Reference Request: Smith Normal Form for maps between free _graded_ modules - MathOverflow

Reference Request: Smith Normal Form for maps between free _graded_ modules

2012-04-11T11:25:20Z

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 http://mathoverflow.net/questions/31275/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.

Answer by Thomas Kahle for Reference Request: Smith Normal Form for maps between free _graded_ modules

2012-04-11T11:59:06Z

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

Answer by Ralph for Reference Request: Smith Normal Form for maps between free _graded_ modules

2012-04-11T19:16:18Z

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.