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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?

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  • $\begingroup$ 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. $\endgroup$
    – jjagmath
    Commented Jun 12 at 3:07

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