Analogue of Smith normal form for matrices over $\mathbb Z[t]$ - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T07:00:12Zhttp://mathoverflow.net/feeds/question/51942http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/51942/analogue-of-smith-normal-form-for-matrices-over-mathbb-ztAnalogue of Smith normal form for matrices over $\mathbb Z[t]$Andreas Thom2011-01-13T10:56:05Z2011-01-21T23:11:22Z
<p>Let $R$ be a principal ideal domain and $A \in M_n R$. It is well known that there exist invertible matrices $Q$ and $S$ and a diagonal matrix $D= {\rm diag}(a_1,\dots,a_n)$ such that </p>
<ul>
<li>$a_i \mid a_{i+1}$ for all $1 \leq i \leq n-1$, and</li>
<li>D=QAS.</li>
</ul>
<p>The matrix $D$ is called Smith normal form of $A$ and is unique. Obviously, this applies to both $\mathbb Z$ and $\mathbb R[t]$.</p>
<blockquote>
<p><strong>Question 1:</strong> Is there any analogue for $R= \mathbb Z[t]$? Is there any classification of matrices over $\mathbb Z[t]$ up to equivalence?</p>
</blockquote>
<p>A related question is the following:</p>
<blockquote>
<p><strong>Question 2:</strong> Is there any classification of $n \times n$-matrices over $\mathbb Z$ up to conjugation?</p>
</blockquote>
<p>(The relation comes from looking at the matrix $t\cdot 1_n - A$ for $A \in M_n \mathbb Z$. Then, classification of $A$ up to conjugation is the same as classification of $t\cdot 1_n-A \in M_n \mathbb Z[t]$ up to equivalence.) </p>
<p>An obvious first invariant is the characteristic polynomial. Even if the matrix is assumed to be symmetric, it is unclear to me what kind of additional information could be added. </p>
<p>In that respect I know of a theorem of Latimer and MacDuffee which says that if the characteristic polynomial $f$ of $A \in M_n \mathbb Z$ is irreducible, then conjugacy classes of integer matrices with the same characteristic polynomial are in bijection with $\mathbb Z[\alpha]$-ideal classes in ${\mathbb Q}(\alpha)$, where $\alpha$ is a root of $f$. However, this seems to depend on the irreducibility of $f$ and I do not know of an extension to the general case. (This is nicely explained in <a href="http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/matrixconj.pdf" rel="nofollow">notes</a> by Keith Konrad.)</p>
<blockquote>
<p><strong>Question 3:</strong> Are there any other positive results going in the direction of the Theorem of Latimer-MacDuffee? What if the characteristic polynomial is a product of two irreducible polynomials?</p>
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<p>Related but maybe easier:</p>
<blockquote>
<p><strong>Question 4:</strong> Is there some characterization (in terms of the characteristic polynomials + additional invariants) of pairs of matrices in $A,B \in M_n\mathbb Z$, such that $A$ and $B$ are conjugate in $M_n \overline{\mathbb F}_p$ for <em>all primes</em> $p$ and in $M_n \mathbb C$? </p>
</blockquote>
<p>And finally</p>
<blockquote>
<p><strong>Question 5:</strong> Is there some characterization (in terms of the characteristic polynomials + additional invariants) of pairs of matrices in $A,B \in M_n\mathbb Z$, such that $A$ and $B$ are conjugate in $M_n \overline{\mathbb F}_p$ for <em>a fixed prime</em> $p$ and in $M_n \mathbb C$? </p>
</blockquote>
<p>(Of course, Question 4 and 5 can be answered by looking at the Jordan decomposition for each of the fields separately. However, the question is, can we do something more conceptual?)</p>
http://mathoverflow.net/questions/51942/analogue-of-smith-normal-form-for-matrices-over-mathbb-zt/51956#51956Answer by Torsten Ekedahl for Analogue of Smith normal form for matrices over $\mathbb Z[t]$Torsten Ekedahl2011-01-13T13:54:36Z2011-01-13T13:54:36Z<p>For Q1 the problem is that one invariant of the matrix is the (isomorpism class
of the) cokernel and any $\mathbb Z[t]$-module generated by $n$ elements and
$n$-relations appears as such an invariant. There simply are too many modules
over a $2$-dimensional ring such as $\mathbb Z[t]$.</p>
<p>As for Q2 you cannot really hope for a classifiction even for the conjugacy classes
of matrices of finite order. In fact already for matrices of order $p^m$, $p$ prime
and $m>2$, the problem is wild which essentially means that any complete
classification is hopeless.</p>
<p>Finally (as far as these comments go) for the part of Q3 where the characteristic
polynomial is the product of two distinct irreducible polynomials $f$ and $g$:
If $M$ is $\mathbb Z^n$ as a module of $\mathbb Z[t]$ through the matrix, then
we have a direct sum part $M'\bigoplus M''\subseteq M$, where $M'$ is the
annihilator of $f$ and $M''$ of $g$. Hence, $M'$ and $M''$ are given by ideals
as in Latimer-McDuffee and can be considered classified. Then $M$ is given by a
submodule $\overline M$ of $M'\bigotimes\mathbb Q/\mathbb Z\bigoplus
M''\bigotimes\mathbb Q/\mathbb Z$. As $M'$ and $M''$ are the kernels of
multiplication by $f$ resp.\ $g$ we also get that $\overline
M\cap M'\bigotimes\mathbb Q/\mathbb Z=0$ and the same for $M''\bigotimes\mathbb
Q/\mathbb Z$. Hence $\overline M$ is the graph of an isomorphism between
submodules $\overline M'\subseteq M'\bigotimes\mathbb Q/\mathbb Z$ and
$\overline M''\subseteq M''\bigotimes\mathbb Q/\mathbb Z$. This means that
$\overline M'$ and $\overline M''$ are killed by both $f$ and $g$ and as they
are relatively prime, the kernel of $g$ on $M'\bigotimes\mathbb Q/\mathbb Z$ (as
well as the kernel of $f$ on $M''\bigotimes\mathbb Q/\mathbb Z$) are finite (and
usually quite small). Hence one can (in principle) determine the possible
$\overline M$ and they have to be considered modulo automorphisms of $M'$ and
$M''$ which are given by units in their endomorphism rings (which are overorders
of $\mathbb Z[t]/(f)$ resp. $\mathbb Z[t]/(g)$).</p>
<p>This sometimes works very well. For instance this is exactly one way of doing
the classification of matrices of order $p$ (where $f=t-1$ and
$g=t^{p-1}+\cdots+t+1$). On the other hand I am pretty sure it is as
hopeless as the general conjugacy problem for general $f$ and $g$.</p>