I am studying the Main Coding Theory Problem.
In the definition, in Hill's book, an $(n,M,d)$-code $C$ is a code of length $n$, containing $M$ codewords and having minimum distance $d$. However, in the proofs of some results like $A_q(n,1)=q^n$, it seems that he uses that $d$ has minimum distance at least $d$.
Later, I found on some online notes the definition where q-ary $(n, M, d)$-code means a code of length $n$ containing exactly $M$ words and with minimum distance at least $d$.
Is there a kind of implication? For example "at least $d$ implies exactly $d$"? Or can we claim this without loss of generality? Could you please explain?
In my minimal experience with coding theory, I never encountered that the $d$ in $(n,M,d)$-code actually refers to $d$ being minimal distance at least $d$.
The implication you are asking for is non-existent. At least $d$ will definitely not imply $d$.
I will give you a proof of $A_q(n,1) = q^n$ where $d$ is the exact minimum.
Recall that $A_q(n,1):= \max \{M \in \mathbb{N}_0: \exists [n,M,1]_q-code\}$.
Denote your alphabet with $A$ (where $|A| = q$). Then any $[n,M',1]$-code lives by definition in $A^n$ and thus $A_q(n,1) \leq q^n$. But $C:=A^n$ itself is a $[n,q^n, 1]$-code, so we get $q^n \leq A_q(n,1)$. This shows that
$$A_q(n,1) = q^n$$
Never did I use a weird interpretation of minimal distance "at least $d$".