theorem：If$ A$ ∈$ M_{m×n}(D)$, D is PID, then$ A$ is equivalent to a matrix
which has the "diagonal" form:
\begin{pmatrix}
d_1 & 0 & \cdots & 0 &\cdots & 0\\
1 & d_2 & \cdots & 0 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots & \cdots & \vdots \\
0 & 0 &\cdots & d_r & \cdots & 0 \\
\vdots & \vdots &\cdots & \vdots & \ddots & \vdots \\
0 & 0 &\cdots &0 & \cdots
& 0 \\
\end{pmatrix}
It is belongs to $ M_{m×n}(D)$.
I can understand the proof about D is ED.this thought is similar with linear algebra. When D is PID,i can't understand The following step：
let $f(a)$ be a number of prime factors of $a $.$A$=$(a_{ij})_{m×n}$,we assume $f(a_{11})$ is minimal number(otherwies take elementary transformation).if $a_{11}$ not division $a_{1k}$, we assume k=2. We want to eliminate the elements in the first row and the first column of the matrix expect $a_{11}$ let $d$=gdc($a_{11}$, $a_{12}$) then we have $xa_{11}$+$ya_{12}$=$d$
Consider a matrix B：
\begin{pmatrix}
x & s & \cdots &0 \\
y & t & \cdots &0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 \\
\end{pmatrix}
s = $a_{12}d^{-1}$ t =$a_{11}d^{-1}$. we can see this matrix is invertible matrix over$ F$ . F is quotient field of $D$. Multiplying $A$ on the right by $B$, we can get a new matrix which frits row is $( d,0,a_{13},\cdots,a_{1n}) $we can Continue to do this will get the matrix we want.But $B$ maybe not invertible matrix over D.
This theorem comes from jacobson Basic algebra page 182~184.
Best Answer
Let's examine more closely the generalization from Euclidean domains to PIDs in Jacobson's proof of the Smith normal form algorithm.
In the Euclidean domain case he is essentially using invertible (elementary) linear transformations whose effect on a row (or column) is to view it as the gcd of its entries, then apply the Euclidean algorithm to reduce the row to one whose only nonzero entry is the gcd of all initial entries $\,a_i.\,$ It's the same as the below method to compute a gcd $\,(a_1, a_2,\ldots, a_n)\,$ of a "row" of $n$ naturals not all $\,0$.
Let $\,a_i\neq 0\,$ be the least entry. If $\,a_i\,$ divides all entries then we are done. Else $\,a_i\nmid a_j\,$ so replace $\,a_j\,$ by $\,a_j' := a_j\bmod a_i\neq 0\,$ which is $< a_i,\,$ so now the row has smaller least element, so by induction we can reduce it to a row with same gcd whose least element divides all others. Finally use the least entry (= gcd) to mod out all the others to $\,0,\,$ then swap the least entry into the first index.
Generally a PID has no "mod" (remainder) so we can't use that method to compute gcds. But it does have linear (Bezout) gcds, so we can use them to achieve the descent to the gcd in a single-step, i.e. instead of using $\,(a,\,b) = (a,\,b\bmod a)\,$ we can do $\,(a,\,\,b) = (d,\, 0)\,$ for $\,d = \gcd(a,b),\,$ This too is an invertible linear transformation since, by Bezout $\,d = a u + b v\,$ for some $\,u,v,\,$ thus
$\qquad\qquad\qquad (a\ \ \,b) \begin{pmatrix} 1 & \!\!\!-q\\ 0 & 1\end{pmatrix} =\, (a\ \ \, \overbrace{b\!-\!aq}^{ b\bmod a})\quad \text{[Euclidean descent step]}$
$\qquad\qquad\qquad (a\ \ \,b) \begin{pmatrix} u &\!\! -b/d \\v & a/d\end{pmatrix} =\, (d\ \ \,0)\ \ \ \quad\text{[PID descent step]}$
Both matrices are unimodular (have $\det = 1)\,$ so they preserve the gcd $\,(a,b)\,$.
So the above row gcd algorithm also works if we replace the mod steps by this step of doing the gcds a pair at a time - which exploits the fact that gcd is associative, so $\,(a,b,c) = ((a,b),c),\,$ hence e.g. in row form $\,(a,b,c) = (d,0,c) = (d,c,0) = (e,0,0),\,$ for $\, e=(d,c)$.
One "size" metric that the Bezout reduction step decreases is the number of prime factors, since if $\,a\nmid b\,$ then the gcd $\,(a,b)\,$ divides $\,a\,$ properly (else $\,a\mid (a,b)\Rightarrow\,a\mid b),\,$ so, by unique factorization, the gcd arises by deleting at least one prime factor from $\,a,\,$ so it has fewer prime factors than $\,a\,$ (counting multiplicity). Said equivalently, the algorithm terminates because it produces a chain of proper factors $\,\ldots a_3\mid a_2\mid a_1\,$ and such chains cannot be infinite (i.e. "divides" is well-founded in a UFD), because each proper divisor has fewer prime factors. See also the Dedekind-Hasse PID generalization of the Euclidean division algorithm .
Jacobson's proof proceeds essentially as above, except we need to extend those $\,2\times 2\,$ matrices to $\,n\times n\,$ matrices with $\,\det = 1\, $ by appending $1$'s on the diagonal and $0$'s off-diagonal.