MHB Field Theory: Nicholson, 6.2 Algebraic Extensions - Example 14 (p. 282) Solution

[Math Amateur]

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

Example 14 on page 282 (see attachment) reads as follows:

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Example 14. Let E \supseteq F be fields and let u, v \in E.

If u and u + v are algebraic over F, show that v is algebraic over F.

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In the solution, Nicholson writes the following:

Solution. Write L = F(u + v) so that L(u) = F(u, v). ... ... etc etc
Can someone please show me (formally and exactly) why L = F(u + v) \Longrightarrow L(u) = F(u, v).

Peter

[This has also been posted on MHF]

 

[caffeinemachine]

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

Example 14 on page 282 (see attachment) reads as follows:

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

Example 14. Let E \supseteq F be fields and let u, v \in E.

If u and u + v are algebraic over F, show that v is algebraic over F.

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

In the solution, Nicholson writes the following:

Solution. Write L = F(u + v) so that L(u) = F(u, v). ... ... etc etc
Can someone please show me (formally and exactly) why L = F(u + v) \Longrightarrow L(u) = F(u, v).

Peter

[This has also been posted on MHF]

We have $L=F(u+v)$. Thus $L(u)=(F(u+v))(u)=F(u+v,u)$.

We claim that $F(u+v,u)=F(v,u)$. Now that $u+v,u\in F(v,u)$. Thus $F(u+v,u)\subseteq F(v,u)$. Also, $u,v\in F(u+v,u)$. This is simply because $u+v-u$ is in $F(u+v,u)$. So we have $F(v,u)\subseteq F(u+v,u)$.
Thus we have $F(u+v,u)=F(v,u)$.

From here it's easy to get $L=F(v,u)$.