Referring to both
- Wikipedia page and
- ECDSA-cert paper
I can understand that, given $\mathcal{E} = \mathcal{E}(a,\,b,\,\mathbb{F}_{2^m})$ as our elliptic curve on $\mathbb{F}_{2^m}$ group
- $G \in \mathcal{E}$ is a generator of the group, i.e. $\exists n: n\times G = O$
- $n$ is the order of point $G$ with respect to the group
- choosen randomly $d_A\in[1,n-1]$ as a private key, $Q_A\triangleq n \times d_A$ is the public one.
Got it, but in the "Key pair validation from Bob" process, I see that $Q_A$ must respect these bounds:
- $Q_A \neq O$ (trivial)
- $Q_A \in \mathcal{E}$ (trivial)
- $Q_A\times n = O$
I can't really understand the last thing. I'm not very acquainted with EC algebra, but comparing $Q_A\times n = O$ with $n\times G = O$ seems the same thing, so I would conclude that $Q_A=G$, which is clearly false.
What am I missing?