# Help me to verify the proof of this theorem, which is proving $k$ª $=$ $\prod_{i=1} ^{\infty}$ $\mathbf K_{p_i}$ by using maximality and minimality

I want to verify my proof is true or false. The exercise what I want to prove is under theorem.

$$\mathbf {Exercise}$$: Let $$\mathbf k$$ be some perfect field and $$\mathbf K_{p_i}$$ is a compositum of all cyclic extensions that have orders are power of $$p_i$$, that $$p_i$$'s are all prime numbers. Then $$\mathbf k$$ª is field compositum of all $$\mathbf K_{p_i}$$.

solution

Let some embedding $$\sigma$$: $$\mathbf k$$ª $$\to$$ ($$\mathbf k$$ª)ª = $$\mathbf k$$ª over $$\mathbf K_{p_1}$$...$$\mathbf K_{p_n}$$ for all primes $$p_i$$. Then, one of them can be identity function, so $$\mathbf k$$ª is smallest normal extension of $$\mathbf K_{p_1}$$...$$\mathbf K_{p_n}$$ containing $$\mathbf K_{p_1}$$...$$\mathbf K_{p_n}$$ . Because $$\mathbf K_p$$ are normal, because cyclic extensions are normal, and Compositum of Normal extensions are normal. So, $$\mathbf k$$ª is a minimal galois extension of $$\prod_{i=1} ^{\infty}$$ $$\mathbf K_{p_i}$$ containing $$k$$ª. all galois extensions are algebraic, so $$\mathbf k$$ª is contained in all galois extension of field compositum. This field is also normal over $$\mathbf K_{p_1}$$... $$\mathbf K_{p_n}$$, Normal extension is algebraic. algebraically closed extension of algebraically closed field is itself. So, all galois extension of compositum field is $$\mathbf k$$ª. And, $$k$$ª $$=$$ $$(\prod_{i=1} ^{\infty}$$ $$\mathbf K_{p_i})$$ª because $$k$$ $$\subseteq$$ $$\prod_{i=1} ^{\infty}$$ $$\mathbf K_{p_i}$$, and $$\prod_{i=1} ^{\infty}$$ $$\mathbf K_{p_i}$$ $$\subset$$ $$k$$ª. So, from this, normal extension of $$\prod_{i=1} ^{\infty}$$ $$\mathbf K_{p_i}$$ containing $$k$$ª is unique, and the unique field is $$\prod_{i=1} ^{\infty}$$ $$\mathbf K_{p_i}$$. So, It says that required result.(Idea of my proof is, $$\prod_{i=1} ^{\infty}$$ $$\mathbf K_{p_i}$$ is algebaically closed field.)

my Question

I proved this by maximalality and minimality. But, Is there some set-theoretical problem? Because in minimality, I proved it only the case of containing $$\mathbf k$$ª. I think I proved this by connecting two answers that "$$\mathbf k$$ª is $$(\prod_{i=1} ^{\infty}$$ $$\mathbf K_{p_i})$$ª", and the minimal normal extension of $$(\prod_{i=1} ^{\infty}$$ $$\mathbf K_{p_i})$$ containing $$\mathbf k$$. Is this proof right?

• Please, use descriptive titles. "Can you tell me other proof? This proof doesn't look like required answer." says nothing about the subject of the question. Commented Jun 12 at 3:07