\begin{array}{ll} SignFDH_{N,d}(M) & & & &&&&&\\ \quad\quad y \leftarrow H_{FDH}(M)\\ \quad\quad \textbf{return } \sigma = y^d \bmod n\\ \end{array}\begin{array}{ll} \operatorname{SignFDH}_{N,d}(M) & & & &&&&&\\ \quad\quad y \leftarrow H_{FDH}(M)\\ \quad\quad \textbf{return } \sigma = y^d \bmod n\\ \end{array}
\begin{array}{l} VerifyFDH_{N,e}(M,\sigma) &\\ \quad\quad y \leftarrow \sigma^e \bmod N&\\ \quad\quad y \leftarrow H_{FDH}(M)&\\ \quad\quad \textbf{if } y = y' \textbf{ then return } 1 \textbf{ else return } 0\\ \end{array}\begin{array}{l} \operatorname{VerifyFDH}_{N,e}(M,\sigma) &\\ \quad\quad y \leftarrow \sigma^e \bmod N&\\ \quad\quad y \leftarrow H_{FDH}(M)&\\ \quad\quad \textbf{if } y = y' \textbf{ then return } 1 \textbf{ else return } 0\\ \end{array}
Can a random text that has taken the $e$-th power be a signature forgery?
No, The forger needs to provide the message $m'$ too. During verification, the hash of the message $SHA-512(m')$$\operatorname{SHA-512}(m')$ will be taken and compared with the one in the signature.
