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Degree of extension invariant upto isomorphism?

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Let $K$ be a field and $F_1$ and $F_2$ be subfields of $K$. Assume that $F_1$ and $F_2$ are isomorphic as fields. Further assume that $[K:F_1]$ is finite and is equal to $n$.

Is it necessary that $[K:F_2]$ is finite and is equal to $n$??
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I have not found this question in a book so I don't know the answer to the above question. I could not construct a counterexample.
If the isomorphism between $F_1$ and $F_2$ can be extended to an automorphism of $K$ (this probably required additional hypothesis) then the result can be proved in the affirmative.
 

PaulRS

Member
Jan 26, 2012
37
Here is an idea:

Let $F = {\mathbb Q}$ (to fix ideas). Consider $F_1 = F(x)$ and $F_2 = F(x^2)$, and finally $K = F(x)$. The field $F_1$ should be isomorphic to $F_2$ since $x^2$ is trascendental over $F$.

Clearly $K/F_1$ has degree 1. However ;).
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Here is an idea:

Let $F = {\mathbb Q}$ (to fix ideas). Consider $F_1 = F(x)$ and $F_2 = F(x^2)$, and finally $K = F(x)$. The field $F_1$ should be isomorphic to $F_2$ since $x^2$ is trascendental over $F$.

Clearly $K/F_1$ has degree 1. However ;).
So $K/F_2$ doesn't have degree $1$ simply because $1$ and $x$ in $K$ are LI over $F_2$. Is that right?
 

PaulRS

Member
Jan 26, 2012
37
So $K/F_2$ doesn't have degree $1$ simply because $1$ and $x$ in $K$ are LI over $F_2$. Is that right?
Right. Otherwise $x\in F_2$ and so $x = f(x^2) / g(x^2)$ for some polynomials $f$ and $g$ ($g\neq 0$), which you can check is nonsense.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Right. Otherwise $x\in F_2$ and so $x = f(x^2) / g(x^2)$ for some polynomials $f$ and $g$ ($g\neq 0$), which you can check is nonsense.
That's great Paul.

I have one more question.

I think that $[K:F_2]$ is finite.
My Argument:
Clearly $K=F_2(x)$. Now define a polynomial $p(y)=x^2y^3-x^4y$ in $F_2[y]$. Clearly $p(x)=0$. Thus $[K:F_2]$ is finite.

Is this okay?
 

PaulRS

Member
Jan 26, 2012
37
That's great Paul.

I have one more question.

I think that $[K:F_2]$ is finite.
My Argument:
Clearly $K=F_2(x)$. Now define a polynomial $p(y)=x^2y^3-x^4y$ in $F_2[y]$. Clearly $p(x)=0$. Thus $[K:F_2]$ is finite.

Is this okay?
Correct. :)

There is a simpler polynomial $p(y)\in F_2[y]$ such that $p(x) = 0$, can you find it?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Correct. :)

There is a simpler polynomial $p(y)\in F_2[y]$ such that $p(x) = 0$, can you find it?
$p(y)=y^2-x^2$. That's just awesome!!