Let $p$ be prime and let $K:= \bigcup_{i=1}^{\infty} GF(p^i)$. Prove $k$ is the algebraic closure of $GF(p)$.
Union of Galois Fields and Algebraic Closure
2
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field-theory
galois-theory
finite-fields
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1No, *you* prove it! In all seriousness, please also include what you've tried, and where you get stuck! – 2017-02-14
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0The first step here is to come up with a working way of interpreting that union. After all, this is quite unlike the union of number fields, where we can think of the individual fields as being subsets of $\Bbb{C}$. Here no such "known" umbrella field is available. Hint 1: Instead of this unwieldy collection of fields think of the nested union $$GF(p)\subset GF(p^{2!})\subset GF(p^{3!})\subset \cdots$$ using the fact that $GF(p^{(k+1)!})$ has a unique subfield isomorphic to $GF(p^{k!})$ (though the identification is not unique at all). – 2017-02-14
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0Hint 2: With that union in place show that any polynomial with coefficients in it actually has coefficients in some finite subfield. Work from there! If everything else fails, search the site. I'm fairly sure we have this covered, but am too lazy/busy to look for a duplicate now. – 2017-02-14