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

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

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 $\displaystyle E \supseteq F$ be fields and let $\displaystyle u, v \in E$.

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 $\displaystyle L = F(u + v) \Longrightarrow L(u) = F(u, v)$.

Peter

2. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as a+ b(u+ v) where a and b are members of F. L(u) is the smallest field that contains all members of L and u. Any such number, then, can be written as (c+ d(u+ v))+ jv where c and d are members of F and j is a member of L, not necessarily a member of F. So any such number is of the form (c+ d(u+ v))+ (e+ f(u+ v))v= c+ d(u+ v)+ ev+ fuv+ fv^2. Of course, uv and v^2 are members of F(u,v) as are c, d(u+v), and ev. So it is clearly true that L(u) is a subset of F(u,v). It remains to be shown that F(u,v) is a subset of L(u).

Any member of F(u,v) is of the form a+ bu+ cv. That is the same as "c+ d(u+ v)+ ev+ fuv+ fv^2= c+ du+ (d+e)v+ fuv+ fv^2" if we take c= a, d= b, e= c- b, and f= 0.

3. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

$\displaystyle 1\cdot (u+v) + (-1)\cdot u = v$ That's about all there is to it.

4. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

Originally Posted by HallsofIvy
L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as a+ b(u+ v) where a and b are members of F. L(u) is the smallest field that contains all members of L and u. Any such number, then, can be written as (c+ d(u+ v))+ jv where c and d are members of F and j is a member of L, not necessarily a member of F. So any such number is of the form (c+ d(u+ v))+ (e+ f(u+ v))v= c+ d(u+ v)+ ev+ fuv+ fv^2. Of course, uv and v^2 are members of F(u,v) as are c, d(u+v), and ev. So it is clearly true that L(u) is a subset of F(u,v). It remains to be shown that F(u,v) is a subset of L(u).

Any member of F(u,v) is of the form a+ bu+ cv. That is the same as "c+ d(u+ v)+ ev+ fuv+ fv^2= c+ du+ (d+e)v+ fuv+ fv^2" if we take c= a, d= b, e= c- b, and f= 0.
Thanks for the help, HallsofIvy!

Just one issue:

You write:

" L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as a+ b(u+ v) where a and b are members of F. ... ... etc"

But F(u + v) is a vector space over F with basis $\displaystyle \{ (u+ v)^0 , (u+ v)^1, (u+ v)^2, ... ... , (u+ v)^{deg(u + v) - 1} \}$

So shouldn't we write:

"L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as $\displaystyle a_0 + a_1 (u + v) + a_2 (u + v)^2 + ... ... + a_{deg(u + v) - 1} (u + v)^{deg(u + v) - 1}$

So writing: "Any such number can be written as a+ b(u+ v) where a and b are members of F." seems to imply that the degree of the minimal polynomial - or the degree of (u + v) is 2?

Can you clarify?

Peter

5. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

Originally Posted by Bernhard
Thanks for the help, HallsofIvy!

Just one issue:

You write:

" L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as a+ b(u+ v) where a and b are members of F. ... ... etc"

But F(u + v) is a vector space over F with basis $\displaystyle \{ (u+ v)^0 , (u+ v)^1, (u+ v)^2, ... ... , (u+ v)^{deg(u + v) - 1} \}$

So shouldn't we write:

"L= F(u+ v) is the smallest field that contains all members of F and the number u+ v. Any such number can be written as $\displaystyle a_0 + a_1 (u + v) + a_2 (u + v)^2 + ... ... + a_{deg(u + v) - 1} (u + v)^{deg(u + v) - 1}$

So writing: "Any such number can be written as a+ b(u+ v) where a and b are members of F." seems to imply that the degree of the minimal polynomial - or the degree of (u + v) is 2?

Can you clarify?

Peter
You can generate every element of $\displaystyle L = F(u+v)$ by all products of the form $\displaystyle a+b(u+v)$. This is obvious since $\displaystyle [0+1(u+v)]^k$ gives the basis you mentioned. So, to show that $\displaystyle L(u) = F(u,v)$, you only need to show that for any $\displaystyle a + bu \in L(u)$ with $\displaystyle a,b \in L$, there exists $\displaystyle x \in F(u,v)$ such that $\displaystyle x = a+bu$. Then, for any $\displaystyle c + du + ev \in F(u,v)$ with $\displaystyle c,d,e \in F$, show that there exists $\displaystyle y \in L(u)$ such that $\displaystyle c+du+ev = y$.

So, since $\displaystyle (u+v) \in L, u \in L(u)$, we know $\displaystyle v \in L(u)$. Since $\displaystyle F \subseteq F(u+v) = L \subseteq L(u)$, we know $\displaystyle c,d,e \in L(u)$. Hence, $\displaystyle c+du+ev \in L(u)$.

6. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

V
Originally Posted by SlipEternal
You can generate every element of $\displaystyle L = F(u+v)$ by all products of the form $\displaystyle a+b(u+v)$. This is obvious since $\displaystyle [0+1(u+v)]^k$ gives the basis you mentioned. So, to show that $\displaystyle L(u) = F(u,v)$, you only need to show that for any $\displaystyle a + bu \in L(u)$ with $\displaystyle a,b \in L$, there exists $\displaystyle x \in F(u,v)$ such that $\displaystyle x = a+bu$. Then, for any $\displaystyle c + du + ev \in F(u,v)$ with $\displaystyle c,d,e \in F$, show that there exists $\displaystyle y \in L(u)$ such that $\displaystyle c+du+ev = y$.

SlipEternal, HallsofIvy, thanks so much for your help on this Example!!

peter
So, since $\displaystyle (u+v) \in L, u \in L(u)$, we know $\displaystyle v \in L(u)$. Since $\displaystyle F \subseteq F(u+v) = L \subseteq L(u)$, we know $\displaystyle c,d,e \in L(u)$. Hence, $\displaystyle c+du+ev \in L(u)$.

7. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 14, page

SlipEternal, HallsofIvy, Thanks so much for your help on this example

PS sorry for the problem with the formatting of this message on the previous post

Peter