TITLE: Does $SL_3(R)$ embed in $SL_2(R)$?
QUESTION [44 upvotes]: Is there any non-trivial ring such that $SL_{3}(R)$ is isomorphic to a subgroup of $SL_{2}(R)$?
$SL_{3}(\mathbb{Z})$ is not an amalgam, and has the wrong number of order $2$ elements to be a subgroup of $SL_{2}(\mathbb{Z})$. 
Is there any non-trivial ring where this occurs? When can this definitely not occur? I am trying to understand if there is any sort of group-theoretically apparent notion of dimension here.

REPLY [23 votes]: (Edit: I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you Andrei Smolensky for repeating correcting my embarrassing mistakes here.)
I am surprised that this old question was not fully answered yet.
The answer is "No" and it is well known in some circles.
In fact, a far more general statement holds:
1) Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ factors via the non-trivial quotient $\text{SK}_1(3,R):=\text{SL}_3(R)/\text{EL}_3(R)$, where $\text{EL}_3(R)$ is the subgroup generated by elementary matrices in $\text{SL}_3(R)$.
Note that $\text{SL}_3(R)$ contains an epimorphic image of $\text{SL}_3(\mathbb{Z})$ (induced by the map $\mathbb{Z}\to R$). It is well known (and easy) that $\mathrm{EL}_3(\mathbb{Z})\simeq \mathrm{SL}_3(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ is contained in $\mathrm{EL}_3(R)$, and in fact $\mathrm{EL}_3(R)$ is generated by $\text{SL}_3(\mathbb{Z})$ as a normal subgroup (as you can observe by playing with commutation relation of elementary matrices). Thus (1) is equivalent to:
2) Let $S$ be a ring (commutative with 1). Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S)$ is trivial.
We now fix a homomorphism as in statement (2) and assume its image is non-trivial. Let $\mathfrak{n}<S$ denote the nilpotent radical. 
It is easy to see that every finitely generated subgroup of the kernel of $\text{SL}_2(S)\to \text{SL}_2(S/\mathfrak{n})$ is nilpotent. By the facts that $\text{SL}_3(\mathbb{Z})$ is finitely generated without nilpotent quotients we deduce that it is mapped non-trivially to $\text{SL}_2(S/\mathfrak{n})$.
Since $\mathfrak{n}$ is the intersection of all prime ideal we deduce that $\text{SL}_3(\mathbb{Z})$ is mapped non-trivially to $\text{SL}_2(S/\mathfrak{p})$ for some prime ideal $\mathfrak{p}<S$.
By letting $k$ be the field of fractions of $S/\mathfrak{p}$ we see that it is enough to prove
the following statement:
3) Let $k$ be a field. Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(k)$ is trivial.
(here I had before an argument I liked, but I had to replace it by a simpler one.)
Here is a nice exercise:
4) Let $k$ be a field. Then for any group homomorphism $\text{H}(\mathbb{Z})\to\text{SL}_2(k)$, where $\text{H}(\mathbb{Z})$ is the integral Heisenberg group, the image of the center (=commutator group) of $\text{H}(\mathbb{Z})$ consists of scalar matrices.
Hint: Assume the image of a generator of the center is not a scalar matrix and show that $\text{H}(\mathbb{Z})$ is in the Borel, in which every nilpotent group is abelian (you may assume that $k$ is algebraically closed here).
Remark: Actually, the image of the center of $\text{H}(\mathbb{Z})$ will be trivial unless $\text{char}(k)=2$.
To finish up with (3), observe that every elementary matrix in $\text{SL}_3(\mathbb{Z})$ is the center of a conjugate of $\text{H}(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ consists of scalar matrices. But $\text{SL}_3(\mathbb{Z})$ is perfect, so this image is trivial.