Let $R$ be a DVR, and $f:R^n\to R^n$ a map. Suppose $f(r\_1,\dots,r\_n)=(s\_1,\dots,s\_n)$, and write $d=\min(v(s\_1),\dots,v(s\_n))$, where $v$ is the valuation on $R$. Knowing $d$, what is the best bound I can get on $\min(v(r\_1),\dots,v(r\_n))$? I'm looking for a bound in terms of the map $f$ and the individual $v(r\_i)$.



How much can a map $R^n\to R^n$, $R$ a DVR, increase the valuation?
Let $\pi$ be a uniformiser, and write $r = (r\_1,\ldots,r\_n)$ and $s = (s\_1,\ldots,s\_n)$. For a vector $x = (x\_1,\ldots,x\_n) \in R^n$, write $v(x) = \min(v(x\_1),\ldots,v(x\_n))$. Note that it equals the largest integer $m$ such that $x \in \pi^mR^n$.



> 
> **Lemma.** *Let $f \colon R^n \to R^m$ be a linear map, let $r \in R^n$ be nonzero, and let $s = f(r)$. Then*
> $$v(r) \leq v(s).$$
> 
> 
> 


*Proof.* If $r$ is divisible by $\pi^m$, then so is $s$. $\square$



> 
> **Corollary.** *Let $f \colon R^n \to R^n$ be a linear map whose cokernel is killed by $\pi^m$, let $r \in R^n$ be nonzero, and let $s = f(r)$. Then*
> $$v(s) - m \leq v(r) \leq v(s).$$
> 
> 
> 


For example, if $\det(f) \neq 0$, then $\operatorname{coker}(f)$ is killed by $\det(f)$, so
$$v(s) - v(\det(f)) \leq v(r) \leq v(s).$$
Conversely, if $\pi^m$ kills $\operatorname{coker}(f)$, then $\det(f)$ is nonzero and divisible by $\pi^m$. If the [Smith normal form](https://en.wikipedia.org/wiki/Smith_normal_form) of $f$ is $\operatorname{diag}(a\_1,\ldots,a\_n)$ with $a\_i | a\_{i+1}$ for all $i < n$, then the smallest $m$ we may take in the lemma is $v(a\_n)$.


*Proof of Lemma.* Because $\operatorname{coker}(f)$ is killed by $\pi^m$, the image of $\pi^m \colon R^n \to R^n$ contains $\operatorname{im}(f)$, so we may factor $\pi^m$ as $f \circ g$ for some $g \colon R^n \to R^n$. This also forces $g \circ f = \pi^m$ (one-sided inverse matrices are two-sided inverses). Applying the lemma to $f$ gives the upper bound, whereas applying the lemma to $g$ gives
$$v(s) \leq v(g(s)) = v(\pi^m r) = m+v(r),$$
which proves the claim. (Alternatively, we could directly read it off from the Smith normal form.) $\square$


**Remark.** In general there is not much more you can say, as is seen by the example
\begin{align\*}
f \colon R^n &\to R^n\\
(x\_1,\ldots,x\_n) &\mapsto (\pi^m x\_1,x\_2,\ldots,x\_n).
\end{align\*}
Indeed, this map has determinant $\pi^m$, and the lower bound is attained for $x = e\_1$ whereas the upper bound is attained for $x = e\_i$ with $i > 1$. The situation does not seem to improve if you allow bounds depending on the $v(r\_i)$, because all the $r = e\_i$ have the same multiset of componentwise valuations 
$$\left\{v(r\_i)\ \Big|\ i \in \{1,\ldots,n\}\right\}.$$


**Remark.** If the determinant of $f$ is $0$, then there is never a lower bound on $v(r) - v(s)$, even if we restrict to $r$ such that $f(r) \neq 0$. For example, if $y \in \ker(f)$ and $x \not \in \ker(f)$, then set $r = \pi^m x + y$. Then
$$v(r) = v(y)\hspace{4em} \text{for } m > v(y).$$
On the other hand, $s = f(r) = \pi^m f(x)$, so
$$v(s) = m+v(f(x)) \to \infty \hspace{4em} \text{as } m \to \infty,$$
so $v(r) - v(s)$ is unbounded below. 


For example, if $f \colon R^2 \to R^2$ is the first coordinate projection, then we took $r = (\pi^m,1)$ and noted that $v(r) = 0$ and $v(s) = m$, so $v(r) - v(s) = -m$ is not bounded below.


There is of course the trivial bound $v(r) \geq 0$, but this is very weak if $v(s)$ is big.


