A file chunk in a group/field
You have to split your file in such way that each block will be smaller then prime $p$ you're using. After that you have $n$ such blocks.

You calculate $\sigma_i$ for every block so $i \in \{1,2,...,n\}$ and $m_i$ means block number $i$

Example:


  Assume you're working in field $Z_p$ with $p=2^{255}-19$ (used in Curve25519 for example) and your file is 254 bytes or $254 \times 8$ bits long. In practice you can split it to blocks of (at most) 254 bits as $2^{254} \ll 2^{255}-19$. You can split your file to 8 parts, each 254bits long. So you have $n$=8, and each $m_1,m2,m3,m4,m5,m6,m7,m8$ are respective blocks from file.


Example2:


  Assume you're working in field $Z_p$ with $p=2^8+1$ and your file is 5 bytes long. You can split it to blocks of 8 bits as $2^8 \lt 2^8+1$. So you have 5 such parts and $n=5$ and $m1,m2,m3,m4,m5$ are respective bytes from file.


Please note that second example is in no way secure as target field is way too small.