Timeline for Is the map $GL_n(\mathbb{Z})\to GL_n(\mathbb{Z}/2\mathbb{Z})$ surjective?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 18, 2021 at 14:24 | comment | added | LSpice | Ah, I saw the Chinese-Remainder-Theorem reduction, but got stuck on the prime-power case, which is embarrassing since I'm a $p$-adic analyst! If one were inclined to overcomplicate things (and probably risk circular reasoning), I suppose that one could think of it in Hensel's-lemma terms of lifting points on smooth schemes from the residue field of a local field up to the ring of integers of the local field. | |
| Mar 18, 2021 at 13:22 | comment | added | David E Speyer | Then, multiplying by transvections, we can make $g_{ik}=g_{ki}=0$ for $2 \leq k \leq n$. Now induct on $n$. | |
| Mar 18, 2021 at 13:22 | comment | added | David E Speyer | Sketch of proof of claim about local rings: Recall that the matrices $\left[ \begin{smallmatrix} 0&1 \\ -1&0 \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} u& 0 \\ 0& u^{-1} \end{smallmatrix} \right]$ are products of transvections (for $u$ any unit). Now, let $g$ be an $n \times n$ determinant $1$ matrix over a local ring $R$. There must be some entry $g_{ij}$ which is not in the maximal ideal, and hence a unit. Using the $2 \times 2$ matrices above, we can move that entry into position $(1,1)$ and make it be $1$. So we can assume that $g_{11}=1$. | |
| Mar 18, 2021 at 13:19 | comment | added | YCor | @LSpice yes. The general case follows from the case when $m$ is a prime power by Chinese theorem. The case $m=p^k$ prime power is an induction on $k$ and one can boil down to a matrix of the form $I+p^{k-1}A$ with $A$ of trace zero, and one easily reduces to $A$ diagonal $(1,-1,0,\dots,0)$ which is easy to deal with. | |
| Mar 18, 2021 at 13:19 | comment | added | David E Speyer | @LSpice Yes. By the Chinese Remainder Theorem, we can reduce to prime powers. The ring $\mathbb{Z}/p^k \mathbb{Z}$ is local and, if $R$ is local, then $SL_n(R)$ is generated by transvections. | |
| Mar 18, 2021 at 12:59 | comment | added | LSpice | @YCor, re, my answer used that $\operatorname{SL}_n(\mathbb Z/m\mathbb Z)$ was generated by transvections, which I only know to be true for $m$ prime. Is it true in general? | |
| Mar 18, 2021 at 7:00 | comment | added | YCor | In general, by the same simple argument as in the answer, the image of $GL_n(Z)\to GL_n(Z/mZ)$ is the seet of matrices with det $\pm 1$, and in particular it's surjective iff $m=2,3,4,6$. | |
| Mar 18, 2021 at 4:54 | vote | accept | Nanjun Yang | ||
| Mar 18, 2021 at 3:51 | answer | added | LSpice | timeline score: 8 | |
| Mar 18, 2021 at 3:49 | history | asked | Nanjun Yang | CC BY-SA 4.0 |