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1d
comment More generalized RSA construction
By "CRT" I mean the Chinese remainder theorem, and $e$ and $d$ are two positive integers chosen so that $ed \equiv 1 \bmod \varphi_K(\mathfrak a)$. As you see (if you check the details), the ingredients in the usual RSA carry over.
1d
comment More generalized RSA construction
Sure. If $K$ is a number field with ring of integers $O_K$, then for any nonzero proper ideal $\mathfrak a$ in $O_K$ the ring $O_K/\mathfrak a$ is finite. Let ${\rm N}(\mathfrak a)$ be its size and $\varphi_K(\mathfrak a)$ be the number of units in this ring. For two different (nonzero) prime ideals $\mathfrak p$ and $\mathfrak q$ in $O_K$, let $\mathfrak a = \mathfrak p\mathfrak q$. Then $\varphi_K(\mathfrak a) = ({\rm N}\mathfrak p - 1)({\rm N}\mathfrak q - 1)$. Check by CRT that for all $x$ in $O_K$, if $ed \equiv 1 \bmod \varphi_K(\mathfrak a)$ that $x^{ed} \equiv x \bmod \mathfrak a$.
Apr
15
comment What is known about the reverse mathematics of algebraic number fields?
"Thm 8 shows $\mathsf{RCA}_0$ itself proves every automorphism of a number field $L$ extends to an automorphism of any larger number field $K/L$." This is false. Take $L = \mathbf Q(\sqrt{2})$ and $K = \mathbf Q(\sqrt[4]{2})$. There are only two automorphisms of $K$, with the effect $\sqrt[4]{2} \mapsto \pm \sqrt[4]{2}$, and both of these fix $\sqrt{2} = \sqrt[4]{2}^2$. Therefore the nontrivial automorphism of $L$ does not extend to an automorphism of $K$. You must be leaving out a Galois hypothesis.
Apr
14
comment How to Taylor series expand at the prime at infinity
You are looking for an analogue of the residue theorem, and that would be the product formula $\prod_v |x|_v = 1$ for $x \in \mathbf Q^\times$.
Apr
14
comment How to Taylor series expand at the prime at infinity
Try the usual decimal expansion, for instance. Or base 2. And so on. There is no canonical base to use.
Apr
12
comment What does this proof of Fermat's little theorem mean for Euler's theorem?
@LevBorisov, by "Euler's theorem for all $a$" do you mean any congruence valid for all $a$ that reduces to Fermat's little theorem when $n$ is prime? Two possibilities are $\sum_{k=0}^{n-1} a^{(k,n)} \equiv 0 \bmod n$ for all $a$ and $\sum_{d \mid n} \varphi(n/d)a^d \equiv 0 \bmod n$ for all $a$.
Apr
6
revised Applications of functional analysis beyond analysis(towards algebra, geometry, number theory…)
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Apr
6
revised Applications of functional analysis beyond analysis(towards algebra, geometry, number theory…)
added 138 characters in body
Apr
6
revised Applications of functional analysis beyond analysis(towards algebra, geometry, number theory…)
added 144 characters in body
Apr
6
answered Applications of functional analysis beyond analysis(towards algebra, geometry, number theory…)
Apr
5
awarded  Nice Answer
Apr
1
comment Examples of math hoaxes/interesting jokes published on April Fool's day?
See groups.google.com/forum/m/#!topic/sci.math/LC71u21hJCE
Apr
1
comment Special topics to include in course in algebraic number theory
Irreducibility over $\mathbf Q$ of truncated exponential polynomials $1+x+\cdots + x^n/n!$. See math.uconn.edu/~kconrad/blurbs/gradnumthy/schurtheorem.pdf.
Mar
26
revised How to use Gronwall's inequality?
edited title
Mar
17
comment Is the absolute Galois group the same as the automorphism group?
@YCor, oxeimon started with "Let G be a profinite group." I took that to mean G starts off as a profinite group and was just indicating that a suitable inverse limit construction recovers the group as a topological group. (There are multiple ways of defining profinite groups.) Admittedly this is not as interesting as theorems that recover the topology from the group structure.
Mar
16
comment Is the absolute Galois group the same as the automorphism group?
@HJRW, reread my comment again: I am only talking about a given profinite group (e.g., I talk about open normal subgroups), not an abstract group without a topology in advance.
Mar
16
comment Is the absolute Galois group the same as the automorphism group?
@oxeimon, the topology on a profinite group comes from taking the inverse limit over all finite quotient groups, working modulo all open normal subgroups. So there is no "choice" of a way to view the group as an inverse limit.
Mar
9
revised Are submodules of free modules free?
added 2 characters in body
Feb
28
comment Can the Dedekind zeta function distinguish between real and imaginary quadratic number fields?
If you know the zeta-function of a number field $K$ "in full" then yes: this function knows the number of real and complex (i.e., non-real) embeddings of $K$. Calling these $r_1$ and $2r_2$, the order of vanishing of $\zeta_K(s)$ at a negative integer $n$ is $r_2$ if $n$ is odd and $r_1 + r_2$ if $n$ is even. The order of vanishing at 0 is $r_1 + r_2 - 1$. In particular, if $K$ is quadratic then $\zeta_K(s)$ is nonzero at negative odd integers for real $K$ and it is zero at negative odd integers for imaginary $K$. Also $\zeta_K(0) = 0$ for real $K$ and $\zeta_K(0) \not= 0$ for imaginary $K$.
Feb
27
comment What goes wrong in a ring that does not have unique factorization?
I think the last paragraph here should have gone first: it is a very concrete illustration of something weird that can happen. In a Dedekind domain the property of being able to write every ratio of nonzero elements in its fraction field in "lowest terms" is in fact equivalent to it being a UFD.