Call a commutative unital ring $R$


* *Hermite* if for all $m, n\in \mathbb{N}$ with $m<n$, and all $f\in R^{m\times n}$ such that transpose($f$) is left invertible (with a matrix with entries from $R$ serving as a left inverse of $f$), there exists a $g\in R^{n\times (n-m)}$ such that $\ker f = g R^{(n-m)\times 1}$. (This is equivalent to the condition that every tall matrix which is left invertible can be completed to a square invertible matrix while working in $R$.)
* *coherent* if for all $n,m\in \mathbb{N}$ and $f\in R^{m\times n}$, there exists a $k\in \mathbb{N}$ and a $g\in R^{n\times k}$ such that $\ker f = g R^{k\times 1}$.


Question: Is there a ring which is not Hermite but is coherent?



Is there a ring which is not Hermite but is coherent?
Let $F$ be a field not of characteristic $2$, and let $R:=F[x,y,u,v,s,t]/(xu+yv=1,xs+yt=0)$. This ring has a reduction system given by $yt\mapsto-xs$ and $xu\mapsto 1-yv$. (Think of this as trying to get rid of $t$'s the most, then $u$'s, and so forth.)


This is a Noetherian ring, hence coherent (if I'm understanding your definition correctly). However, it is not Hermite. Take $f=(x,y)$. Then $(x,y)\cdot \begin{pmatrix}u\\ v\end{pmatrix}=1$, so the transpose of $f$ is left invertible. However, $\ker(f)$ contains the vectors $h\_1=\begin{pmatrix}-y\\ x\end{pmatrix}$ and $h\_2=\begin{pmatrix}s\\ t\end{pmatrix}$. It suffices to show that these vectors are not mutliples of a single vector in the kernel.


First, I leave it as an exercise that if $\alpha,\beta\in R$ are any elements in reduced form, then $\deg(\alpha\beta)=\deg(\alpha)\cdot\deg(\beta)$.


Suppose, by way of contradiction, that $g=\begin{pmatrix}\alpha\\ \beta\end{pmatrix}$ generates the kernel. We may assume $\alpha,\beta$ are in reduced form, and they are definitely not constant. From the fact of the previous paragraph, we must have $\deg(\alpha)=\deg(\beta)=1$. It is easy to check that the only possibilities are linear combinations of the two vectors $h\_1,h\_2$ above. But no $F$-combination of them works. (It would work in characteristic $2$ however, to just use $h\_1$! So there is something to show here, but I'll leave the details to you.)


I may have made a mistake somewhere, so you need to double check this work.


