TITLE: Why must the determinant of a matrix with with integer entries be an integer?
QUESTION [7 upvotes]: Why must the determinant of a matrix with integer entries be an integer? 
Note: I know what a determinant of a matrix is, not sure how to explain this question.
Is that because if the matrix is made with integers the determinant has to be an integer as well?

REPLY [18 votes]: One of the various formulas for the determinant of an $n\times n$ matrix $A$ is
$$\det(A) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{i=1}^n A_{i,\sigma_i} $$
where as usual $A_{i,j}$ denotes the entry of $A$ in the $i$th row and $j$th column, and $\mathrm{sgn}(\sigma)$ is either $+1$ or $-1$, depending on $\sigma$. Thus, if every $A_{i,j}$ is an integer, then $\det(A)$ is just a huge sum of products of integers, and is therefore an integer itself.

REPLY [12 votes]: This is precisely because $\mathbb{Z}$ is a ring - it is closed under addition and multiplication.  That is to say, any time you multiply two integers, you get an integer.  Any time you add two integers, you get an integer.  So if the entries of your matrix are all integers, the determinant is just a sum of products of integers.