MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

9 deleted 1063 characters in body

One approach: bash it out: take the "universal" representation of $S_4$

It seems according to a local $k$ algebra where $2 = 0$, and show Tim that the image is Artinian.

Slightly more subtle approach: Let $G = S_4$. We have a representation:$$\rho: G \rightarrow \SL_2(k),$$and we are interested in the universal deformation to a complete local$k$-algebra, which we want to show is Artinian. If $X$ denotes the universal deformation ring, then byuniversalitythis won't work, the image of since$S_4$ lands inside injects into $\SL_2(\F[[x]])$ via the image of some map$X$(12) \rightarrow A$,and so it suffices to show that$X$has finite length. Now, hopefully,$X = k[x]/x^2$(on the principle that finite groups should not have infinite deformation rings by some kind of rigidity). Let$V$be the$2$-dimensional space with an action of$G$via$\rho$, andlet$W = mapsto \mathrm{Ad}(V)$. Then the tangent space to$X$(as a complete local$k$-algebra) is isomorphic to$H^1(G,W)$, left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)$$the relation ideal has H^2(G,W) relations.So it would be nice if they both have dimension 1, since thenX (1234) \simeq k[x]/x^n for some n. An unreliable computation: by inflation-restriction,$$0 mapsto \rightarrow H^1(\Z/2\Z,W^{\Z/3\Z}) left( \rightarrow H^1(S_3,W)begin{matrix} 1+x+x^2 & 1+x^2 \rightarrow H^1(\Z/3\Z,W).$$The last term is zero\ x^2 & 1+x+x^2 \end{matrix} \right)$$ Hence this answer, and$W^{\Z/3\Z}$is the regular representation for$S_3$,so$H^1(S_3,W) = 0$. Then, by inflation restrictionthe time being, $$H^1(S_4,W) \hookrightarrow H^1(K,W)^{S_3}.$$is a complete fail. 8 added 8 characters in body$\newcommand{\SL}{\mathrm{SL}}\newcommand{\m}{\mathfrak{m}}\newcommand{\F}{\mathbf{F}}\newcommand{\Z}{\mathbf{Z}}$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. One approach: bash it out: take the "universal" representation of$S_4$to a local$k$algebra where$2 = 0$, and show that the image is Artinian. Slightly more subtle approach: Let$G = S_4$. We have a representation: $$\rho: G \rightarrow \SL_2(k),$$ and we are interested in the universal deformation to a complete local$k$-algebra, which we want to show is Artinian. If$X$denotes the universal deformation ring, then by universality, the image of$S_4$lands inside the image of some map$X \rightarrow A$, and so it suffices to show that$X$has finite length. Now, hopefully,$X = k[x]/x^2$(on the principle that finite groups should not have infinite deformation rings by some kind of rigidity). Let$V$be the$2$-dimensional space with an action of$G$via$\rho$, and let$W = \mathrm{Ad}(V)$. Then the tangent space to$X$(as a complete local$k$-algebra) is isomorphic to$H^1(G,W)$, and the relation ideal has$H^2(G,W)$relations. So it would be nice if they both have dimension$1$, since then$X \simeq k[x]/x^n$for some$n$. An unreliable computation: by inflation-restriction, $$0 \rightarrow H^1(\Z/2\Z,W^{\Z/3\Z}) \rightarrow H^1(S_3,W) \rightarrow H^1(\Z/3\Z,W).$$ The last term is zero, and$W^{\Z/3\Z}$is the regular representation for$S_3$, so$H^1(S_3,W) = 0$. Then, by inflation restriction, $$H^1(S_4,W) \hookrightarrow H^1(K,W)^{S_3}.$$ 7 added 415 characters in body$\newcommand{\SL}{\mathrm{SL}}\newcommand{\m}{\mathfrak{m}}\newcommand{\F}{\mathbf{F}}\newcommand{\Z}{\mathbf{Z}}$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 $$(y) y + \m^k = R,$$ 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. One approach: bash it out: take the "universal" representation of$S_4$to a$k$algebra where$2 = 0$, and show that the image is Artinian. Slightly more subtle approach: Let$G = S_4$. We have a representation: $$\rho: G \rightarrow \SL_2(k),$$ and we are interested in the universal deformation to a complete local$k$-algebra, which we want to show is Artinian. If$X$denotes the universal deformation ring, then by universality, the image of$S_4$lands inside the image of some map$X \rightarrow A$, and so it suffices to show that$X$has finite length. Now, hopefully,$X = k[x]/x^2$(on the principle that finite groups should not have infinite deformation rings by some kind of rigidity). Let$V$be the$2$-dimensional space with an action of$G$via$\rho$, and let$W = \mathrm{Ad}^0(V)$. mathrm{Ad}(V)$. Then the tangent space to $X$ (as a complete local $k$-algebra) is isomorphic to $H^1(G,W)$, and the relation ideal has $H^2(G,W)$ relations. So it would be nice if they both have dimension $1$, since then $X \simeq k[x]/x^n$ for some $n$.

An unreliable computation: by inflation-restriction, $$0 \rightarrow H^1(\Z/2\Z,W^{\Z/3\Z}) \rightarrow H^1(S_3,W) \rightarrow H^1(\Z/3\Z,W).$$ The last term is zero, and $W^{\Z/3\Z}$ is the regular representation for $S_3$, so $H^1(S_3,W) = 0$. Then, by inflation restriction, $$H^1(S_4,W) \hookrightarrow H^1(K,W)^{S_3}.$$

6 added 11 characters in body
5 deleted 2 characters in body
4 added 1609 characters in body
3 deleted 22 characters in body
2 added 312 characters in body