Timeline for Inducing surjections on $GL_n(-)$?

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Jan 6, 2017 at 14:31	comment	added	მამუკა ჯიბლაძე		@Andrei It suffices to have it for scalar diagonal matrices. Still not obvious, though
Jan 4, 2017 at 13:58	comment	added	Andrei Smolensky		On the other hand, it must be surjective "on the determinants".
Jan 4, 2017 at 9:24	comment	added	Andrei Smolensky		@abx Why is a diagonal matrix the image of a diagonal matrix? Consider $R=\mathbb{C}[x, y, a, b]/(xb+ya-1)$ and then quotient out $(x-a, y-b)$. The matrix $\begin{pmatrix}x&0\\0&y\end{pmatrix}$ is the image of $\begin{pmatrix}(x+a)/2&(x-a)/2\\(y-b)/2&(y+b)/2\end{pmatrix}$, but, I believe, not the image of a diagonal matrix. This is not a counter-example to the claim, just a note that the proof is not obvious.
Jan 2, 2017 at 20:53	comment	added	abx		A necessary condition is that $A^\rightarrow B^$ is surjective (consider diagonal matrices).
Jan 2, 2017 at 20:19	comment	added	მამუკა ჯიბლაძე		A quick consideration: if $A\to B$ is surjective then ${Mat}_{n\times n}(A)\to{Mat}_{n\times n}(B)$ is surjective too, and $GL_n(A)$ is the same as $GL_1(\operatorname{Mat}_{n\times n}(A))$. So if you believe that it need not be true for $n=1$, then... :)
Jan 2, 2017 at 20:00	history	edited	YCor		edited tags
Jan 2, 2017 at 19:52	history	asked	BillScroggs	CC BY-SA 3.0