11,064 reputation
11230
bio website math.brown.edu/~lubinj
location Minnesota
age 77
visits member for 2 years, 6 months
seen 47 mins ago

Aged mathematician


46m
revised General methods for solving p-adic inequality $|a^2+b|_{p}<p^{-k}$ for $a\in \mathbb{Z}$
clarification
52m
answered General methods for solving p-adic inequality $|a^2+b|_{p}<p^{-k}$ for $a\in \mathbb{Z}$
14h
comment Roots of polynomial in $F_3[x]$
$\alpha^2=-\alpha+1$ and $\alpha^4=-1$; neither of these is a root of $X^3+X+1$.
15h
comment Finding the value of trigonometric functions
The important thing, beyond @Seth’s suggestion, is to remember that in view of the periodicity of $\tan$ and $cos$, you can add or subtract multiples of $\pi$ from the argument of $\tan$ without changing the value, similarly, you can add or subtract multiples of $2\pi$ from the argument of $\cos$.
15h
comment Is the reasoning/algebra for my proof correct? (musical tuning theory proof)
That would have been my proof, but you got there first! (I would have pointed out that @SanathDevalpurkar’s proof was perfectly fine, but that this argument is much more basic.)
15h
comment how to find congruence class modulo a polynomial
The unspoken requirement is to represent each expression in $\alpha$ as a linear combination of $1$ and $\alpha$, via Euclidean division. The standard, mindless, way to find the inverse of $2\alpha+1$ would be to work through the Euclidean Algorithm for gcd. In a case like this, it’s far easier to write down the successive powers of $\alpha$ (I’m assuming that there are $8$ distinct ones here), and pick out the exponent for $\alpha+1$, and then you know the reciprocal of $\alpha+1$, presto pronto.
15h
comment Characterizing the Galois group using permutations of roots
I think that for this, one must be very careful with quantifications. I’d want to see a precise statement, with all quantifiers before the various clauses.
19h
answered Characterizing the Galois group using permutations of roots
1d
comment What does root of unity in $\mathbb{Z}_p$ look like?
Other than $1$ and $-1$, the roots of unity are not rational numbers. You can prove that.
1d
comment In $Z/3Z[x]$ find the GCD of $x^5+2x^3+x^2+x+1$ and $x^4+2x^3+x+1$
Books are sometimes wrong. Your teachers are sometimes wrong. It’s an imperfect world, populated by imperfect human beings.
1d
revised $2$-adic inverse of $3$
Expanded considerably on my answewr, explaining a technique for $p$-adic hand computation.
2d
answered $2$-adic inverse of $3$
2d
comment Philosophical question in real analysis
Yes. It’s not the compactness of the interval that’s important, but the fact that it’s a connected set with more than one point.
2d
comment Criterion that $R[\sqrt{D}]$ be a Dedekind ring.
Sorry, sorry, sorry, I meant to say $D\equiv1\pmod4$. And as all too often happens, I managed to skip over your essential hypothesis that $2$ should be a unit. My deepest apologies.
2d
comment Criterion that $R[\sqrt{D}]$ be a Dedekind ring.
I suppose you mean that $D$ is not divisible by the square of a prime idea of $R$. The test cases are $\mathbb Z[\sqrt D\>]$ with $D\equiv3\pmod4$. Have you examined these?
Apr
13
comment Maximal Ideal of $\mathbb{Z}[i]$
And you may find it helpful to use the Euclidean Division that you have in $\mathbb Z[i]$.
Apr
13
answered What does root of unity in $\mathbb{Z}_p$ look like?
Apr
13
comment The number of rational solutions to the cubic analogue of Pell's equation
Well, I was supposing that the first two you found would span a rank-two free-abelian group ($\mathbb Z$-module), and that the third would be dependent on those (with respect to multiplication, of course!). I didn’t check my guess out, though.
Apr
11
comment Degree of closure of $\mathbb{Q}_p$
You might as well make things easier for yourself by looking not at all $p^{1/n}$ but at a subset: all $p^{1/2^m}$. The corresponding fields clearly form a totally ordered chain, and since $[\mathbb Q_p(p^{1/2^m})\colon\mathbb Q_p]=2^m$, you’re home free.
Apr
11
answered Is convex hull of a finite set of points in $\mathbb R^2$ closed?