Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking for references, if there is any, for this problem:

Characterize all elements $a \in M_n(\mathbb{Z})$ for which we have $\mathbb{Q}[a] \cap M_n(\mathbb{Z})=\mathbb{Z}[a].$

Here, by $C[a]$ I mean the ring of polynomials in $a$ with coefficients in a ring $C.$

Thank you

share|cite|improve this question
One direction is easy, that $\mathbb{Q}[a] \cap M_n(\mathbb{Z})$ contains $\mathbb{Z}[a]$ for any $a \in M_n(\mathbb{Z})$. So the question amounts to for which $a$ the left-hand expression doesn't yield any extra integer matrices. A necessary condition is that $a$ has entries with greatest common divisor $1$. – hardmath Oct 19 '12 at 12:41
Let $a = \begin{pmatrix} 1&2 \\ 4&3 \end{pmatrix}$. Then $\mathbb{Q}[a]$ contains $\begin{pmatrix} 1&1 \\ 2&2 \end{pmatrix}$ but this is not in $\mathbb{Z}[a]$. Thus it is not sufficient that $a$ has entries with GCD $1$, although this is a necessary condition for $\mathbb{Q}[a] \cap M_n(\mathbb{Z})=\mathbb{Z}[a]$. – hardmath Oct 21 '12 at 15:25
I don't know why you think that the gcd of entries of those $a$ that satisfy the property in the problem must be one? For example, if $I$ is the identity matrix and $a=mI, \ m \in \mathbb{Z},$ then $a$ satsifies the property in the problem, doesn't it? – Yashar Oct 22 '12 at 11:15
Yes, that's true. Let me restate it as if $a \notin \mathbb{Z}[I]$, so that we have a simple proper extension, then the gcd of $a$'s entries must be 1 in order for the rational polynomials in $a$ not to produce an additional integer matrix. – hardmath Oct 22 '12 at 12:49

Let $n\times n$ integer matrix $a \notin \mathbb{Z}[I]$ have (monic) minimal polynomial $f(x) \in \mathbb{Z}[x]$ of degree $d$.

Construct a $d\times n^2$ matrix $P$ containing the entries of $\{I,a,..,a^{d-1}\}$ expressed as consecutive rows.

Claim: $\mathbb{Q}[a] \cap M_n(\mathbb{Z})=\mathbb{Z}[a]$ if and only if the elementary divisors of (the Smith normal form of) $P$ are units.

To motivate this let's look at two examples:

Example 1: Let $a = \begin{pmatrix} 1&2 \\ 4&3 \end{pmatrix}$, whose minimal polynomial is $x^2 - 4x - 5$. Thus we look at the matrix whose rows represent the powers of $a$ less than the degree of this minimal polynomial:

$$ \begin{pmatrix} 1&0&0&1 \\ 1&2&4&3 \end{pmatrix}$$

and reduce it to Smith normal form:

$$ \begin{pmatrix} 1&0&0&0 \\ 0&2&0&0 \end{pmatrix}$$

The fact that the last elementary divisor is greater than 1 corresponds to a linear combination of $I$ and $a$ with coprime coefficients, say:

$$\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} + \begin{pmatrix} 1&2 \\ 4&3 \end{pmatrix}$$

which is twice (and the smallest multiple of) $\begin{pmatrix} 1&1 \\ 2&2 \end{pmatrix}$ in $\mathbb{Z}[a]$. But that integer matrix does occur in $\mathbb{Q}[a]$ and thus we don't have $\mathbb{Q}[a] \cap M_n(\mathbb{Z})=\mathbb{Z}[a]$.

Example 2: Consider $a = \begin{pmatrix} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{pmatrix}$, whose minimal polynomial is $x^2 - x - 2$. Thus we look at the Smith normal form of:

$$ \begin{pmatrix} 1&0&0&0&1&0&0&0&1 \\ 0&1&1&1&0&1&1&1&0 \end{pmatrix} $$

which reduces to:

$$\begin{pmatrix} 1&0&0&0&0&0&0&0&0 \\ 0&1&0&0&0&0&0&0&0 \end{pmatrix} $$

The fact that all elementary divisors are 1 corresponds to the fact that all the nonzero combinations of $I$ and $a$ we can form in $\mathbb{Z}[a]$ can be chosen to get entries with greatest common divisor 1, and thus no smaller multiple in $\mathbb{Q}[a]$ exists than what we already find in $\mathbb{Z}[a]$.

