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I am trying to show that the kernel of the natural map $SL(n, \mathbb{Z}) \to SL(n, \mathbb{Z}/3\mathbb{Z})$ has no torsion. That is, if $A$ is in the kernel then $A = I$ or $A^n \neq I$ for all $n \in \mathbb{N}\setminus\{0\}$.

If $A$ is in the kernel, it is of the form $I + 3B$ where $B$ is an integer matrix. Beyond that, I don't have any ideas. None of the canonical forms seem to be helpful as these are integer valued matrices.

As always, any hints would be very much appreciated.

Michael Albanese
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  • I've replaced $\mathbb{Z}_3$ by $\mathbb{Z}/3$. Given that the notation $\mathbb{Z}_p$ is used for the ring of $p$-adic integers and that one really studies the linear groups over such rings, I really wonder why many authors and teachers still write $\mathbb{Z}_n$ when they really mean $\mathbb{Z}/n$ (pretending that it is shorter... well it's just confusing). – Martin Brandenburg Aug 17 '14 at 06:02
  • I prefer the notation $\mathbb{Z}/3\mathbb{Z}$ to $\mathbb{Z}/3$. – Michael Albanese Aug 17 '14 at 14:53

2 Answers2

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Suppose that $A=I+3B$ is torsion. Then, note that it's eigenvalues must all be of unit length. But, the eigenvalues of $I+3B$ are precisely those of the form $1+3\lambda$ for $\lambda$ an eigenvalue of $B$. So, in particular, we see that for any eigenvalue $\lambda$ of $B$ that

$$|3\lambda|=|1+3\lambda-1|\leqslant |1+3\lambda|+1=2$$

So, $|\lambda|\leqslant \frac{2}{3}$. But, note that since all the conjugates of $\lambda$ must also be roots of the characteristic polynomial of $B$ (since the characteristic polynomial is in $\mathbb{Z}[x]$), the same argument shows that they must also have norm less than or equal to $\frac{2}{3}$. So, if $m(x)$ denotes the minimal polynomial for $\lambda$ over $\mathbb{Q}$, then it's constant term is an integer (since $\lambda$ is an algebraic integer), but certainly less than $\frac{2}{3}$ in modulus, being (up to sign) the product of the conjugates of $\lambda$, and so must be $0$. But, since $m(x)$ is irreducible, and monic, this implies that $m(x)=x$. So $\lambda=0$. So, all the eigenvalues of $B$ are $0$.

But, then this implies that all the eigenvalues of $A$ are $1$, but since $A$ is diagonalizable, since it's finite order, $A=I$.

Alex Youcis
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  • What do you mean by the conjugates of $\lambda$? – Michael Albanese Aug 17 '14 at 03:05
  • @MichaelAlbanese Galios conjugates. – Alex Youcis Aug 17 '14 at 03:05
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    (Given this argument fails for $2$ instead of $3$, I'm tempted to find a counterexample in the case of $\Bbb Z_2$, or an improved argument). – Pedro Aug 17 '14 at 03:05
  • For anyone who wants to know why a finite order matrix is diagonalisable, see this question. – Michael Albanese Aug 17 '14 at 03:34
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    Pedro, what about diag(1,-1,1,-1). – Alex Zorn Aug 17 '14 at 03:39
  • @AlexYoucis the counterexample as you've written it doesn't have determinant 1. The thing I wrote down generalizes for n a multiple of 4, but I don't see how you find one for other n. – Alex Zorn Aug 17 '14 at 07:57
  • Wait, nevermind. You can just do diag(-1,-1,1,1,1...) and that should work for $n \geq 2$. – Alex Zorn Aug 17 '14 at 07:59
  • @AlexZorn I apologize, I shouldn't have commented so hastily. :) – Alex Youcis Aug 17 '14 at 08:12
  • Very nice answer! I just wanted to point out something small. The fact that "if $A$ has finite order, then $A$ is diagonalizable" is not true in general over the ground field; for example, if $A = \left( \begin{array}{cc} -1 & 1 \ -1 & 0 \end{array} \right)$, then $A$ satisfies $A^{3} = 1$ but $A$ is not diagonalizable over $\mathbb{R}$. So it was easier for me to conclude your argument by saying that: If $A\neq I$, since $A$ has finite order, the minimal polynomial of $A$ divides $X^{n}-1$, so it has at least two distinct eigenvalues. But all eigenvalues of $A$ are 1, contradiction! – Prism Mar 15 '17 at 23:46
  • One more thing: The same argument should prove that the kernel of $GL(n, \mathbb{Z})\to GL(n, \mathbb{Z}/3\mathbb{Z})$ has no torsion, correct? I am just curious where (if anywhere) we used anything special about $SL(n, \mathbb{Z})$. Thanks in advance for your time! :) – Prism Mar 16 '17 at 00:24
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Here's a proof that's basis free. It works with $3$ replaced by any odd prime.

Let $A$ be a finite order element of the kernel. The first thing to realize is that $A$ has order a power of $3.$ To see this one observes that since $\langle A \rangle$ is finite and $\bigcap \ker(SL_n(\mathbb{Z}) \rightarrow SL_n(\mathbb{Z}/3^d)) = 1,$ it must be the case $\langle A \rangle$ injects into $ \Gamma_d := \ker(SL_n(\mathbb{Z}/3^d) \rightarrow SL_n(\mathbb{Z}/3))$ for some large integer $d.$ Since $ \Gamma_d$ is a $3$-group the claim follows.

We now replace $A$ with some power and assume $A$ has order dividing 3.

Let $R$ be the ring $\mathbb{Z}_3[X]/(X^3 - 1)$ and $M$ be the $R$-module which is free over $\mathbb{Z}_3$ (the 3-adic integers) of rank $n$ where $X$ acts by $A.$ The ring $R$ is 1 dimensional. It has two minimal prime ideals: $$P_1 := (X -1) \text{ and } P_2 := (X^2 +X +1)$$ and one maximal ideal $\mathfrak{m} := (3, X - 1).$ One observes $$P_1 + P_2 = \mathfrak{m}.$$ To say $A =1$ is to say $R$ acts on $M$ through $R/P_1.$ This occurs if and only if $R/P_2$ acts on $R/P_2\otimes_R M $ through $R/P_1 + P_2 =R/\mathfrak{m}.$

The ring $R/P_2$ is a PID (in fact it's a DVR). Therefore by the classification of modules over a PID,

$$M \otimes_R R/P_2 \cong \bigoplus_{i=1}^N R/(\mathfrak{m}^{e_i} + P_2)$$ where $e_i$ is a positive number or infinity which depends on $i$.

Since $A$ is trivial modulo $3,$ one has $R$ acts on $M/3M$ through $R/(3)+ P_1 = R/\mathfrak{m}.$

But $R/((3) + \mathfrak{m}^{e_i} + P_2) \cong R/\mathfrak{m}$ only if $e_i = 1$. (This fact is not true if 3 is replaced by 2.) It follows $$R/P_2\otimes_R M \cong (R/\mathfrak{m})^N$$ and the claim follows.

user26857
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jspecter
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