MHB Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - pages 281-282

[Math Amateur]

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 etcMy problem is as follows:

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

Would appreciate some help.

Peter

 

[caffeinemachine]

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)

------------------------------------------------------------------------------------------------------------------------

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).

------------------------------------------------------------------------------------------------------------------------

The proof begins as follows:

" Proof. Here F(u) \supseteq F(v) \supseteq F ... ... etc etcMy problem is as follows:

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

Would appreciate some help.

Peter

We have $v\in F(u)$. We know that $F(v)=\{f(v)/g(v):f(x),g(x)\in F[x], g(v)\neq 0\}$. Note that $f(v)\in F(u)$. Why? Simply because $F(u)$ is a field and hence is closed under addition and multiplication. Note that $f(v)$ is a sum of terms like $av^j$ where $a\in F$. Now both of $a$ and $v$ are in $F(u)$. Thus $av^j$ is in $F(u)$ too. And thus $f(v)\in F(u)$.
Same goes for $g(v)$ and hence $F(v)\subseteq F(u)$.

Another way of looking at is is that $F(v)$ is the smallest field containing $v$ and $F$. But $v$ and $F$ are both contained in $F(u)$. Thus $F(v)$ is contained in $F(u)$.