Degree of the kernel of a module map $R^n\rightarrow R^n$ for an Euclidean domain $R$ Let $R$ be an Euclidean domain with the degree function $d$. Let $A\in R^{n\times n}$ be an $n\times n$-matrix with entries in $R$ such that det$(A)=0$. As a module map $A:R^n\rightarrow R^n$, there always exists a kernel element $v\in R^n$ since det$(A)=0$.
Assuming $d(A_{ij})\leq m$ for all $i,j$, is there an explicit bound $k(m,n)$ such that there exists a kernel element $v\in R^n$ satisfying $d(v_i)\leq k(m,n)$?
Edit : $v$ is assumed to be nonzero.
 A: I guess that in order to provide an explicit bound $k(m,n)$ we need to know more about $d$, because it can grow arbitrary fast, that is if any strictly monotone map $f:\Bbb N\setminus\{0\}\to \Bbb N\setminus\{0\}$ a function $f(d)$ is also a degree function on $R$. 
On the other hand, recently I have proved a similar result for $(R,d)=(\Bbb Z,|\cdot|)$ (see below). Maybe it can be generalized to other cases of $(R,d)$.
Proposition. Let $K$ and $N$ be positive integers, $V=\{v_1,\dots, v_k\}\subset [0,K]^N$ be a linearly dependent over $\mathbb R$ system of vectors with integer entries. There exist integers $f_1,\dots, f_k$ which are not all zeroes such that $|f_i|\le (kK)^{k-1}$ for each $i$ and $f_1v_1+\dots+f_kv_k=0$.
Proof. Let $W$ be a maximal linearly independent subset of a set $V$. Since the set $V$ is linearly
dependent, $|W|\le k-1$. For each $i\in [N]$ let
$e^i=(e^i_1,\dots,e^i_N)\in\mathbb R^N$ be $i$-th standard orth, that is $e^i_i=1$ and $e^i_j=0$ for each $j\ne i$. Let
$B_0=\{e^1,\dots,e^n\}$ be the standard basis of the linear space $\mathbb R^N$.
By [L, Ch. III, Theorem 2], there exists a basis $B$ of the space $\mathbb R^N$
such that $W\subset B\subset W\cup B_0$. Let $C=B_0\setminus (B\setminus W)$ and
$p_{C}:\mathbb R^N\to \langle C\rangle$ be the orthogonal
projection, that is $p_{C}(x)=\sum\{x_ie^i:x_i\in\mathbb R$, $e^i\in C\}$
for each vector $x=(x_1,\dots,x_N)\in \mathbb R^N$. Thus
$\ker p_{C}=\{x\in \mathbb R^N:p_{C}(x)=0\}=\langle B_0\setminus C\rangle=
\langle B\setminus W\rangle$. We have $\ker p_{C}\cap \langle W\rangle=\langle B\setminus
W\rangle\cap\langle W\rangle=0$, because otherwise the set $B$ is linearly dependent.
Thus the restriction $p_{C}|\langle W\rangle$ of the map $p_{C}$ on the set $\langle W\rangle$
is injective.
Put $K'=(kK)^{k-1}$. Define a map $f$ from the subset $D^k$ of points of the set $[0, K']^k$ with all
integer coordinates to $\langle W\rangle\cap \mathbb Z^N\subset \mathbb R^N$ as follows.
Let $d=(d_1,\dots,d_k)\in D^k$. Put $f(d)=p_C(dv)$, where $dv=d_1v_1+\dots d_kv_k$.
Since $d_i\in [0, K']$ and $v_i\in [0,K]^N$ for each $i\in [k]$, each coordinate of a vector
$dv$ (and, hence, of the vector $f(d)=p_C(dv)$ too) is at most $kK'K$. Since
$$|C|=|B_0\setminus (B\setminus W)|=|B_0|-|B\setminus W|=|B_0|-(|B|-|W|)=$$
$$N-(N-|W|)=|W|\le k-1,$$
$|f(Q)|\le (kK'K+1)^{k-1}$. We have $|D^k|>|f(Q)|$, because $(1+(kK)^{k-1})^{\frac 1{k-1}}>(1+(kK)^k)^{\frac 1{k}}$,
because when $a>1$ is a constant and $x>0$ a function $(1+a^x)^{\frac 1x}$ decreases.
Therefore the function $f$ is not injective. So there exist distinct elements $d=(d_1,\dots,d_k)$
and $d'=(d'_1,\dots,d'_k)$ of $D^k$ such that $p_C(dv)=f(d)=f(d')=p_C(d'v)$. Since $dv$ and $dv'$
belong to $\langle W\rangle$ and the restriction $p_{C}|\langle W\rangle$ is injective,
$dv=d'v$. It remains to put $f_i=d_i-d'_i$ for each $i\in [k]$. $\square$
References
[L] Serge Lange, Algebra, Addison-Wesley, 1965 (Russian translation, Moskow, Mir, 1968).
