9,378 reputation
2247
bio website
location
age
visits member for 4 years, 4 months
seen 2 hours ago

3h
comment What is the discrete log used for?
Yes, but the point is that when working in modular arithmetic, exponentiation can be done without the output getting explosively large (beyond the size of the modulus) and can be carried out efficiently.
8h
comment What is the discrete log used for?
From your comment about "presumably small numbers" it is not clear if you realize that the equation $b^k = g$ is not in the integers, but is really a congruence to some modulus: $b^k \equiv g \bmod p$.
1d
awarded  Revival
1d
comment Automophism of G and Haar measure
The new measure $\mu(A) = \alpha(\lambda(A))$ on $G$ is also a Haar measure, so there's a positive number $c$ such that $\mu = c\alpha$. That is, for all measurable subsets $A$ in $G$, $\alpha(\lambda(A)) = c\alpha(A)$. That's the sense in which $\lambda$ turns $\alpha$ into $c\alpha$. For example, if $G = \mathbf R$, $\alpha = dx$, and $\lambda(x) = 2x$ then $\lambda$ transforms $\alpha$ into $2\alpha$. For generally, if $\lambda_t(x) = tx$ for a nonzero real number $t$, then $\lambda_t$ transforms $\alpha$ into $|t|\alpha$.
2d
awarded  Nice Answer
2d
answered Have any definitions in mathematics been redefined
2d
awarded  Caucus
Dec
15
comment For $G$ an abelian group and $H$ a subgroup, is $[G : H]$ the smallest positive integer $n$ such that $ng \in H$ for all $g \in G$?
The term you should look up is the "exponent" of a group, which can be smaller than the order of the group.
Dec
14
revised Show that $\mathbb Q[x]/(x^2+2)$ and $\mathbb Q[x]/(x^2-2)$ are not isomorphic.
added 28 characters in body
Dec
13
comment Why are every structures I study based on Real number?
If you replace your first two conditions by a condition called nondegeneracy ($\langle u,v\rangle = 0$ for all $v$ implies $u = 0$) then you have the concept of a nondegenerate symmetric bilinear form, which does not need the scalar field to be ordered at all.
Dec
13
comment Why are every structures I study based on Real number?
Your definition of the length of a vector as the square root of an inner product takes you outside the original scalar field unless the positive elements of the field all have square roots in the field. Most ordered fields do not have that property! Can you think of such an ordered field besides the real numbers?
Dec
12
comment Cyclotomic extension of $\mathbb{F}_p((T))$
Of course the latter is a splitting field of some polynomial over the former: it is a finite extension (of degree $n$) and all finite extensions are splitting fields of some polynomial. Your question really has nothing to do with Laurent series fields. Can you view $\mathbf F_{p^n}$ as a splitting field of some concrete polynomial over $\mathbf F_p$?
Dec
12
comment Cyclotomic extension of $\mathbb{F}_p((T))$
It is not even true for the field $\mathbf F_p$ (forget Laurent series) or for $\mathbf Q$, or any field at all if $n > 1$: the polynomial $x^n-1$ is not irreducible if $n>1$, so adjoining an $n$th root of unity to a field will give an extension of degree at most $n-1$, or more precisely at most $\varphi(n)$.
Dec
12
comment What are the elements of $Z[x]/\langle x^2 - 1\rangle$
This is misleading. The Chinese remainder theorem does not work the way you suggest it does when the coefficient ring is $\mathbf Z$ rather than a field. The natural ring homomorphism from ${\mathbf Z}[x]/(x^2-1)$ to ${\mathbf Z} \times {\mathbf Z}$ by $f(x) \bmod x^2-1 \mapsto (f(-1),f(1))$ is injective but not surjective. For example, if $f(x) \bmod x^2-1 \mapsto (1,0)$ then from $f(1) = 0$ we get $f(x) = (x-1)g(x)$ in ${\mathbf Z}[x]$, and that implies $f(-1) = -2g(-1)$ is even, so $f(-1)$ can't be $1$. The correct image is all pairs $(a,b)$ where $a \equiv b \bmod 2$.
Dec
11
answered Prove GCD of polynomials is same when coefficients are in a different field
Dec
8
comment Solvability of Artin-Schreier Polynomial
No, but not everything in the world exists (yet?) online. Go to a library and look at a physical copy. If your library does not have one, they can get one for you by interlibrary loan.
Dec
8
comment Solvability of Artin-Schreier Polynomial
See section 8.6 part C of David Cox's Galois Theory (2nd edition) for solvability in characteristic p.
Dec
8
comment Solvability of Artin-Schreier Polynomial
Being solvable by radicals doesn't require the splitting field itself to be a radical extension, but only to lie in a radical extension.
Dec
2
comment Assume we have $\mathbb{Z}_{p}[x]$ with $p$ being a prime. Prove that $x^{p-1}-1=(x-1)(x-2)…(x-(p-1))$
@AndréNicolas: in ${\mathbf Z}_3[x]$, $x^2$ and $2x^2-1$ each have degree $2$ and their difference $x^2-1$ has 2 distinct roots but is not the zero polynomial (Pedro made a comment while I was about to write the same thing, so I give an example instead.)
Nov
28
comment How do I go with proving that the coefficient of each terms of $\prod^{k=n}_{1}{1-x^k}$ is either 1,-1 or 0?
This is a famous result called the pentagonal number theorem. Googling that term will show you a formula for the coefficients that are $1$ and $-1$, which will provide a pattern for them.