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This might be a stupid question but nevertheless I want to be sure.

Let $A$ be an $n \times n$ matrix such that \begin{equation} A = \begin{pmatrix} a_{1,1} & a_{1,2} & \dots & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n,n} \\ \end{pmatrix} \end{equation} Is it true that \begin{equation} \det(A) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma_i} \end{equation} even if the elements of A are from a finite field (or a commutative ring)? Also, does the regular way of calculating the determinant of a matrix with elements $a_{i,j} \in \mathbb{R}$ (using determinant of minors) carry over to the case where $a_{i,j} \in \mathbb{F}$, a finite field or if $a_{i,j} \in R$, a commutative ring?

By "regular way" I mean picking a row or a column and then iterating through it element by element (from $i = 1$ to $n$), finding the determinant of the minor result from eliminating the row and column of that element then multiplying it by $(−1)^i \times$ the value of that (row/column) element, then sum over all results of this iteration.

Finally, I would really appreciate it if someone can provide a reference (book or lecture notes) on the subject.

Thanks.

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    $\begingroup$ What is the "regular way"? And yes, the formula is the same for any commutative ring. I prove lots of basic results using the "sum over permutations" formula in Chapter 6 of Notes on the combinatorial fundamentals of algebra; at the beginning of that chapter I also give references to various books. $\endgroup$ – darij grinberg Oct 18 '18 at 21:04
  • $\begingroup$ By "regular way" I mean picking a row or a column and then iterating through it element by element (from i = 1 to n), finding the determinant of the minor result from eliminating the row and column of that element then multiplying it by $(-1)^i \times$ the value of that (row\column) element, then sum over all results of this iteration. $\endgroup$ – Yahia Shabara Oct 18 '18 at 21:11
  • $\begingroup$ Yes, you can do that. $\endgroup$ – darij grinberg Oct 18 '18 at 21:15
  • $\begingroup$ Thanks a lot for the answer and the reference you provided.. looks like exactly what I'm looking for!! $\endgroup$ – Yahia Shabara Oct 18 '18 at 21:17
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    $\begingroup$ Indeed, the formulas for evaluating a determinant are $\Bbb Z$-formulas, so they are universal. $\endgroup$ – Lubin Oct 18 '18 at 22:12
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Yes, not only does it work but some authors use it to prove Nakayama's Lemma. It can also be used to prove the commutative ring version of the Cayley-Hamilton theorem. See for example the early pages of Matsumura or of Atiyah-MacDonald.

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