I am currently studying perfection in the context of Galois Theory. For a field $K$ of characteristic $p$ and algebraic closure $K'$, we define $$ K^{\text{perf}} = \{ a \in K' : \text{there exists } n \in \mathbb{N} \text{ such that } a^{p^{n}} \in K \} $$
I know that this field is perfect because the Frobenius endomorphism is an isomorphism (a field is perfect if and only if the Frobenius endomorphism is an isomorphism), and by construction it's the smallest perfect field containing $K$ (because we are adjoining the $p^{\text{th}}$ roots, and then the roots of the roots, and so on).
But now I would like to compute both $[K^{\text{perf}}: K]$ and $[K^{\text{perf}}: K]_{s}$.
My claim is that $[K^{\text{perf}}: K]_{s}$ = $[K': K]_{s}$, because any homomorphism $K^{\text{perf}}\rightarrow K'$ can be extended to $K' \rightarrow K'$ (as the target is algebraically closed). Conversely, any $K' \rightarrow K'$ can be restricted to $K^{\text{perf}} \rightarrow K'$. We therefore encounter a bijection between $\operatorname{Hom}(K^{\text{perf}}, K')$ and $\operatorname{Hom}(K',K')$, which proves the claim. Is this right?
As for $[K^{\text{perf}}: K]$, I am way more lost. I believe that $[K^{\text{perf}}: K] \leq [K':K]$ since $K'$ contains $K^{\text{perf}}$. I've also noticed that that $K^{\text{perf}}/K$ is not separable, so the extension degree is bigger than the separable degree. But I can't seem to get an explicit computation of $[K^{\text{perf}}: K]$. What should the final answer be or how to obtain it?
Since the irreducible equation for any $\alpha\in K^{\text{perf}}$ is of the form $X^{p^n}-a$ where $a\in K, K^{\text{perf}}/K$ is purely inseprarable, i.e. $[K^{\text{perf}}:K]_s=1$. This is not necessarily the same as $[K':K]_s$. The restriction and extension in your argument may not be one-to-one.
$K'$ is separable and perfect over $K^{\text{perf}}$. This is a special case of Corollary $6.12$ in Serge Lang's Algebra Chapter V, which is a corollary of Proposition $6.11$. See the accepted answer in every algebraic extension of a perfect field is separable and perfect.
Hence $[K':K]=[K':K^{\text{perf}}]_s[K^{\text{perf}}:K]_i$, and $[K^{\text{perf}}:K]=[K^{\text{perf}}:K]_i$.