Proof of Claim:

Let $\mathscr{M} = M_n(\mathbb{Z})$. For matrix $a \in \mathscr{M}$ define $\mathscr{N} = \mathbb{Z}[a]$. Clearly $\mathscr{M}$ is a free $\mathbb{Z}$-module of rank $n^2$, and thus (since $\mathbb{Z}$ is PID) submodule $\mathscr{N}$ is also free. By Cayley-Hamilton $a$ is integral over $\mathbb{Z} \cong \mathbb{Z}[I]$, and because $\mathbb{Z}$ is integrally closed (in $\mathbb{Q}$), the (monic) minimal polynomial of $a$ has integer coefficients. If $d$ is the degree of the minimal polynomial of $a$, then $\mathscr{N}$ has basis $\{1,a,\ldots,a^{d-1}\}$ and so has rank $d$.

The $d \times n^2$ matrix $P$ described in the claim expresses the above basis of $\mathscr{N}$ in terms of the standard basis for $\mathscr{M}$. Thus the proof of the Stacked Basis Thm. identifies $P$'s elementary divisors $q_1 \mid \ldots \mid q_d$ with the nonzero integer multiples $q_i$ of basis elements such that:

$$\text{(i) } \{\mu_1,\ldots,\mu_{n^2}\} \text{ is a basis for } \mathscr{M}$$ $$\text{(ii) } \{q_1 \mu_1,\ldots,q_d \mu_d\} \text{ is a basis for } \mathscr{N}$$

Let $\overline{\mathscr{N}} = \mathbb{Q}[a] \cap \mathscr{M}$, so our claim is $\mathscr{N} = \overline{\mathscr{N}}$ if and only if elementary divisors $q_i, 1 \leq i \leq d$ are all units.

Lemma: Let $b \in \mathscr{M}$. Then:

$$b \in \overline{\mathscr{N}} \iff \exists r \in \mathbb{Z}^+ \text{s.t. } rb \in \mathscr{N}.$$

Proof: By clearing denominators in basis coefficients, $b \in \mathbb{Q}[a] \iff \exists r \in \mathbb{Z}^+ \text{s.t. } rb \in \mathbb{Z}[a]$. Then take intersections with $\mathscr{M}$.

Accordingly the set $\{\mu_1,\ldots,\mu_d\} \subset \overline{\mathscr{N}}$. This set is also linearly independent by (i) above. From this we can show one side of the claim:

$$\mathscr{N} = \overline{\mathscr{N}} \implies q_i \text{ is a unit}, 1 \leq i \leq d $$

That is, if $\mathscr{N} = \overline{\mathscr{N}}$, then in particular $\mu_d \in \mathscr{N}$, and we have the integer combination:

$$\mu_d = \sum_{i=1}^d c_i q_i \mu_i$$

By linear independence, $1 = c_d q_d$, and $q_d$ is a unit. By their divisibility relations, all $q_i$ are units, $1 \leq i \leq d$.

To show the other direction:

$$q_i \text{ is a unit}, 1 \leq i \leq d \implies \mathscr{N} = \overline{\mathscr{N}} $$

We know $\mathscr{N} \subset \overline{\mathscr{N}}$, so it suffices to prove the reverse inclusion. Given any $b \in \overline{\mathscr{N}}$, express $b$ in terms of the basis for $\mathscr{M}$:

$$b = \sum_{i=1}^{n^2} c_i \mu_i$$

By the Lemma there exists integer $r \gt 0$ s.t. $rb \in \mathscr{N}$:

$$rb = \sum_{i=1}^{n^2} r c_i \mu_i$$

All $q_i$'s are units, so (ii) above says $\{\mu_1,\ldots,\mu_d\}$ is a basis for $\mathscr{N}$. Then by uniqueness of coefficients, we have $r c_i = 0$ for all $i \gt d$. As r is nonzero, this implies $c_i = 0$ for all $i \gt d$. Therefore $b \in \mathscr{N}$, which proves the sought reverse inclusion, $\overline{\mathscr{N}} \subset \mathscr{N}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.