# How do i prove the Leibniz formula of the determinant over a commutative ring?

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?

• Have you tried using induction on $n$? Sep 27 '14 at 10:08
• 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. 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)