Let $K$ a field with characteristic $p>0$. Show that $\{x \in K : x^{p^n} =x \}$ is a subfield. Let $K$ a field with characteristic $p>0$. I've shown that for every positive $n$ the set $\{ x^{p^n} : x \in K \}$ is a subfield of $K$, I did this by showing that $F:K\to K: x \mapsto x^{p^n}$ is a field homommorphism.
Now I try to show that for every positive $n$ the set $\{x \in K : x^{p^n} =x  \}$ is a subfield with at most $p^n$ elements. 
We tought about that the polynomial $X^{p^n}-X=0$ in $K[X]$ has not more then $p^n$ solutions in $K$. But I'm not sure how to show that this is a subfield.
 A: For any two fields $K,L$ and field homomorphisms $\varphi,\psi \colon K \to L$, the set
$$I(\varphi,\psi) = \{ x\in K : \varphi(x) = \psi(x)\}$$
is a subfield of $K$. This is a generalisation of the often-used fact that the set of fixed points of a field endomorphism is a subfield of its domain [and this special case is what is used here; nevertheless, the more general fact is not harder to prove].
Proof: By the definition of field homomorphisms, we have $0,1\in I(\varphi,\psi)$. For $x,y\in I(\varphi,\psi)$, we have
\begin{gather}
\varphi(x+y) = \varphi(x) + \varphi(y) = \psi(x) + \psi(y) = \psi(x+y),\\
\varphi(x\cdot y) = \varphi(x)\cdot \varphi(y) = \psi(x)\cdot \psi(y) = \psi(x\cdot y),
\end{gather}
hence $x,y\in I(\varphi,\psi) \implies x+y,x\cdot y\in I(\varphi,\psi)$. Further, for $x\in I(\varphi,\psi)$ we have
$$\varphi(-x) = -\varphi(x) = -\psi(x) = \psi(-x),$$
whence $x\in I(\varphi,\psi) \implies -x\in I(\varphi,\psi)$, which, together with the above shows that $I(\varphi,\psi)$ is a subring of $K$.
Finally, for $x\in I(\varphi,\psi)\setminus \{0\}$ we have
$$\varphi(x^{-1}) = \varphi(x)^{-1} = \psi(x)^{-1} = \psi(x^{-1}),$$
so the subring $I(\varphi,\psi)$ is a subfield.
In particular, $\operatorname{Fix}(F) = I(F,\operatorname{id}_K) = \{x\in K : x^{p^n} = x\}$ is a subfield of $K$ for every field $K$ of characteristic $p$. Since $\operatorname{Fix}(F)$ is the set of zeros of the polynomial $P(X) = X^{p^n} - X$, it follows that $\operatorname{Fix}(F)$ has at most $\deg P = p^n$ elements.
