(adsbygoogle = window.adsbygoogle || []).push({}); 1. Find the group

[tex]G = Gal(\mathbb{F}_{p^{n^2}}/\mathbb{F}_{p^n})[/tex]

There is a 1-1 correspondence between elements of this group, and intermediate fields. An intermediate field is a finite field of order m, where p^{n}| m | p^{n²}. The m that satisfy this are p^{k}for each k in {n, n+1, ..., n²-1, n²}. So the group in question has order n²-n+1. I'm very iffy on this material, so where should I go from here? Do I find

[tex]Gal(\mathbb{F}_{p^{k+1}}/\mathbb{F}_{p^k})[/tex]

for n < k < n², and then G will be somehow made up of these Galois groups?

2. Is x²=2 solvable in 7-adic numbers?

For all k, x² = 2 (mod 7^{k}) has two solutions, since 7^{k}is congruent to either 1 or 7 (mod 8), depending on whether k is prime. Moreover, for each k, x² = 2 (mod 7^{k}) has two distinct solutions, y and -y (which are incongruent since all powers of 7 are odd), and if j < k, then y (mod 7^{j}) and -y (mod 7^{j}) are solutions to x² = 2 (mod 7^{j}).

So if y_{1}is a solution to x² = (mod 7), then there exists y_{2}such that y_{2}= y_{1}(mod 7) and y_{2}² = 2 (mod 7²). Inductively, if y_{i}is a solution to x² = (mod 7^{i}), then there exists y_{i+1}such that y_{i+1}= y_{i}(mod 7^{i}) and y_{i+1}² = 2 (mod 7^{i+1}). Likewise, starting with -y_{1}I can get a different sequence. The two sequences should give rise to 7-adic numbers y and -y which satisfy y² = 2 as 7-adic numbers, right?

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# Algebra problems

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