Determining consistency of a general overdetermined linear system

When $m=3$ and $\mathbb F$ is infinite, there are no other obstructions besides the determinant. When $\mathbb F$ is finite, there are many others : for example if we put $\chi_{\mathbb F}(X)=\prod_{t\in {\mathbb F}^*} (X-t)$, $\chi(t)$ is zero iff $t$ is nonzero, so that the following $n$ polynomials are all obstructions :

$$ w_i=b_i\prod_{j=1}^n\chi_{\mathbb F}(a_{ij}) $$

For the infinite case, one can use the following lemma :

Generalized Euclidean division. Let $A$ and $B$ be two polynomials in ${\mathbb F}[X_1,X_2,\ldots,X_n,Y]$. Let $a={\sf deg}_Y(A)$, $b={\sf deg}_Y(B)$, and let $L$ be the leading coefficient of $B$ with respect to $Y$ (so that $L\in{\mathbb F}[X_1,X_2,\ldots,X_n]$ and $B-LY^b$ has degree $<b$ in $Y$). Then if $a \geq b$, there are two polynomials $Q,R\in {\mathbb F}[X_1,X_2,\ldots,X_n,Y]$ such that $L^{a-b+1}A=QB+R$ and ${\sf deg}_Y(R)<b$.

Proof. Let ${\mathbb K}={\mathbb F}(X_1,X_2,\ldots,X_n)$. We can view $A$ and $B$ as members of ${\mathbb K}[Y]$, and perform ordinary euclidian division ; this yields $Q^{\sharp},R^{\sharp}\in {\mathbb K}[Y]$ such that $A=Q^{\sharp}B+R^{\sharp}$. Since the division process involves $a-b+1$ divisions by $L$, we see that $Q^{\sharp}$ and $R^{\sharp}$ are of the form $\frac{Q}{L^{a-b+1}}$ and $\frac{R}{L^{a-b+1}}$ with $Q,R\in {\mathbb F}[X_1,X_2,\ldots,X_n,Y]$. This concludes the proof of the lemma.

Let us now explain how this can be used when $m=3$. Let $I$ be the ideal (in the ring ${\mathfrak R}={\mathbb F}(A_{11},A_{12},A_{13},A_{21},A_{22},A_{23},B_1,B_2,B_3)$ of all obstructions. In particular, the determinant

$$ \Delta=(A_{12}A_{23}-A_{13}A_{22})B_1+ (A_{13}A_{21}-A_{11}A_{23})B_2+ (A_{11}A_{22}-A_{12}A_{21})B_3 \tag{1} $$

is a member of $I$. Let $P\in I$, and let $p={\sf deg}_{B_3}(P)$. By the generalized Euclidean division property above, there are polynomials in $Q,R$ in $\mathfrak R$ such that $(A_{11}A_{22}-A_{12}A_{21})^p P=\Delta Q+R$, such that $R$ does not contain the variable $B_3$ (note that we need $p\geq 1$ in order to apply the lemma ; but if $p=0$, we can simply take $Q=0,R=P$). Then $R\in I$. Consider the set

$$ W=\bigg\lbrace (a_{11},a_{12},a_{13},a_{21},a_{22},a_{23},b_1,b_2) \in {\mathbb F}^{8} \ \bigg| \ a_{11}a_{22}-a_{12}a_{21} \neq 0\bigg\rbrace \tag{2} $$

Since $\mathbb F$ is infinite, $W$ is a Zariski-dense open subset of ${\mathbb F}^{8}$. We have a natural map $\phi : W \to V(I)$, defined by

$$ \phi(a_{11},a_{12},a_{13},a_{21},a_{22},a_{23},b_1,b_2)= \bigg(a_{11},a_{12},a_{13},a_{21},a_{22},a_{23},b_1,b_2, -\frac{(a_{12}a_{23}-a_{13}a_{22})b_1+ (a_{13}a_{21}-a_{11}a_{23})b_2}{a_{11}a_{22}-a_{12}a_{21}}\bigg) \tag{3} $$

For any $w\in W$, we have $R(\phi(w))=0$ since $R\in I$. We deduce $R(w)=0$ for all $w\in W$. Since $W$ is Zariski-dense, $R$ is zero eveywhere. So $R$ must be the zero polynomial, $(A_{11}A_{22}-A_{12}A_{21})^p P=\Delta Q$. Since $A_{11}A_{22}-A_{12}A_{21}$ and $\Delta$ have no common factors, we see that $\Delta$ divides $P$.