# Matrix over a ring with given kernel

Let $R$ be a (commutative, unital,) Noetherian ring and let $M$ be an $m\times n$ matrix over $R$. The columns of $M$ span a submodule $\widetilde M$ of $R^m$, but in general $\widetilde M$ will not be free. Define, as usual, the kernel of $M$ to be $\ker(M) = \{v \in R^n \mid Mv = 0\}$ and denote by $M^t$ the transpose of $M$. The kernel is a submodule of $R^n$; it is therefore finitely generated and so $\ker(M) = \widetilde K$ for some matrix $K$ with $n$ rows.

Question: Let $M$ be an $m\times n$ matrix over $R$ and let $\widetilde K = \ker(M^t)$. Can we show that $\ker(K^t) \subseteq \widetilde M$?

Remarks:

• Note that $K^t M = (M^t K)^t = 0$ so $\widetilde M \subseteq \ker(K^t)$. The opposite inclusion mentioned in the question will give $\widetilde M = \ker(K^t)$, which is what I want, namely that the matrix $K^t$ has kernel spanned by the given matrix $M$.
• The proof above that $\widetilde M \subseteq \ker(K^t)$ only uses the fact that $\widetilde K \subseteq \ker(M^t)$, but we know more, namely that $\widetilde K = \ker(M^t)$. I haven't been able to turn this extra knowledge into a proof though.
• When all the modules in play are free and all the matrices give bases for the modules (echelon forms over fields for example), then the result follows easily by counting dimensions.
• For what it's worth, I've tested the result for lots ($\sim 10^4$) of small random matrices over $R = \mathbb{Z}/n\mathbb{Z}$ for $n$ composite with many small factors (e.g. $n = 6, 12, 15, 24, 30$) hoping to find a counterexample, but the result has always been verified.
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