How to show this is subfield for the given field.

Asked Nov 08 '19 at 09:29

Active Nov 08 '19 at 09:29

Viewed 21 times

Let the $F$ be the finite field over $\mathbb{Z_2} s.t. [F;Z_2]= 60$

$[F;K_1] = 5,[F; K_2] = 10$. Show $K_2 \subset K_1$

My attempt)

By the tower rule, $[K_1 ; \mathbb Z_2] = 5$ and $[K_2; \mathbb Z_2] = 6$

Then $[K_1 \cap K_2]$ $\vert gcd(6,12)$. Hence $[K_1 \cap K_2] \in \{1,2,3,6\}$

So all I have to just show the $[K_2 ; K_1 \cap K_2]= 1 $ by proving the $[K_1 \cap K_2 ; Z_2] = 6$.

But I couldn't find any way to show $[K_1 \cap K_2 ; Z_2] = 6$. Any help would be appreciated.

asked Nov 08 '19 at 09:29

se-hyuck yang

1

If the degrees are $m$ and $n$, then $K_1\supset K_2$ if and only if $m\mid n$ (see section on finite fields). – Dietrich Burde Nov 08 '19 at 09:37
Does proposition you suggested still hold the general case, every finite extension field? – se-hyuck yang Nov 08 '19 at 09:47
1

Yes, see the duplicate. – Dietrich Burde Nov 08 '19 at 09:51
Hmm the proposition you suggested only hod the finite field. The case $K_1 = Q(\sqrt 2) $and$ K_2=Q(\sqrt 3)$ but$ K_1 $is not the subfield of the $K_2 $though their degree is equal. – se-hyuck yang Nov 08 '19 at 10:03
Yes, but you said we have a field over $\Bbb F_2$ of degree $60$. So we talk about finite extension fields of the prime field $\Bbb F_p$, right? – Dietrich Burde Nov 08 '19 at 10:04
Yes we did. But nothing specially, I just have a curious when the general case, which is finitely extension not finite field. If we talk only for the finite field then the proposition is true. – se-hyuck yang Nov 08 '19 at 10:11
For infinite, $m\mid \infty$ is not sensible. – Dietrich Burde Nov 08 '19 at 10:12

0 Answers0