Show that the kernel of the map $SL(n, \mathbb{Z}) \to SL(n, \mathbb{Z}/3\mathbb{Z})$ has no torsion.

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.

Solution 1:

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, $A=I$.

Solution 2:

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.