Over a commutative unital ring, does $\det A = 0$ imply that $A\mathbf{x} = 0$ has a non-zero solution? This is a standard theorem in linear algebra over fields that is still true when working over an integral domain.


  For a commutative unital integral domain $R$, let $A$ be an $n\times n$ matrix with entries in $R$. The system of linear equations $A\mathbf{x}=0$ has a non-zero solution if and only if $\det(A) = 0$.


The forward direction of this statement is not true though if we are working over a ring $R$ that is not an integral domain, a counter example being 
$$
R = \boldsymbol{Z}_6 
\qquad 
A = \begin{pmatrix} 3 &amp; 0  \\ 0 &amp; 3 \end{pmatrix} 
\qquad 
\mathbf{x} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}
\,.
$$

But what about the other direction? Does $\det(A) = 0$ imply that $A\mathbf{x} = 0$ has a non-zero solution over an arbitrary commutative unital ring? I've played around for a bit, and have yet to find a counterexample.
Let $A$ be square of size
$n \times n$, and $\det A=0$. Then $A\,\textrm{adj}(A)=0$ where $\textrm{adj}(A)$ is the adjugate of $A$.
As long as $\textrm{adj}(A)\neq 0$ then taking some column of $\textrm{adj}(A)$ gives you $x$ with
$Ax=0$.

But what if adj$(A)=0$? In this case all minors of $A$ vanish. Find
a largest square submatrix of $A$ with nonzero determinant. We can assume
that it's size $r$ by $r$ and fills the top left corner of $A$.
Let $B$
be the top-left $r+1$ by $r+1$ submatrix of $A$. Then $\det B=0$ but
adj$(B)\ne0$. Let $y$ be the $(r+1)$-th column of adj$(B)$ and $z$ be the column
vector of height $n$ got by appending zeroes below $y$. Then $z\ne0$
(due to the non-vanishing of the top left $r$ by $r$
determinant) and I claim that $Az=0$.

The top $r+1$ entries of $Az$ are certainly zero. This is the identity
$B\,$adj$(B)=0$. If we look at another entry, say the $s$-th then
it is zero, essentially by replacing the bottom row of $B$ by the first $r+1$-th
entries of the $s$-th row of $A$, and noting that the new matrix has zero
determinant, as it is obtained from a submatrix of $A$ of size $r+1$
by $r+1$ by elementary row operations. As all submatrices of $A$ of this
size have vanishing determinant, this completes the proof that $Az=0$.
Consider $\pmatrix{2&amp;3\cr 2&amp;3}$ in $\mathbb{Z}/6$ its determinant is $0$, and $2x+3y=0$ implies that $2x=3y$. But the multiples of $2$ are $\{0,2,4\}$ and the multiples of $3$ are $\{0,3\}$.