# Determinant of a matrix over the integers mod n

1. May 4, 2006

### jdstokes

Hi,

I'm curious if the following statement is true for all prime numbers n,

$\det_{\mathbb{Z}_n}M = (\det_{\mathbb{R}}M)\mod n$

where $\det_F M$ is the determinant of M over the field F.

Thanks.

James

Last edited: May 4, 2006
2. May 4, 2006

### AKG

Z isn't a field. But if you mean that if you take a matrix, and compute its determinant by multiplying, adding, and subtracting the entries as integers and then reduce mod n, versus if you do all the arithmetic mod n, then the answer is 'yes'.

3. May 4, 2006

### jdstokes

Thanks for the correction, I guess what I meant to say was $\mathbb{R}$.

4. Jul 17, 2006

### mathwonk

no, what you meant to saY WAS THATyour notation denoted the determinant over the ring F.

since you cannot consider elements of R as if they were in Z/n unless they are integers.

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