Given a field K with characteristic $p$ ($p$ prime) show that $K$ contains exactly one subfield with $p$ elements

Question

can anybody give me a hint on how to prove this statement? I am given a field $K$ with char($K$) = $p$ and $p$ prime. I need to show that $K$ contains exactly one subfield with $p$ elements. I understand that in order to prove this I need to prove that such a subfield exists and in a separate step prove that only one can exist. I also know that the cardinality of $K$ must be $p^n$ for some $n \geq 1$. I tried proving the existence via induction over $n$ but got nowhere. Any help is appreciated. Thank you!

Answer 1

To prove that $K$ has a subfield of order $p$, observe that the map $f: \mathbb{Z} \to K$ defined by $$f(m) = m\cdot 1,$$ is a homomorphism and the kernel is $p\mathbb{Z}$. Thus, the image of $f$ is a subfield of $K$ isomorphic to $\mathbb{Z}_p$. Thus $K$ has subfield $F$ of order $p$.

Note that $F$ has 1. It also has $p$ elements. Now, $k\cdot 1$, $1\leq k \leq p$, are all distinct elements and hence are the elements of $F$. As any subfield of $K$ contains 1, it must also contain $F$ and hence, there is only one subfield of order $p$.