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edited Dec 6 2010 at 1:32 |
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)$$the relation ideal has $H^2(G,W)$ relations.So it would be nice if they both have dimension $1$, since then$X $(1234) \simeq k[x]/x^n$ for some $n$. An unreliable computation: by inflation-restriction,$$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.
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edited Dec 6 2010 at 0:59 |
$\newcommand{\SL}{\mathrm{SL}}$
$\newcommand{\m}{\mathfrak{m}}$
$\newcommand{\F}{\mathbf{F}}$
$\newcommand{\Z}{\mathbf{Z}}$
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,
$$H^1(S_4,W) \hookrightarrow H^1(K,W)^{S_3}.$$
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edited Dec 6 2010 at 0:44 |
$\newcommand{\SL}{\mathrm{SL}}$
$\newcommand{\m}{\mathfrak{m}}$
$\newcommand{\F}{\mathbf{F}}$
$\newcommand{\Z}{\mathbf{Z}}$
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,
$$H^1(S_4,W) \hookrightarrow H^1(K,W)^{S_3}.$$
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edited Dec 5 2010 at 23:09 |
$\newcommand{\SL}{\mathrm{SL}}$
$\newcommand{\m}{\mathfrak{m}}$
$\newcommand{\F}{\mathbf{F}}$
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 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
$k$ 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
$Wk$-algebra, 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 $W = \mathrm{Ad}^0(V)$. 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.
We So it would be done nice if these they both have dimension $1$. Will report back on this soon.1$, since then
$X \simeq k[x]/x^n$ for some $n$.
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edited Dec 5 2010 at 22:40 |
$\newcommand{\SL}{\mathrm{SL}}$
$\newcommand{\m}{\mathfrak{m}}$
$\newcommand{\F}{\mathbf{F}}$
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 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
$k$ 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
$W(k)$-algebra, Wk$-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 $W = \mathrm{Ad}^0(V)$. 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.
We would be done if these both have dimension $1$. Will report back on this soon.
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edited Dec 5 2010 at 22:35 |
Claim: Suppose Assume that $S_4$ \SL_3(R)$ is a subgroup of $\SL_2(S)$. \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 subgroupof $\SL_3(A \oplus B)$, 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 sucha decomposition when $2 = 0$in $S$. Proof: Assume . Hence we may assume that $2 \ne 0$in , and thus that$S$. Then there exists S_4$ is a Noetherian ring subgroup of $R \SL_3(R)$, and hence of $\SL_2(R)$. If $S \subset S$in which R$ contains a subring of the form $2 A \ne oplus B$ with $2A = 0$and such , thenso does $R$.Thus, WLOG, assume that $S_4$ R$ is contained in generated by the entries of $\SL_2(R)$ (Obvious).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). Arguing as in Silence Dogood's answer: http://mathoverflow.net/questions/47710/does-s-4-inject-into-sl2-r-for-some-commutative-ring-r/47761#47761 there exists an Artinian quotient$R/I$ of By construction, $R$ such that $S_4$ embeds in $\SL_2(R/I)$ and is Noetherian. If $2 x \ne 0$ in $R/I$.Proof: (When applying Step 2 of SD's answerR$ is any element, add and $2$ to \m$ is a maximal ideal containing the list annihilator of elements considered; by assumption, $2$ is non-zero in $R$, and hence x$, then $2$ x$ is non-zero in the Artinian quotient localization $R/I$.)Since Artinian rings are semi-local, R_{\m}$. Hence there exists an injectionof $S_4$ into a product of groups $\SL_2(A)$ for a finite number of Artinian rings \m$ such that $A$.The map from K \rightarrow \SL_2(R_{\m})$ is injective. (Choose $K$ x$ to at least one of these factors must be a non-zero , and hence as remaked above,we obtain an injectionmatrix entry of $S_4 \rightarrow \SL_2(A)$g-1$ for some local Artinian ring $g \in K$.) Let $A = (A,\m,k)$.