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.
Best Answer
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.