Reputation
Next tag badge:
99/100 score
32/20 answers
Badges
2 25 56
Impact
~334k people reached

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.