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 Apr 14 comment P-adic expansions of Integers Finite in case the integer is nonnegative; for negative integers, the coeffs of $p^n$ will eventually all be $p-1$. Apr 14 comment Find $[\mathbb{Q(\alpha)}: \mathbb{Q}]$ where $\alpha=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9$ Knowing (or guessing) that $\alpha$ is quadratic over $\Bbb Q$, you get the minimal polynomial for $\alpha$ by looking at $1$, $\alpha$, and $\alpha^2$ and finding a $\Bbb Q$-linear relation of linear dependence. Apr 14 comment Find $[\mathbb{Q(\alpha)}: \mathbb{Q}]$ where $\alpha=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9$ There’s only one subgroup of order five because the original group is cyclic. These have precisely one group of each order dividing the order of the main group. Apr 14 answered Find $[\mathbb{Q(\alpha)}: \mathbb{Q}]$ where $\alpha=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9$ Apr 14 comment Proving that a Galois group is cyclic This is very confusing till you get the hang of it. @anon has given you the key. Apr 13 answered Numbers in $\mathbb{Q}_p$ can be written uniquely as $\sum_{i=k}^\infty \alpha_i p_i$ Apr 13 comment Numbers in $\mathbb{Q}_p$ can be written uniquely as $\sum_{i=k}^\infty \alpha_i p_i$ Yes, @DietrichBurde, that’s probably it. Maybe I’ll try a guide to a complete proof. Apr 13 comment Numbers in $\mathbb{Q}_p$ can be written uniquely as $\sum_{i=k}^\infty \alpha_i p_i$ It’s not clear what your question is. Are you saying that you understand the completion but don’t understand why every element can be written in unique $p$-ary expansion, or are you saying that you don’t understand why, if you define $\Bbb Q_p$ to be the set of $p$-ary expansions, you get a $p$-adically complete space? Apr 13 comment A normal closure of an arbitrary field extension Agreeing with @CaptainLama, I’d say that for infinite extensions, you’ll usually need to appeal to Zorn, and say that if an overfield of $L$ is not yet normal, you find an element in it with a $K$-conjugate not in it. Adjoin that element. There is your induction step. For explicitly-given extensions, maybe the set of all real $n$-th roots of all primes, over $\Bbb Q$, there may well be techniques peculiar to your situation (here, adjoin as well all roots of unity). Apr 13 comment Bézout's Identity of polynomials? I guess we gave each other a plus-one. Mutual admiration. Apr 13 answered Bézout's Identity of polynomials? Apr 13 answered Formula for calculating the $α$ and $β$ in $gcd(a, b) = αa + βb$ Apr 12 comment splitting field of $x^n-1$ over $\mathbb{Q}$ It’s true, but part of the proof is rather advanced, namely the proof that the degree of the extension is no smaller than $\phi(n)$. This is equivalent to the $\Bbb Q$-irreducibility of the $n$-th cyclotomic polynomial $\Phi_n(X)$. In case $n$ is a prime power, the proof is not at all hard, but when $n$ is divisible by two odd primes, or $4$ and an odd prime, things get sticky. Apr 11 answered Galois extentions Apr 11 answered Finding last 2 digits of $2016^{500}$ without repeated squaring Apr 11 comment If a function is odd/even, then its best polynomial approximation is also odd/even. Just to reiterate my own comment and support the others, what could it possibly mean to say that a function defined on $[0,b]$ is odd or even? Apr 10 comment If a function is odd/even, then its best polynomial approximation is also odd/even. I'm an outsider to this field, but if $a\ne-b$, I don't believe the claim. Apr 10 answered How to find $W^{\perp}$ in the following polynomial inner product space? Apr 10 comment How to find $W^{\perp}$ in the following polynomial inner product space? Right method, bad integration. Apr 10 comment Quick help on why this extension is of degree $2$ Right. Nor $i$. Just two things in the basis.