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)


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