Let \(V\) be a finite-dimensional vector space over \(\mathbb Q\), and let \(M\in\operatorname{Aut}(V)\) fix no nonzero vector. Suppose \[ M^p=I \] for a prime \(p\). Show that \[ p-1\mid\dim_{\mathbb Q}V. \]
Solution
If \(V=0\), then \[ \dim_{\QQ}V=0 \] is divisible by \(p-1\). Assume henceforth that \(V\neq0\).
1. The linear map\[ M-I:V\longrightarrow V \]is invertible.
Its kernel is \[ \ker(M-I) = \{v\in V:Mv=v\}. \] By hypothesis, \(M\) fixes no nonzero vector, so this kernel is \(\{0\}\). Thus \(M-I\) is injective. Since \(V\) is finite-dimensional, every injective endomorphism of \(V\) is surjective and hence invertible.
2. Let\[ \Phi_p(x) = 1+x+\cdots+x^{p-1}. \]Then\[ \boxed{ \Phi_p(M)=0. } \]
The polynomial identity \[ x^p-1 = (x-1)\Phi_p(x) \] gives \[ M^p-I = (M-I)\Phi_p(M). \] By hypothesis, \(M^p=I\), so \[ (M-I)\Phi_p(M)=0. \] Step 1 says that \(M-I\) is invertible. Multiplying by its inverse gives \[ \Phi_p(M)=0. \]
3. The polynomial \(\Phi_p(x)\) is irreducible over \(\QQ\).
The substitution \[ x\longmapsto x+1 \] is an automorphism of \(\QQ[x]\), so it preserves irreducibility. Using \[ \Phi_p(x) = \frac{x^p-1}{x-1}, \] one obtains \[ \begin{aligned} \Phi_p(x+1) &= \frac{(x+1)^p-1}{x}\\ &= \sum_{k=1}^{p} \binom pk x^{k-1}. \end{aligned} \] Its leading coefficient is \(1\). Since \(p\) is prime, \[ p\mid\binom pk \] for every \(1\leq k\leq p-1\). Its constant term is \[ \binom p1=p, \] which is not divisible by \(p^2\). Eisenstein’s criterion with the prime \(p\) therefore shows that \(\Phi_p(x+1)\) is irreducible over \(\QQ\). Consequently \(\Phi_p(x)\) is irreducible over \(\QQ\).
4. The quotient\[ K = \QQ[x]/(\Phi_p(x)) \]is a field satisfying\[ \boxed{ [K:\QQ]=p-1. } \]
By step 3, the polynomial \(\Phi_p\) is irreducible over the field \(\QQ\), so its principal ideal is maximal and the quotient is a field. Moreover, \[ \deg\Phi_p=p-1. \] The residue classes of \[ 1,x,\ldots,x^{p-2} \] form a \(\QQ\)-basis of the quotient, so its degree over \(\QQ\) is \(p-1\).
5. The vector space \(V\) has a natural structure of vector space over\(K\).
For a residue class \[ [q(x)]\in K \] and \(v\in V\), define \[ [q(x)]\cdot v = q(M)v. \] This is well defined: if \[ q(x)-r(x) \in (\Phi_p(x)), \] then \[ q(x)-r(x) = s(x)\Phi_p(x) \] for some \(s(x)\in\QQ[x]\), and step 2 gives \[ \bigl(q(M)-r(M)\bigr)v = s(M)\Phi_p(M)v = 0. \] The field and vector-space axioms follow from the corresponding polynomial identities under evaluation at \(M\).
6. The dimension of \(V\) over \(\QQ\) satisfies\[ \dim_{\QQ}V = (p-1)\dim_KV. \]
Let \[ m=\dim_KV. \] This dimension is finite because \(V\) is already finite-dimensional over \(\QQ\). Choose a \(K\)-basis \[ v_1,\ldots,v_m \] of \(V\) and a \(\QQ\)-basis \[ e_1,\ldots,e_{p-1} \] of \(K\). Then the \[ m(p-1) \] vectors \[ e_i v_j \qquad 1\leq i\leq p-1,\quad 1\leq j\leq m \] form a \(\QQ\)-basis of \(V\). Hence \[ \dim_{\QQ}V = m(p-1). \]
7. Therefore\[ \boxed{ p-1\mid\dim_{\QQ}V. } \]
Step 6 expresses \(\dim_{\QQ}V\) as an integer multiple of \(p-1\).
8. Q.E.D.
Step 7 is the required divisibility statement.