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Would the Smith Normal Form of the following matrix over $\mathbb Q[x]$

$$\begin{pmatrix}   (x+a)(x+b) & 0 & 0 &0 \\  0 & (x+c)(x+d) & 0 & 0 \\   0  &0 & x^3(x+a) & 0  \\   0 & 0 & 0& x^2(x+b)  \end{pmatrix}$$

 simply be

$$\begin{pmatrix}   f(x) & 0 & 0 &0 \\  0 & f(x) & 0 & 0 \\   0  &0 & f(x) & 0  \\   0 & 0 & 0& f(x)  \end{pmatrix}$$

where $f(x)= x^3(x+a)(x+b)(x+c)(x+d)$?

I am not sure because that would make the question quite trivial.

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The Smith normal form for a matrix has to have the same determinant as the original matrix (up to multiplication by a unit), and your suggested form does not. –  Geoff Robinson Mar 15 '12 at 23:14
    
@GeoffRobinson: Right, thanks. Is there an effective way of finding the SNF? –  helms Mar 15 '12 at 23:34
    
Typically Gaussian elimination is used for finding SNF Wikipedia page on SNF is a good pointer. –  user2468 Mar 15 '12 at 23:44

1 Answer 1

No.

Let $s_k$ denote the $k$th entry on the diagonal of Smith form (i.e., the $k$th invariant factor). Then $$s_k = \frac{d_k}{d_{k-1}},$$ where $d_k = \gcd$ of all $k \times k$ minors of the original matrix (aka $k$th determinantal divisor). I'm assuming $a,b,c,d$ are distinct here. So the Smith form is: $$ \begin{pmatrix} 1 & & & {\large 0} \\ & 1 & & \\ & & x^2(x+a)(x+b) & \\ {\large 0} & & & x^3(x+a)(x+b)(x+c)(x+d) \end{pmatrix} $$

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More here as well around Theorem 6.7: numbertheory.org/courses/MP274/smith.pdf –  user2468 Mar 15 '12 at 23:46
    
Note above, $d_0 = 1.$ –  user2468 Mar 15 '12 at 23:46
    
Thank you very much! –  helms Mar 16 '12 at 0:03

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