3
$\begingroup$

I feel like this should be easy, but I cannot quite find a literature reference for this: We know (i.a. from the Kaplansky reference in Does Smith normal form imply PID?) that sufficient for Smith normal form as well as Hermite normal form to work is that the underlying ring be a PID.

I am interested in the case where the ring is $k[t]$, for some field $k$, and all modules involved are $\mathbb N$-graded with the "obvious" grading of $k[t]$. For a matrix $M$ representing a map between two graded $k[t]$-modules $S\to T$, it seems obvious to me that Smith normal form is computable, and about as efficient as one might hope over any ring. The presence of a grading seems to imply one should take some minute care — but the care needed seems to be almost non-existent.

Has anyone dealt with this sort of setting in the literature already? I'd rather have a good reference for this than develop everything in analogy with well-known results myself.

Improve this question
$\endgroup$
7
  • $\begingroup$ What kind of "map" are you interested in ? $\endgroup$
    – Ralph
    Commented Apr 11, 2012 at 12:02
  • $\begingroup$ Graded $k[t]$-linear. $\endgroup$
    – Mikael Vejdemo-Johansson
    Commented Apr 11, 2012 at 12:15
  • $\begingroup$ ... of degree zero ? $\endgroup$
    – Ralph
    Commented Apr 11, 2012 at 12:38
  • $\begingroup$ $S,T$ are free of finite rank (with various degrees)? $\endgroup$
    – Martin Brandenburg
    Commented Apr 11, 2012 at 13:06
  • $\begingroup$ $S$, $T$ are free of finite rank. They both have homogenous bases, but the basis elements need not all have the same degree as each other. $\endgroup$
    – Mikael Vejdemo-Johansson
    Commented Apr 11, 2012 at 13:51

2 Answers 2

Reset to default
2
$\begingroup$

I don't know a reference either, but if I understand the question properly, it's rather easy and perhaps one don't even need to have a reference to postulate the existence of SNF in the OP's situation.

According to one of the OP's comments below the question, there are homogeneous bases of $S$, $T$. Order these according to the degree of the basis elements. Since a degree zero map preserves degrees, its matrix is block diagonal with entries in $k$. Now one can apply SNF to the single blocks and obtains a diagonal matrix: $$ \begin{pmatrix}L_1 & & \newline & \ddots & \newline & & L_n\end{pmatrix} \begin{pmatrix}M_1 & & \newline & \ddots & \newline & & M_n\end{pmatrix} \begin{pmatrix}R_1 & & \newline & \ddots & \newline & & R_n\end{pmatrix} = \begin{pmatrix}D_1 & & \newline & \ddots & \newline & & D_n\end{pmatrix} $$ By construction, the matrices $L$, $R$ (that also have entries in $k$) represent $k[t]$-linear maps of degree zero.

Improve this answer
$\endgroup$
0
$\begingroup$

$k[t]$ is a PID when $k$ is a field.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I know this, and as I mentioned, I expect this to actually be rather easy — which is why it's a reference request, not a question; but in my setting, the rows of the matrix have inherent degrees, and I worry somewhat about keeping everything compatible with the grading. $\endgroup$
    – Mikael Vejdemo-Johansson
    Commented Apr 11, 2012 at 12:15
  • $\begingroup$ I guess I don't see what you question is. The smith normal form of a given matrix exists and can be computed using elimination. If you are asking for the complexity, then that is measured usually in ring operations, so it depends on how fast you can do ring operations. I don't understand the meaning of the sentence "The presence of a grading seems to imply one should take some minute care". $\endgroup$
    – Thomas Kahle
    Commented Apr 11, 2012 at 13:20
  • $\begingroup$ The presence of a grading means, for instance, that you cannot use all the tricks available for doing Gaussian reduction on the kinds of matrices I'm interested in — you could not reduce rows in lower degrees using rows in higher degrees. I suspect that similar restrictions are in place for computing Smith normal forms, but also that the restrictions do not actually limit things very much. $\endgroup$
    – Mikael Vejdemo-Johansson
    Commented Apr 11, 2012 at 13:50
  • 3
    $\begingroup$ @Thomas: -1 since you don't address the question. When you don't understand it (whatever is the reason), leave a comment first. $\endgroup$
    – Martin Brandenburg
    Commented Apr 11, 2012 at 18:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .