Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
You should always make it clear that you have in mind commutative rings... – Mariano Suárez-Alvarez Feb 14 '11 at 21:46
up vote 2 down vote accepted

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}

share|cite|improve this answer
Great! Thank you very much... – PseudoNeo Feb 14 '11 at 22:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.