For which numbers $l \leq |K|$ field $K$ has subfield, which has $l$ elements?

Asked Jun 04 '19 at 19:08

Active Jun 04 '19 at 19:29

Viewed 46 times

char$(K) = p$, $|K|= p^n$, $|K^{\times}| = p^n-1.$

We know that $K^{\times}$ is cyclic group. Let $H$ be a cyclic subgroup of $K^{\times}$. Subgroup of cyclic group is also cyclic, so $H$ is cyclic and $|H|\mid|K^{\times}|$.

Let $|H| = r,\ r\mid p^n-1, \ p^n-1 = r q,\,\, q \in \mathbb{Z}$. Each element in $H$ has inverse element (from definition on cyclic group). I just add $0$ and we have subfield. So, $l = r + 1$, where $r \mid p^n-1$.

How to show that $r = p^k - 1$ for $k \leq n$?

edited Jun 04 '19 at 19:29

asked Jun 04 '19 at 19:08

Aleksandra

2

If $H$ is a subgroup of $K^\times$ its order must be coprime to $p$, so $r$ cannot equal a power of $p$. – Angina Seng Jun 04 '19 at 19:12
@LordSharktheUnknown youre right, $r = p^k - 1$, my mistake – Aleksandra Jun 04 '19 at 19:29
4

If $L$ is a subfield of $K$, then $K$ is a vector space over $L$, so $|K|$ must be a power of $|L|$. – Derek Holt Jun 04 '19 at 20:24

For which numbers $l \leq |K|$ field $K$ has subfield, which has $l$ elements?

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