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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…) added 869 characters in body 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? 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.