Suppose $F$ is a field. I want to know whether the map $GL_n(GW(F))\to GL_n(W(F))$ is surjective, where $GW$ means Grothendieck-Witt and $W$ means Witt. In the case $F$ is algebraic closed, it reduces to the surjectivity of $GL_n(\mathbb{Z})\to GL_n(\mathbb{Z}/2\mathbb{Z})$. I know the case $n=1$ is true.
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3$\begingroup$ 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 $m=2,3,4,6$. $\endgroup$– YCorCommented Mar 18, 2021 at 7:00
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1$\begingroup$ @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. $\endgroup$– David E SpeyerCommented Mar 18, 2021 at 13:19
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1$\begingroup$ @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. $\endgroup$– YCorCommented Mar 18, 2021 at 13:19
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1$\begingroup$ 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 $g_{11}=1$. $\endgroup$– David E SpeyerCommented Mar 18, 2021 at 13:22
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1$\begingroup$ Then, multiplying by transvections, we can make $g_{ik}=g_{ki}=0$ for $2 \leq k \leq n$. Now induct on $n$. $\endgroup$– David E SpeyerCommented Mar 18, 2021 at 13:22
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1 Answer
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For the question in your title, yes: $\operatorname{GL}_n(\mathbb Z/2\mathbb Z) = \operatorname{SL}_n(\mathbb Z/2\mathbb Z)$ is generated by transvections, and these obviously lift to $\operatorname{GL}_n(\mathbb Z)$.
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2$\begingroup$ This actually answers the surjectivity of $GL_n(GW(F))\to GL_n(W(F))$ since $GW(F)$ is determined by a fiber square $\mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}\leftarrow W(F)$! $\endgroup$ Commented Mar 18, 2021 at 4:54