Let $R$ be a commutative ring.

My definition for the determinant over $M_n(R)$ is defined inductively as $\det_{n+1}(A)=\sum_{j=1}^n (-1)^{1+j}A_{1j} \det_n(\tilde{A_{ij}})$.

(Here, $(-1)$ denotes its sign, it does not mean the additive inverse of a unity)

($\tilde{A_{ij}}$ is given by removing $i$-th row and $j$-th column of $A$)

With this definition, I have proved the following:

Let $A\in M_n(R)$

Then, $\forall x\in R$, $\det(xA)=x^n \sum_{\sigma\in S_n} \operatorname{sgn}(\sigma)\prod_{i=1}^n A_{i,\sigma(i)}$.

However, I want to show that $\det(A)$ is exactly $\sum_{\sigma\in S_n} \operatorname{sgn}(\sigma)\prod_{i=1}^n A_{i,\sigma(i)}$. (I think this is true)

If $R$ has a unity, then this is trivial. However, if it does not contain a unity, how do I show this?

  • $\begingroup$ Have you tried using induction on $n$? $\endgroup$ – Henry Sep 27 '14 at 10:08
  • $\begingroup$ What about $\det_1(A)$ which essentially demands $\det_0(A)$? Means, without identity matrix $I$ and defining that $\det(I)=1$, I dont think so, its possible. $\endgroup$ – David Feb 13 '18 at 10:36

How about injecting $R$ into a ring with unity?

(though I can't for the life of me figure out how one would prove the formula for $\det{xA}$ without going through $\det{A}$ first)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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