2
$\begingroup$

Just finished my semester in advanced abstract algebra and there was one question that I could not answer.

Let $R$ be commutative and let $(e_1,...,e_n)$ be a base for $R^{(n)}$. Put $f_i=\sum_{j=1}^n a_{ij}e_j$ where $A=(a_{ij})\in M_n(R)$. Show that the $f_i$ form a base for a free submodule $K$ of $R^{(n)}$ if and only if $\det A$ is not a zero-divisor.

$\endgroup$

marked as duplicate by user26857, Potato, Arthur, Davide Giraudo, azimut Dec 14 '13 at 13:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ "$\Leftarrow$" If $Ax=0$, then $0=A^{ad} A x = \det(A) x$, hence $x=0$. The proof of "$\Rightarrow$" is more complicated (too lazy to write it down, just a hint: Laplace expansion). $\endgroup$ – Martin Brandenburg Dec 13 '13 at 22:31
  • $\begingroup$ www2.im.uj.edu.pl/actamath/PDF/30-215-218.pdf $\endgroup$ – user26857 Dec 14 '13 at 9:30
2
$\begingroup$

Let $r_1, \dots, r_n$ be elements of $R$ such that $\sum_i r_if_i = 0$. Our goal is to show that $r_i = 0$ for all $i$.

Write $\vec{r} = \langle r_1, \dots, r_n \rangle^t$ (so a column vector), one quickly sees (using that $e_1, \dots, e_n$ is a basis) that this is equivalent to $A\vec{r} = \vec{0}$. Even though $\det(A)$ may not be invertible and hence we can not have $A^{-1}$, one can still find the Adjugate matrix of $A$, call it $\tilde{A}$, so that $\tilde{A}A = \det(A) \cdot I$. This gives us $\det(A) \cdot I \cdot \vec{r} = \vec{0}$ making $\vec{r} = 0$.

If you were not aware of the adjugate matrix, then this is probably the wrong way to go.

$\endgroup$

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