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.
(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.
(Edit: the arguments below build on observations by several other people in this thread.)
Case: $2 = 0$ in $R$. As has been observed elsewhere, in this case $G = \text{SL}_3(\mathbb{F}_2)$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $4$ and trace $r$. Then $g^4 = 1$ and $g^2 = rg + 1$, hence $g^4 = r^2 g^2 + 1$ and $r^2 g^2 = 0$. This gives $r^4 g^4 = 0$, hence $r^4 = 0$. But $r^2 g^2 = r^3 g + r^2 = 0$, hence $r^4 g = r^3 = 0$, hence $r^3 g = r^2 = 0$, and this is both necessary and sufficient.
The elements of $R$ which square to zero form a nilpotent ideal $I$. It follows that if $G$ embeds into $\text{SL}_2(R)$, the image of any elements of order $4$ in $\text{SL}_2(R/I)$ must, by the above computation, have order dividing $2$. In particular the image of $G$ is not isomorphic to $G$, so it must be trivial since $G$ is simple. Hence the image of $G$ in $\text{SL}_2(R)$ consists only of matrices congruent to the identity $\bmod I$. But any such matrix has trace squaring to zero, hence order dividing $4$, which contradicts the existence of elements of order $7$ in $G$. So no such embedding exists.
Case: $2$ is nilpotent in $R$. It is still true in this case that $SL_3({\mathbb F}_2)$ embeds into $\text{SL}_3(R)$, because the two complex 3-dimensional irreducible representations of $SL_3({\mathbb F}_2)=\text{PSL}_2({\mathbb F}_7)$ are realisable over ${\mathbb Z}_2$ (they need $(1\pm\sqrt{-7})/2$ which are in ${\mathbb Z}_2$), and $R$ is a ${\mathbb Z}_2$-algebra. However, as explained in the follow-up question, $SL_3({\mathbb F}_2)$ is not a subgroup of $\text{SL}_2(R)$ for any $R$.
Case: $2$ is not a zero divisor in $R$. As has been observed elsewhere, in this case $G = S_4$ embeds into $\text{SL}_3(R)$. Let $g \in \text{SL}_2(R)$ be an element of order dividing $2$ and trace $r$. Then $g^2 = 1$ and $g^2 - rg + 1 = 0$, hence $rg = 2$. Squaring gives $r^2 = 4$, so $r$ is also not a zero divisor. It follows that $g$ must be a scalar multiple of the identity, hence central. But $S_4$ contains elements of order $2$ which are not central. (As does $S_3$. Hence this argument also shows, as in Tim Dokchitser's answer, that $S_3$ does not embed into $\text{SL}_2(R)$ in this case.)
In particular the above arguments cover the case that $2$ is invertible, as well as the case that $R$ is an integral domain. The remaining case is that $2 \neq 0$, and it is a zero divisor but is not nilpotent.
One may hope that the reduction to Artinian rings as in the follow-up question and the fact that we know the answer both when $2$ is invertible and for ${\mathbb Z}_2$-algebras can actually finish this off.
Here is a brute force proof that $SL_3(R)$ does not embed into $SL_2(R)$ for any commutative ring $R$ with 1 where 2 is invertible, inspired by Kevin's universality remarks and Alex's observation that $S_3$ in my comment does not land in $SL_2(R)$.
The claim is that the symmetric group $S_3$ cannot be embedded in $SL_2(R)$ for any $R$: suppose it can be, take 2 general matrices $M=\begin{pmatrix}a&b\cr c&d\end{pmatrix}$ and $T=\begin{pmatrix}e&f\cr g&h\end{pmatrix}$ and consider the relations $M^3=id=T^2$, $M^2T=TM$ and $det M=det T=1$. These are 4+4+4+1+1=14 polynomial relations in 8 variables $a,...,h$, and a Groebner basis computation shows that the ideal they generate is $\langle h^2-1,g,f,h-e,d-1,c,b,a-1\rangle$; in Mathematica this is
$\gt$ M = {{a, b}, {c, d}}; T = {{e, f}, {g, h}}; id = {{1, 0}, {0, 1}};
$\gt$ GroebnerBasis[{M.M.M - id, T.T - id, M.M.T - T.M, Det[T] - 1, Det[M]-1 }]
{-1+h^2,g,f,-e+h,-1+d,c,b,-1+a}
In other words, the relations imply that $a=d=1$ and $b=c=0$ for any $R$, so $M=1$, contradicting the assumption that $S_3\to SL(2,R)$ is injective.
