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Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page 282

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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 [TEX] E \supseteq F [/TEX] be fields and let [TEX] u, v \in E [/TEX].

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 [TEX] L = F(u + v) \Longrightarrow L(u) = F(u, v) [/TEX].

Peter

[This has also been posted on MHF]
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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 [TEX] E \supseteq F [/TEX] be fields and let [TEX] u, v \in E [/TEX].

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 [TEX] L = F(u + v) \Longrightarrow L(u) = F(u, v) [/TEX].

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 $L=F(v,u)$.