Finding the Intersection of Subfields in Finite Fields

[mathmari]

Hey!

Let $\mathbb{F}_{p^m}$ and $\mathbb{F}_{p^n}$ be subfields of $\overline{\mathbb{Z}}_p$ with $p^n$ and $p^m$ elements respectively.

To find the field $\mathbb{F}_{p^m} \cap \mathbb{F}_{p^n}$ I have done the following:

Let $\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} = \mathbb{F}_{p^d}$.

$\mathbb{F}_{p^a} \leq \mathbb{F}_{p^b} \Leftrightarrow a \mid b$

We have that $\mathbb{F}_{p^d} \leq \mathbb{F}_{p^n} \Rightarrow d \mid n$ and $\mathbb{F}_{p^d} \leq \mathbb{F}_{p^m} \Rightarrow d \mid m$

So, $d \mid \gcd(n,m)$. We have that $\gcd(n, m) \mid m \Rightarrow \mathbb{F}_{p^{\gcd(n,m)}} \leq \mathbb{F}_{p^m}$ and $\gcd(n, m) \mid n \Rightarrow \mathbb{F}_{p^{\gcd(n,m)}} \leq \mathbb{F}_{p^n}$Does it stand that if $A \leq B$ and $A \leq C$, then $A\leq (B\cap C)$ ?? (Wondering)

If it stands, then we have that $\mathbb{F}_{p^{\gcd(n, m)}} \leq \mathbb{F}_{p^d} \Rightarrow \gcd(n, m) \mid d$



So, $d \mid \gcd(n, m)$ and $\gcd(n, m) \mid d$.

That means that $d=\gcd(n, m)$.

Therefore, $\mathbb{F}_{p^n} \cap \mathbb{F}_{p^m} = \mathbb{F}_{p^{\gcd(n, m)}}$

Is this correct?? (Wondering) Or could I improve something ?? (Wondering)

 

[jvicens]

Hello there! Your approach to finding the intersection of the two subfields is correct. You correctly used the fact that the subfields are ordered by divisibility, and that the intersection must also be a subfield. You also correctly used the fact that the greatest common divisor of two numbers divides both of them.

As for your question about whether $A \leq B$ and $A \leq C$ implies $A \leq (B \cap C)$, the answer is yes. This is because any element in $A$ must also be in both $B$ and $C$, since $A$ is a subset of both $B$ and $C$. Therefore, $A$ is also a subset of their intersection, which is $B \cap C$.

Overall, your reasoning is sound and your conclusion is correct. Great job! Let me know if you have any further questions. Keep up the good work!