On the other hand, using $$ M=\begin{pmatrix}0&0&1\cr 1&0&0\cr 0&1&0\end{pmatrix},\qquad T=\begin{pmatrix}0&-1&0\cr -1&0&0\cr 0&0&-1\end{pmatrix}, $$ we can embed $S_3$ into $SL(3,R)$ for any ring $R$.
P.S. Hopefully, there is a better proof that $S_3$ does not embed into $SL(2,R)$!
Edit: The Groebner basis computation works over $Z[1/2]$, so this only works for rings $R$ with 2 invertible. Kevin & John: thank you for pointing this out!
Define a sequence of groups $G_i$ and associated group rings $R_i=\mathbb{Q}[G_i]$. To start put $G_0=\mathbb{Q}$. Then define $G_{i+1}=SL_3(R_i)$. The group $SL_3(R_i)$ is a subgroup of $SL_2(R_{i+1})$ because $G_{i+1}$ is a subgroup of $SL_2(R_{i+1})$ (as the group of certain diagonal matrices). Similarly, $G_i$ is a subgroup of $G_{i+1}$, and hence $R_i$ is a subring of $R_{i+1}$. Then $R=\bigcup_i R_i$ is the ring you want.
My guess that it should not be possible because $SL_2(R)$ would not have a 3-soluble nonnilpotent subgroup. I am not sure whether it is true that any soluble non-nilpotent subgroup would lie in Borel subgroup but I imagine that this is right...
EDIT: I misquoted the book, and the answers here show it isn't fixable.
Let $A,B \in SL_2(R)$. This book by Brumfiel and Hilden has the following two facts in it:
Taking $A^2 = 1$ and $A = B$, the first fact implies that $Tr(A) = \pm 2$, and then the second fact implies that $A$ is conjugate (and hence equal) to $\pm Id$. Since there is an embedding $S_4 \hookrightarrow SL_3(R)$ (which is described in the other answers), we can finish by noting that the images of the transpositions are not central in $SL_3(R)$.
I would guess that there's a more elementary way to prove that involutions in $SL_2(R)$ are central, but I don't know one at the moment.
$\newcommand{\SL}{\mathrm{SL}}$ $\newcommand{\m}{\mathfrak{m}}$ $\newcommand{\F}{\mathbf{F}}$ $\newcommand{\Z}{\mathbf{Z}}$
Assume that $\SL_3(R)$ is a subgroup of $\SL_2(R)$. We wish to obtain a contradiction.
Here is the strategy. Suppose that $R$ contains a subring of the form $A \oplus B$ where $2A = 0$. Then $SL_3(\F_2)$ is a subgroup of $\SL_3(A)$, which is a subgroup of $\SL_3(A \oplus B)$, which is a subgroup of $\SL_3(R)$. Hence, under our assumption on $R$, $\SL_3(\F_2)$ is a subgroup of $\SL_2(R)$, and this is ruled out by Silence Dogood's answer. $R$ trivially admits such a decomposition when $2 = 0$. Hence we may assume that $2 \ne 0$, and thus that $S_4$ is a subgroup of $\SL_3(R)$, and hence of $\SL_2(R)$.
