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:
- Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ has an abelian image.
Recall that $\mathrm{EL}_3(R)<\text{SL}_3(R)$ (which denotes the subgroup generated by elementary matrices) is normal with an abelian cokernel. Thus, to prove statement (1) it is enough to show that $\mathrm{EL}_3(R)$ is in the kernel of any homomorphism as above. Observe (by playing with commutation relations of elementary matrices) that the normal closure of the image of $\mathrm{EL}_3(\mathbb{Z})\to \mathrm{EL}_3(R)$ (induced by the map $\mathbb{Z}\to R$) is $\mathrm{EL}_3(R)$. Finally recall that $\mathrm{EL}_3(\mathbb{Z})\simeq \text{SL}_3(\mathbb{Z})$. Thus it is enough for us to prove the following statement:
- Let $S$ be a ring (commutative with 1). Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S)$ is trivial.
(By the way, I could have started at this point: statement (2) already answers the asked question.)
We now fix a homomorphism as in statement (2) and assume its image is non-trivial. It is standard that there is a maximal ideal $m\lhd S$ and an integer $k$ such that the image of $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m^k)$ is already non-trivial. Note that the the kernel of $\text{SL}_2(S/m^k)\to\text{SL}_2(S/m)$ is nilpotent and that $\text{SL}_3(\mathbb{Z})$ has no non-trivial nilpotent quotients. It follows that $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m)$ is non-trivial. We are left to prove the following statement:
- Let $k$ be a field. Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(k)$ is trivial.
Now there are many ways to proceed, and mine is not better than yours (please let me know your quick proof in a comment), but let me shoot with all the guns. There is a beatiful argument of Tits which I want to use. Since $\text{SL}_3(\mathbb{Z})$ is finitely generated, the matrix elements of its image generate a finitely generated domain in $k$, and this one could be embedded in a local field. Moreover, if the image of $\text{SL}_3(\mathbb{Z})$ in $\text{SL}_2(k)$ is infinite, this new embedding could be chosen such that the image of $\text{SL}_3(\mathbb{Z})$ is actually unbounded. Thus we get the following:
- In (3) we can take $k$ to be a local field and assume that the image of the homomorphism is either unbounded or finite.
If the image of the homomorphism is unbounded we get a contradiction to Margulis' super-rigidity. So we may assume that the image is finite. In case $\text{char}(k)=0$ we may replace it with $\mathbb{C}$, thus assume that the image is in $\mathrm{SU}(2)$, which has no non-abelain nilpotent group, unlike any image of $\text{SL}_3(\mathbb{Z})$, and get a contradiction. So we get $k=\mathbb{F}_q((t))$ for some prime power $q$. The finite image of $\text{SL}_3(\mathbb{Z})$ is contained (up to conjugation) in the maximal compact $\text{SL}_2(\mathbb{F}_q[[t]])$. Arguing as we did after statement (2), with $S=\mathbb{F}_q[[t]]$ and $m=(t)$, we get:
- In (3) we may assume $k$ is finite.
Since every proper algebraic subgroup of $\text{SL}_2$ is solvable and every non-trival quotient of $\text{SL}_3(\mathbb{Z})$ is not, the image is Zariski-dense. By the CSP the homomorphism factors through $\text{SL}_3(\mathbb{Z}/n)\to\text{SL}_2(\mathbb{F}_q)$ for some $n$. By the simplicity of $\SL_2(\mathbb{F}_q)$ (we know it is not solvable, so $q\geq 5$), the Zariski-density of the image and the CRT $n$ is a prime power. By similar reasons the image cannot have a nilpotent normal subgroup, so $n$ is a prime. We are left to prove:
- For a prime $p$ and a prime power $q$, any group homomorphism $\text{SL}_3(\mathbb{F}_p)\to\text{SL}_2(\mathbb{F}_q)$ is trivial.
Now we should be a bit careful: recall the isomorphism $\text{SL}_3(\mathbb{F}_2)\simeq \mathrm{PSL}_2(\mathbb{F}_7)$. I haven't check here things carefully. I think that every nilpotent non-abelian subgroup of $\text{SL}_2(\mathbb{F}_q)$ has abelinzation of order $2$. This forces $p=2$. I will come back here an edit the rest after I'll have a cup of coffee.
