$\newcommand{\SL}{\mathrm{SL}}$
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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.

isa map from that ring $R$ above to $S$, reducing questions like CH for any ring, to CH for that one universal ring. One notices this when doing the first exercise I suggested: checking CH in the 2x2 case by bashing out the algebra. One has a bunch of equations in $A,B,C,D$ all of which magically work out, but instead of thinking of $A,B,C,D$ as varying through the ring one can think of them as poly variables. $\endgroup$ – Kevin Buzzard Nov 26 '10 at 10:27