If $S \subset R$ contains a subring of the form $A \oplus B$ with $2A = 0$, then so does $R$. Thus, WLOG, assume that $R$ is generated by the entries of $g-1$ where $g \in S_4 \subset \SL_2(R)$. Let $K \subset S_4$ denote the Klein $4$-subgroup. Then any map $S_4 \rightarrow G$ is injective if and only if the restriction $K \rightarrow G$ is non-zero (Obvious). $K$ is the only non-trivial normal subgroup of $S_4$ which is a $p$-group (Obvious). By construction, $R$ is Noetherian. If $x \in R$ is any element, and $\m$ is a maximal ideal containing the annihilator of $x$, then $x$ is non-zero in the localization $R_{\m}$. Hence there exists an $\m$ such that $K \rightarrow \SL_2(R_{\m})$ is non-zero, so $S_4 \rightarrow \SL_2(R_{\m})$ is injective. (Choose $x$ to be a non-zero matrix entry of $g-1$ for $g \in K$.) Let $A = R_{\m}$, and let $k = A/\m$. Consider the projection map $S_4 \rightarrow \SL_2(k)$, and let $H$ denote the kernel. Let $g$ be an element of $H$ which is not the identity (if such an element exists). By the Krull intersection theorem (as in SD's answer), there exists a minimal integer $n$ such that $$g - 1 \equiv 0 \mod \m^n, \qquad (g - 1) \not\equiv 0 \mod \m^{n+1}.$$ If $i$ is co-prime to the characteristic of $k$, then it is a unit in $A$, and $$g^i - 1 = (1 + (g-1))^i - 1 \equiv i (g-1) \mod \m^{n+1} \not\equiv 0 \mod \m^{n+1}.$$ It follows that the order of $g$ is some power of the characteristic (or is trivial if $\mathrm{char}(k) = 0$), and hence $H$ is a $p$-group. Hence either $S_4$ injects into $\SL_2(k)$, or $k$ has characteristic $2$ and $H = K$. The former does not occur. We shall prove that $2 = 0$ in $A$. The image of $S_4$ in $\SL_2(k)$ is $S_3$. $S_4$ contains an element $M$ of order $2$ which maps to an element of order $2$ in $S_3$ (for example, any $2$-cycle). The matrix $M$ has order two, and hence satisfies the polynomial $M^2 - 1 = 0$. Yet $M$ also has determinant one, and thus also satisfies the polynomial $M^2 - \mathrm{trace}(M) M + 1 = 0$, by Cayley--Hamilton. It follows that $\mathrm{trace}(M) M = 2 \ne 0$ (by assumption). Yet $M$ has at least one entry that is a unit, and thus $(\mathrm{trace}(M)) = (2)$ in $A$, and it follows easily that the image of $M$ is a scalar matrix in $\SL_2(k)$. Since $k$ has characteristic $2$, this implies that the image of $M$ in $\SL_2(k)$ is trivial ($v^2 = 1$ implies that $v = 1$), a contradiction. Hence $2 = 0$ in $A$.
We have now shown that $2 = 0$ in $A = R_{\m}$. Suppose we can show in addition that $A$ has finite length, that is $A/\m^k = A$ for some $k$. Assume this is so. Let $x_1, \ldots, x_n$ be generators of $\m^k \subset R$. By definition, $x_i$ maps to zero in the localization map $R \rightarrow R_{\m} = A$. Thus there exists an element $y_i \notin \m$ such that $y_i x_i = 0$. Let $y = y_1 \times \ldots \times y_n$. Since $y_i \notin \m$, the product $y \notin \m$. It follows that $$y + \m^k = R,$$ as the ideal on the LHS is not contained in any maximal ideal. On the other hand, $y$ annihilates $\m^k$ by construction. Thus, by the Chinese remainder theorem, $$R = R/y \m^k = R/y \oplus R/\m^k = R/y \oplus A.$$
Since $2 = 0$ in $A$, this shows that $R$ has the required decomposition. Thus we will be done if we can show that $A$ has finite length. Equivalently, we are done if we can show that the non-unit elements of $A$ are nilpotent.
It seems according to Tim that this won't work, since $S_4$ injects into $\SL_2(\F[[x]])$ via the map $$(12) \mapsto \left( \begin{matrix} 0 & 1 \\\ 1 & 0 \end{matrix} \right)$$ and $$(1234) \mapsto \left( \begin{matrix} 1+x+x^2 & 1+x^2 \\\ x^2 & 1+x+x^2 \end{matrix} \right)$$
Hence this answer, for the time being, is a complete fail.
