Ask Your Question

Revision history [back]

click to hide/show revision 1
initial version

Since the ring is finite of order $3^3$ and the matrix is $4 \times 4$, one approach is to try every possible vector; there are only $(3^3)^4 = 531441$ possibilities.

Since the ring is finite of order $3^3$ and the matrix is $4 \times 4$, one approach is to try every possible vector; there are only $(3^3)^4 = 531441$ possibilities.

A more refined approach is to introduce a vector with $3 \cdot 4 = 12$ undetermined coefficients, e.g. by making the base ring the quotient by $t^3$ of the univariate polynomial ring in $t$ over a polynomial ring in $12$ variables over $\mathbb{F}_3$.

Considering the matrix over this base ring, you can multiply with this vector with undetermined coefficients, take the components, and set the coefficients of powers of $t$ (of which the higher ones were automatically eliminated) equal to zero; this is an ordinary linear system over $\mathbb{F}_3$.

Since the ring is finite of order $3^3$ and the matrix is $4 \times 4$, one approach is to try every possible vector; there are only $(3^3)^4 = 531441$ possibilities.

A more refined approach is to introduce a vector with $3 \cdot 4 = 12$ undetermined coefficients, e.g. by making the base ring the quotient by $t^3$ of the univariate polynomial ring in $t$ over a polynomial ring in $12$ variables over $\mathbb{F}_3$.

Considering the matrix over this base ring, you can multiply with this vector with undetermined coefficients, take the components, and set the coefficients of powers of $t$ (of which the higher ones were automatically eliminated) eliminated, due to the quotient by $t^3$) equal to zero; this is an ordinary $12 \ times 12$ linear system over $\mathbb{F}_3$.

Since the ring is finite of order $3^3$ and the matrix is $4 \times 4$, one approach is to try every possible vector; there are only $(3^3)^4 = 531441$ possibilities.

A more refined approach is to introduce a vector with $3 \cdot 4 = 12$ undetermined coefficients, e.g. by making the base ring the quotient by $t^3$ of the univariate polynomial ring in $t$ over a polynomial ring in $12$ variables over $\mathbb{F}_3$.

Considering the matrix over this base ring, you can multiply with this vector with undetermined coefficients, take the components, and set the coefficients of powers of $t$ (of which the higher ones were automatically eliminated, due to the quotient by $t^3$) equal to zero; this is an ordinary $12 \ times \times 12$ linear system over $\mathbb{F}_3$.