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). Since $A$ is ArtinianBy 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. Hence the 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,proving the claim. If Hence $2 \ne = 0$ in $R$, then A$. We have now shown that $S_4$ 2 = 0$ in $A = R_{\m}$. Suppose we can show inaddition that $A$ has finite length, that is a subgroup of $\SL_3(R)$ but not A/\m^k = A$ for some $k$. Assume this is so.Let $x_1, \ldots, x_n$ be generators of $\SL_2(R)$, by \m^k \subset R$. By definition, $x_i$ maps to zeroin the argument abovelocalization map $R \rightarrow R_{\m} = A$. If Thus there exists an element$2 y_i \notin \m$ such that $y_i x_i = 0$in . Let $R$, then y = y_1 \times \ldots\times y_n$. Since $\mathrm{PSL}_3(\F_2)$ is a subgroup ofy_i \notin \m$, the product $\SL_3(R)$ but not a subgroup of y \notin \m$. It follows that$\SL_2(R)$ (see SD's answer referenced above and $(y) + \m^k = R,$$as the subsequent comments)ideal on the LHS is not contained in any maximal ideal. EDIT: BuggerOn the other hand,$y$ annihilates $\m^k$ by construction. Thus, by the Chinese remainder theorem,$$R = R/y \m^k = R/y \oplus R/\m^k = R/y \oplus A.$$ Since $2 = 0$ in $A$, this shows that $R$ has the required decomposition.Thus we will teach me to be a smart assdone if we can show that $A$ has finite length.There's an issueEquivalently, namely thatthis really only proves we are done if we can show that there exists a maximal ideal the non-unit elements of $\m$ A$are nilpotent. One approach: bash it out: take the "universal" representation of $R$ such thatS_4$ to a $k$ algebra where $2 = 0$in , and show that the image is Artinian. Slightly more subtle approach: Let $R_{\m}$. (Having G = S_4$. We have a representation:$2 $\rho: G \ne 0$ rightarrow \SL_2(k),$$and we are interested in the universal deformation to a complete local$R/I = A W(k)$-algebra, which we want to show is Artinian. If $X$ denotes the universal deformation ring, then byuniversality, the image of $S_4$ lands inside the image of some map $X \oplus B$ doesn't imply rightarrow A$,and so it suffices to show that $2 X$ 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 $W = \ne 0$ in mathrm{Ad}^0(V)$. Then the tangent space to $A$). Will think X$(as a little morecomplete local $k$-algebra) is isomorphic to$H^1(G,W)$, and the relation ideal has $H^2(G,W)$ relations. We would be done if these both have dimension $1$. Will report back on this soon.
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edited Dec 5 2010 at 20:40 |
$\newcommand{\SL}{\mathrm{SL}}$
$\newcommand{\m}{\mathfrak{m}}$
$\newcommand{\F}{\mathbf{F}}$
Claim: Suppose that $S_4$ is a subgroup of $\SL_2(S)$. Then
$2 = 0$ in $S$.
Proof: Assume that $2 \ne 0$ in $S$. Then there exists a Noetherian ring
$R \subset S$
in which $2 \ne 0$ and such that $S_4$ is contained in $\SL_2(R)$ (Obvious).
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).
Arguing as in Silence Dogood's answer:
http://mathoverflow.net/questions/47710/does-s-4-inject-into-sl2-r-for-some-commutative-ring-r/47761#47761
there exists an Artinian quotient
$R/I$ of $R$ such that $S_4$ embeds in $\SL_2(R/I)$ and $2 \ne 0$ in $R/I$.
Proof: (When applying Step 2 of SD's answer, add $2$ to the list of
elements considered; by assumption, $2$ is non-zero in $R$, and hence $2$ is
non-zero in the Artinian quotient $R/I$.)
Since Artinian rings are semi-local, there exists an injection
of $S_4$ into a product of groups $\SL_2(A)$ for a finite number of Artinian rings $A$.
The map from $K$ to at least one of these factors must be non-zero, and hence
as remaked above,
we obtain an injection
$S_4 \rightarrow \SL_2(A)$
for some local Artinian ring $A = (A,\m,k)$.
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). Since $A$ is Artinian, 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.
Hence 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,
proving the claim.
If $2 \ne 0$ in $R$, then $S_4$ is a subgroup of $\SL_3(R)$ but not of
$\SL_2(R)$, by the argument above. If $2 = 0$ in $R$, then $\mathrm{PSL}_3(\F_2)$ is a subgroup of
$\SL_3(R)$ but not a subgroup of $\SL_2(R)$ (see SD's answer referenced above and the subsequent comments).
EDIT: Supposedly 22 people thought this question was interestingBugger, but apparently nobody thinks the solution is... possibly because
it involves elementary commutative algebrathat will teach me to be a smart ass. Is there anyone here
besides Buzzard who can tell when There's an argument is correct?
