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 $$F( \alpha )$$ over $$F$$ is equal to the degree of the minimum polynomial ... ... that is

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

... ... BUT ... ...... ... Proposition 12 states that ... "if $$\alpha$$ is an element of an extension of degree $$n$$ over $$F$$, then $$\alpha$$ satisfies a polynomial of degree at most $$n$$ over $$F$$ ... ... "Doesn't Proposition 11 guarantee that the polynomial (the minimum polynomial) is actually of degree equal to $$n$$?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