I have a module M over some commutative ring R. A square matrix $\Omega$ of size $m$ with coefficients in $R$ defines a $R$-linear map from $M^m$ to itself.

Proposition. If $x = \left( \begin{matrix} x_1 \\ \vdots \\ x_m \end{matrix} \right) \in M^m$ is in the kernel of $\Omega$, then $\det \Omega$ annihilates all the $x_i$.

I think I have a proof of this fact, roughly going as follows.

Let's call "pseudomatrix" a matrix whose coefficients are all in R except for a column, where the coefficients are in M. The remark is that the formulas for the determinant of a matrix and the product of two matrices still make sense in that context: the determinant of a pseudomatrix is an element of M and the product of a matrix and a pseudomatrix is a pseudomatrix. One even has that if A is a matrix and $B$ a pseudomatrix, then $\det(AB) = \det(A)\det(B)$. (I think that no proof is needed here: this formula is true in the classical setting, so it can be seen as a formal identity and can be specified in our setting). I can then easily construct a pseudomatrix whose first column is $x$ and whose determinant is $x_i$, and the proposition follows.

I'm convinced that this proof should be easily expressed in an "algebraic nonsensical" language, using words such as "$\otimes_R M$" and "$\bigwedge$" instead of my ugly hybrid "pseudomatrices" but I've been unable to do the translation. Can you help me?

  • $\begingroup$ You should always make it clear that you have in mind commutative rings... $\endgroup$ Feb 14, 2011 at 21:46

1 Answer 1


The matrix $\Omega$ and its adjugate $\mho$ determine maps $\Omega:R^m\to R^m$ and $\mho:R^m\to R^m$ such that $\mho\circ\Omega=\det\Omega\cdot\mathrm{id}_{R^m}$.

Now suppose $m\in M^m=R^m\otimes_R M$ is such that what you write $\Omega m$ and I will write $(\Omega\otimes\mathrm{id}_M)(m)$ is zero. Then \begin{align} 0 &= (\mho\otimes\mathrm{id}_M)\bigl((\Omega\otimes\mathrm{id}_M)(m)\bigr) \\ &= ((\mho\circ\Omega)\otimes\mathrm{id}_M)(m) \\ &= \det\Omega\cdot m \end{align}

  • $\begingroup$ Great! Thank you very much... $\endgroup$
    – PseudoNeo
    Feb 14, 2011 at 22:02

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