Ah well ... donner de la confiture aux cochonsissue, namely that
this really only proves that there exists a maximal ideal $\m$ of $R$ such that
$2 = 0$ in $R_{\m}$. (Having $2 \ne 0$ in $R/I = A \oplus B$ doesn't imply that $2 \ne 0$ in $A$). Will think a little more...
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edited Dec 5 2010 at 15:59 |
$\newcommand{\SL}{\mathrm{SL}}$
$\newcommand{\m}{\mathfrak{m}}$
$\newcommand{\F}{\mathbf{F}}$
Claim: Suppose that $S_4$ is a subgroup of $\SL_2(S)$. Then
$2 = 0$ in $S$.
Proof: Assume that $2 \ne 0$ in $S$. Then there exists a Noetherian ring
$R \subset S$
in which $2 \ne 0$ and such that $S_4$ is contained in $\SL_2(R)$ (Obvious).
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).
Arguing as in Silence Dogood's answer:
http://mathoverflow.net/questions/47710/does-s-4-inject-into-sl2-r-for-some-commutative-ring-r/47761#47761
there exists an Artinian quotient
$R/I$ of $R$ such that $S_4$ embeds in $\SL_2(R/I)$ and $2 \ne 0$ in $R/I$.
Proof: (When applying Step 2 of SD's answer, add $2$ to the list of
elements considered; by assumption, $2$ is non-zero in $R$, and hence $2$ is
non-zero in the Artinian quotient $R/I$.)
Since Artinian rings are semi-local, there exists an injection
of $S_4$ into a product of groups $\SL_2(A)$ for a finite number of Artinian rings $A$.
The map from $K$ to at least one of these factors must be non-zero, and hence
as remaked above,
we obtain an injection
$S_4 \rightarrow \SL_2(A)$
for some local Artinian ring $A = (A,\m,k)$.
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). Since $A$ is Artinian, 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.
Hence 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,
proving the claim.
If $2 \ne 0$ in $R$, then $S_4$ is a subgroup of $\SL_3(R)$ but not of
$\SL_2(R)$, by the argument above. If $2 = 0$ in $R$, then $\mathrm{PSL}_3(\F_2)$ is a subgroup of
$\SL_3(R)$ but not a subgroup of $\SL_2(R)$ (see SD's answer referenced above and the subsequent comments).
EDIT: Supposedly 22 people thought this question was interesting,
but apparently nobody thinks the solution is... possibly because
it involves elementary commutative algebra. Is there anyone here
besides Buzzard who can tell when an argument is correct?
Ah well ... donner de la confiture aux cochons ...
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answered Dec 4 2010 at 15:06 |
$\newcommand{\SL}{\mathrm{SL}}$
$\newcommand{\m}{\mathfrak{m}}$
$\newcommand{\F}{\mathbf{F}}$
Claim: Suppose that $S_4$ is a subgroup of $\SL_2(S)$. Then
$2 = 0$ in $S$.
Proof: Assume that $2 \ne 0$ in $S$. Then there exists a Noetherian ring
$R \subset S$
in which $2 \ne 0$ and such that $S_4$ is contained in $\SL_2(R)$ (Obvious).
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).
Arguing as in Silence Dogood's answer:
http://mathoverflow.net/questions/47710/does-s-4-inject-into-sl2-r-for-some-commutative-ring-r/47761#47761
there exists an Artinian quotient
$R/I$ of $R$ such that $S_4$ embeds in $\SL_2(R/I)$ and $2 \ne 0$ in $R/I$.
Proof: (When applying Step 2 of SD's answer, add $2$ to the list of
elements considered; by assumption, $2$ is non-zero in $R$, and hence $2$ is
non-zero in the Artinian quotient $R/I$.)
Since Artinian rings are semi-local, there exists an injection
of $S_4$ into a product of groups $\SL_2(A)$ for a finite number of Artinian rings $A$.
The map from $K$ to at least one of these factors must be non-zero, and hence
as remaked above,
we obtain an injection
$S_4 \rightarrow \SL_2(A)$
for some local Artinian ring $A = (A,\m,k)$.
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). Since $A$ is Artinian, 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.
Hence 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,
proving the claim.
If $2 \ne 0$ in $R$, then $S_4$ is a subgroup of $\SL_3(R)$ but not of
$\SL_2(R)$, by the argument above. If $2 = 0$ in $R$, then $\mathrm{PSL}_3(\F_2)$ is a subgroup of
$\SL_3(R)$ but not a subgroup of $\SL_2(R)$ (see SD's answer referenced above and the subsequent comments).
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