# Finite fields and isomorphism

For each prime number p, let $F_p$ denote the field of integers modulo p. Now let K be any finite field. a) Prove that K contains a subfield isomorphic to $F_p$ for some prime number p b) Prove that the intersection of all of the subfields of K will be isomorphic to $F_p$ for some prime number p c) Prove that the cardinality of K is equal to a power of p for some prime number p

I do not understand the question. Can someone explain me, please? To me, $F_p = Z_p$ and $char Z_p= p$ and the finite field K has characteristic = another prime, namely $p_1$. I do not see the link between them.

Take $1 \in K$. Then $1+1, (1+1)^{-1}$ , etc. are in K. If K is infinite, this gives you a copy of the rationals. If not, you get a finite field, which is $\mathbb Z_p$
The question means that if we have a finite fied $K$ then there exists a prime $p$ such that $K$ is isomorphic to $F_p$.
First of all,the characteristic of $K$ must be a prime $p_1$ hence the subfield: $$K_{p_1}=\{1_k,2.1_k,\cdots,(p-1).1_k\}$$ Is isomorphic to $F_{p_1}$.
It says that, if you take finite field $F$, it contains a prime field (isomorphic to $\mathbf Z/p\mathbf Z$ also denoted $\mathbf F_p$ in the context of finite fields) for some $p$. You simply have to map $\mathbf Z$ to $F$ by sending $n$ to $n\cdot 1_F$ and consider the kernel of this homomorphism.
• Do you mean that if $F$ has characteristic $p$, then the map you described has kernel $(p)$ so it maps $\mathbb Z$ onto a subset of $F$ isomorphic to $\mathbb Z/(p)$, and that subset being a finite integral domain is itself a field, hence a (prime) subfield of $F$? – user437309 Apr 14 '18 at 3:07
• @user437307: Yes, exactly. It's a field also simply because it is isomorphic to the field $\mathbf Z/p\mathbf Z$. – Bernard Apr 14 '18 at 8:46