Finite matrix rank bounds under regular specialization

If $r={\mathrm{rank}}_{\mathbb{Q}}(M_{q=1})$ is zero, the desired rank inequality is immediate. If $r>0$, choose $r$ independent columns of the specialized matrix. Gaussian elimination on that $m\times r$ submatrix shows its row space also has dimension $r$, so choose $r$ independent rows; the resulting square submatrix is invertible and has nonzero determinant by A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit. Conversely a nonzero $r\times r$ minor gives $r$ independent columns. Thus some $r\times r$ minor of $M_{q=1}$ has determinant nonzero in the residue field. Its determinant in $R$ specializes to that nonzero value, so it is a nonzero element of $R$ and remains nonzero in its fraction field $F$. Hence ${\mathrm{rank}}_F(M_F)\ge r$. Since both matrices have the same number of columns, rank–nullity gives ${\dim }_F\ker (M_F)\le {\dim }_{\mathbb{Q}}\ker (M_{q=1})$.

For regular endomorphisms $g_1(q),\dots ,g_s(q)$ of a fixed finite free tensor space, vectorize $X$ and form the finite matrix of $X\mapsto ([X,g_1(q)],\dots ,[X,g_s(q)])$. Its kernel is exactly the simultaneous commutant of the $g_i$, which equals the commutant of the algebra they generate. Applying step 1.1 bounds the generic commutant dimension above by the specialized commutant dimension. This argument uses only the selected regular matrices; it does not specialize arbitrary rational functions.

Let $A_0$ be the specialized image, generated by the specialized finite family, in an endomorphism space of finite dimension. The set of all words in those generators spans $A_0$; from this spanning set choose a finite subset that is a basis, by adding a word only when it increases the dimension, which can happen at most $\dim A_0$ times. If $\dim A_0=0$, the image lower bound is immediate; otherwise the same words in the regular lifts form a matrix over $R$ whose specialization has full column rank. Step 1.1, applied to a nonzero square minor of that column matrix, shows the lifted words are $F$-linearly independent. Hence the generic image dimension is at least ${\dim }_{\mathbb{Q}}{A}_0$. The argument is identical for any other finite regular generating family.