# Thread: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

1. ## Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

I am reading Nicholson: Introduction to Abstract Algebra Section 6.2 Algebraic Extensions.

On page 282 the Corollary to Theorem 5 states the following: (see attachment for Theorem 5 and the Corollary)

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Corollary. Let $E \supseteq F$ be fields and let $u \in E$ be algebraic over F .

If $v \in F(u)$, then v is also algebraic over F and ${deg}_F(v)$ divides ${deg}_F(u)$.

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The proof begins as follows:

" Proof. Here $F(u) \supseteq F(v) \supseteq F$ ... ... etc etc

My problem is as follows:

How do you show formally and explicitly that $F(u) \supseteq F(v) \supseteq F$

Would appreciate some help.

Peter

2. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

Since $v \in F(u)$, if we consider $F(u)$ as a vector space over $F$ with basis $\{u^0,u^1,\ldots, u^{\deg(u)-1}\}$, then $\{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ must span a linear subspace (since $v$ can be represented as a linear combination of elements of the basis for $F(u)$). And $\{1\}$ spans a linear subspace of both spaces. So, $\span(\{1\}) \subseteq \span(\{1,v^1, \ldots, v^{\deg(u)-1}\}) \subseteq F(u)$. Since $F(v)$ can be viewed as the linear space spanned by the basis of powers of $v$, we have $F \subseteq F(v) \subseteq F(u)$.

3. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

Originally Posted by SlipEternal
Since $v \in F(u)$, if we consider $F(u)$ as a vector space over $F$ with basis $\{u^0,u^1,\ldots, u^{\deg(u)-1}\}$, then $\{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ must span a linear subspace (since $v$ can be represented as a linear combination of elements of the basis for $F(u)$). And $\{1\}$ spans a linear subspace of both spaces. So, $\span(\{1\}) \subseteq \span(\{1,v^1, \ldots, v^{\deg(u)-1}\}) \subseteq F(u)$. Since $F(v)$ can be viewed as the linear space spanned by the basis of powers of $v$, we have $F \subseteq F(v) \subseteq F(u)$.
Thanks SlipEternal, that is pretty clear, but just one issue ...

You write:

" ... ... if we consider $F(u)$ as a vector space over $F$ with basis $\{u^0,u^1,\ldots, u^{\deg(u)-1}\}$, then $\{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ must span a linear subspace ... ..."

How do we know for sure that $\{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ is a subspace ... why is it not possible that $\{u^0,u^1,\ldots, u^{\deg(u)-1}\} \subseteq \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ - that is $\{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ is a superspace of F(u) - that is $F(u) \subseteq F(v)$

Can you clarify?

Peter

4. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

Originally Posted by Bernhard
Thanks SlipEternal, that is pretty clear, but just one issue ...

You write:

" ... ... if we consider $F(u)$ as a vector space over $F$ with basis $\{u^0,u^1,\ldots, u^{\deg(u)-1}\}$, then $\{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ must span a linear subspace ... ..."

How do we know for sure that $\{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ is a subspace ... why is it not possible that $\{u^0,u^1,\ldots, u^{\deg(u)-1}\} \subseteq \{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ - that is $\{v^0, v^1, \ldots, v^{\deg(u)-1}\}$ is a superspace of F(u) - that is $F(u) \subseteq F(v)$

Can you clarify?

Peter
You are given that $v \in F(u)$, so since $F(u)$ is a field, $v^i \in F(u) \forall i\in \mathbb{Z}$. Hence, every element of $\{1,v,\ldots, v^{\deg(u)-1}\}$ is expressible as a linear combination of elements of $\{1,u,\ldots, u^{\deg(u)-1}\}$.