# Servaes

Graduate student mostly dabbling in algebraic geometry.

 Mar 5 answered Computing basis of a field extension Feb 26 comment Proper decimal fraction for $\frac{4n+1}{n(2n-1)}$By noting that $\gcd(4n+1,2n-1)$ divides $(4n+1)+(2n-1)=6n$, and using the fact that $4n+1$ is odd and $\gcd(4n+1,n)=1$. Feb 26 comment Proper decimal fraction for $\frac{4n+1}{n(2n-1)}$For case $(1)$: Note that $2$ divides $p+1$ by conruence mod $3$. But then $$3\times5^q=2n-1=2^{p+1}-1=(2^r-1)(2^r+1),$$ where $r=\tfrac{p+1}{2}$. These factors are coprime, so either $r=1$ or $r=2$, and hence $n=2$ or $n=8$, respectively. Feb 26 comment Proper decimal fraction for $\frac{4n+1}{n(2n-1)}$For case $(2)$ it is not necessary to use such heavy machinery: If $2n=5^q+1$ then $2n\equiv2\pmod4$ so $n=1$. Feb 26 comment Proper decimal fraction for $\frac{4n+1}{n(2n-1)}$For cases $(3)$ and $(4)$: By congruence mod $2$ we have $q=0$ and hence $n=2$ (a contradiction) resp. $n=1$. Feb 26 comment Proper decimal fraction for $\frac{4n+1}{n(2n-1)}$It is also worth noting that $n$ and $4n+1$ are coprime. Feb 26 comment Proper decimal fraction for $\frac{4n+1}{n(2n-1)}$Does 'proper decimal fraction' mean the same as 'finite decimal expansion'? Feb 25 comment pigeonhole principle on a circleOr equivalently; what is the least amount of disks of radius $1$ required to cover a disk of radius $10$. Jan 27 comment Is $\mathbb{Z}[x]/(f(x),p)\cong(\mathbb{Z}/p\mathbb{Z})[x]/(f(x))$?Yes, this is true. One way to prove this is to give an isomorphism. Can you think of a map between the two rings that would do the trick? Note that in general for a commutative ring $R$ with ideals $I,J\lhd R$ it is true that $(R/I)/J\cong R/(I+J)\cong(R/J)/I$. Try constructing the isomorphisms. Jan 27 revised Is $\mathbb{Z}[x]/(f(x),p)\cong(\mathbb{Z}/p\mathbb{Z})[x]/(f(x))$?Added parentheses for clarity. Jan 27 comment Suppose x and y are coprime integers and z is a natural number. Prove that If xy is a zth power then x and y are both zth powers.What have you tried so far? Jan 14 comment Decompose $A=D+N$ with $DN=ND$, $N$ nilpotent, $D$ diagonalizableNow you have $A= PJP^{-1}=P(D+N)P^{-1}=PDP^{-1}+PNP^{-1}$. Is $PDP^{-1}$ diagonalizable? Is $PNP^{-1}$ nilpotent? Do these two matrices commute? Once you have checked this, can you use these facts to find a simple expression for $A^n$ for any $n\geq1$? Jan 8 comment 2014 Money Challenge - EquationI wish I could upvote more than once. Nov 18 comment Taking the automorphism group of a group is not functorial.No, that is not what I am saying. I am showing that $\operatorname{Aut}$ does not work functorially on all of Grp. I do not exclude the possibility that it works functorially almost everywhere in Grp, though I certainly do not expect this to be true for any sensible definition of 'almost everywhere'. Nov 18 comment Taking the automorphism group of a group is not functorial.@Derek Holt: Your example would do the trick if the identity on $\operatorname{Aut}(H)$ cannot factor over $\operatorname{Aut}(NH)$. In this case it can, though as far as I know there is no 'nice' way for it to factor. Your general idea is a good one though. I have understood $NH\cong\operatorname{GA}(1,8)$, the group of affine transformations of $\Bbb{F}_8$. The group $\operatorname{GA}(1,32)$ does work; we have $\operatorname{GA}(1,32)=N\rtimes H$ with $|N|=32$, $|H|=31$. Then $\operatorname{Aut}(H)$ is cyclic of order $30$, and $|\operatorname{Aut}(NH)|=NHT$ with $T$ cyclic of order $5$. Nov 17 comment Taking the automorphism group of a group is not functorial.That would be ideal, though greater cardinality is stronger than what is needed; $\#\operatorname{Aut}(G)$ not being divisible by $\#\operatorname{Aut}(H)$ would suffice, I think. I'm not familiar enough with groups to have a large array of examples/candidates at hand. My first ideas were $V_4\subset D_4$ and $S_6\subset S_7$, and some variations on this, but they aren't quite retractions. Nov 17 comment Taking the automorphism group of a group is not functorial.There is no obvious candidate for what the induced maps should be. The point is to prove that for any choice of induced maps, the association is not functorial. Nov 17 asked Taking the automorphism group of a group is not functorial. Nov 14 accepted For a compact covering space, the fibres of the covering map are finite. Nov 14 comment For a compact covering space, the fibres of the covering map are finite.Aha, I failed to see that the fibre of such a point $x_i$ is disjoint from all $V_k^l$ with $k\neq i$. The last part I can do. Thank you for your help!