Fields over which a matrix is not invertible

Question

I am trying to find the fields over which the matrix:

$\left(\begin{matrix} 1 & 2 & 3 \\ 0 & -1 & 2 \\ 1 & 0 & -2 \end{matrix}\right) $

is not invertible. I have calculated the determinant as 9. I know it is not invertible if detA=$0$. I believe the fields $F_1, F_3$ and $F_9$ both satisfy this (i.e. the factors of 9) but are there any more? Can it be proved these are the only ones?

Answer 1

The determinant is $9,$ and $9=0$ in a field $F$ if and only if the characteristic of $F$ divides $9,$ i.e. if and only if $F$ has characteristic $3.$ For example, the algebraic closure of $\mathbb F_3$ satisfies this.
Hope this helps.
Edit
We have a ring homomorphism $\phi:\mathbb Z\rightarrow F,$ defined as $\phi(n)=n\cdot1_F, \forall n\in\mathbb Z.$
And the kernel $K$ is an ideal of $\mathbb Z;$ since $\mathbb Z$ is a principal ideal domain, this is generated by one single element, say $K=(n_0).$ Then we define $char(F)=n_0.$ And $9=0$ in $F$ if and only if $9\in K,$ so...?
Below is an elementary argument:
The characteristic of a field $F$ is defined as the smallest positive integer $n_F$ such that $n_F=0$ in $F,$ and, if such an integer does not exist, define $n_F=0.$
As an exercise, show that, if $k=0$ in $F$ for some integer $k,$ then $n_F\mid k.$ And the converse is obviously true, namely,

An integer $k$ satisfies $k=0$ in $F$ if and only if $n_F\mid k.$

This should be an easy exercise, using Euclidean division.

Answer 2

The field with one element is not actually a field.

On the other hand, $\mathbb F_3$, $\mathbb F_9$, $\mathbb F_{27}$ and so forth all have characteristic 3 and therefore satisfy the equation $9=0$, so those work. Similarly all infinite fields of characteristic 3 will work too.