# Smith normal form of graded modules. (Major edit)

Ok, this is a major rewriting of my previous entry which no one answered.

Let us have two graded $F[t]$-modules M and N with bases $m_1, \ldots, m_m$ and $n_1, \ldots, n_n$, respectively, and $F$ is a field ($F[t]$ is a PID). Note that $m_i$ and $n_j$ are homogeneous elements. We are now interested in calculating the image of a certain 0-degree graded homomorphism from $M$ to $N$. Note that $$M \simeq F[t]\deg(m_1)\oplus\cdots\oplus F[t]\deg(m_m)$$ and similarly for $N$. The grading is the standard grading, i.e. $t(c_0, c_1, \ldots) = (0,c_0, c_1, \ldots)$.

Now let $\partial$ be our homomorphism and the basis elements of $M$ and $N$ with degrees in () are given by: $$ab(1), bc(1), cd(2), ad(2), ac(3)$$ for $M$ and $$a(0), b(0), c(1), d(1)$$ for $N$.

Then $\partial$ is defined by $$\partial(ab) = t^{\deg(ab)-\deg(b)}b - t^{\deg(ab)-\deg(a)}a = tb-ta$$ and exactly the same for the rest, $$\partial(bc) = c - tb$$ $$\partial(cd) = td - tc$$ $$\partial(ad) = td - t^2a$$ $$\partial(ac) = t^2c - t^3a$$

Now sorting the basis elements of $N$ in descending order $d,c,b,a$ we can represent $\partial$ by

$$\begin{pmatrix} * & ab & bc & cd & ad & ac \\ d & 0 & 0 & t & t & 0 \\ c & 0 & 1 & t & 0 & t^2 \\ b & t& t & 0 & 0 & 0 \\ a & t & 0 & 0 & t^2 & t^3 \end{pmatrix}$$

Using column operations we can keep homogeneous bases and reduce the matrix to column echelon form

$$\begin{pmatrix} * & cd & bc & ab & z_1 & z_2 \\ d & t & 0 & 0 & 0 & 0 \\ c & t & 1 & 0 & 0 & 0 \\ b & 0& t & t & 0 & 0 \\ a & 0 & 0 & t & 0 & 0 \end{pmatrix}$$

where $z_1 = ad - cd - t\cdot bc - t\cdot ab$ and $z_2 = ac - t^2\cdot bc - t^2\cdot ab$ form homogenous basis for the kernel.

Note that we have that for an entry in the matrix that the degree of the element at that position + the degree of the row basis element = the degree of column basis element.

Now the author of the paper argues: The pivots in column-echelon form are the same as the diagonal elements in Smith normal form. Moreover, the degree of the basis elements on pivot rows is the same in both forms.

With proof: Because of our sort, the degree of row basis elements is monotonically decreasing from the top rown down. Within each fixed column $j$ the degree of the column basis element is constant equal to $c$ and therefore $\deg\partial_{i,j} = c - \deg(\text{row}~i)$. Therefore, the degree of the elements in each column is monotonically increasing with row. We may eliminate non-zero elements below pivots using row operations that do not change the pivot elements or the degrees of the row basis elements. We then place the matrix in diagonal form with row and column swaps. $\square$

How is it possible to do row operations WITHOUT altering the degree and keeping a homogeneous basis element? Am I missing something obvious? Note that this proof shall hold for any such $\partial$ where the degree of the row + degree of element is equal to the degree of the column. Another example would be $$\begin{pmatrix} * & ab \\ a & t \\ b & t^2 \end{pmatrix}$$ where $\deg(ab) = 3, \deg(a) = 2, \deg(b) = 1$. How would even that be possible...

What we are really interested in is the image of $\partial$. This becomes $$\deg(d)tF(t)\oplus \deg(c)F(t)\oplus \deg(b)tF(t)$$ according to the statement and matrix above.

I do, however, believe that the result is true and I think I can give a proof for it:

Assume that we have column-echelon-form. Then the degree of the homogenous elements along each column increases as we go top rown down. We may also assume that along each row the degree of the pivot element is greater than the other elements on the row, if not, use column operations to remove the element with greater degree. This gives us a matrix of the form (assuming just 2 elements for simplicitiy)

$$\begin{pmatrix} * & m_1 & m_2 \\ n_1 & t^{\beta_1^1} & 0 \\ n_2 & t^{\beta_2^1} & t^{\beta_2^2} \end{pmatrix}$$ where $\beta_2^2 \geq \beta_2^1 \geq \beta_1^1$. Then use $n_1$ to remove first coordinate of $n_2$. The image then becomes:

$$m_1 \to t^{\beta_1^1}n_1\cdot f(t)$$ where $f(t) \in F(t)$. $$m_2 \to t^{\beta_2^2}(n_2 - n_1\cdot t^{\beta_2^1 - \beta_1^1})g(t)$$ where $g(t) \in F(t)$. Writing in terms of the basis elements of the codomain we have image equal to

$$n_1(t^{\beta_1^1}f(t) - t^{\beta_2^2 +\beta_2^1-\beta_1^1}g(t))$$ $$n_2t^{\beta_2^2}$$

and since $\beta_2^2 \geq \beta_1^1$ we have that this is isomorphic to $$\deg(n_1)t^{\beta_1^1}F(t)\oplus \deg(n_2)t^{\beta_2^2}F(t)$$ and the proof generalizes in an obvious way to higher dimensions. Non-pivot rows are skipped.

Source: http://comptop.stanford.edu/preprints/persistence1.pdf Chapter 4.1

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Ok.. The problem should be more accessible now. – M.B. Jun 26 '11 at 5:26