (x) = \text{ deg } \alpha
\)

... ... BUT ... ...... ... Proposition 12 states that ... "if \(\displaystyle \alpha\) is an element of an extension of degree \(\displaystyle n\) over \(\displaystyle F\), then \(\displaystyle \alpha\) satisfies a polynomial of degree at most \(\displaystyle n\) over \(\displaystyle F\) ... ... "Doesn't Proposition 11 guarantee that the polynomial (the minimum polynomial) is actually of degree equal to \(\displaystyle n\)?Can someone please explain in simple terms how these statements are consistent?Help will be appreciated ...

Peter

 

[Euge]

It appears you overlooked something simple. The second sentence of Proposition 12 considers an arbitrary extension of degree $n$, not $F(\alpha)$. For a concrete example, the number $\sqrt{3}$ is algebraic over $\Bbb Q$ of degree $2$, and it belongs to the extension $Q(\sqrt{2},\sqrt{3})$, of degree $4$ over $\Bbb Q$.

 

[Math Amateur]

It appears you overlooked something simple. The second sentence of Proposition 12 considers an arbitrary extension of degree $n$, not $F(\alpha)$. For a concrete example, the number $\sqrt{3}$ is algebraic over $\Bbb Q$ of degree $2$, and it belongs to the extension $Q(\sqrt{2},\sqrt{3})$, of degree $4$ over $\Bbb Q$.

Thanks Euge ... appreciate your help ...

... now reading text again carefully ...

Peter