Let $v: K/k\to \Lambda$ be a valuation on a field $K$ extension of $k$, with values in an ordered group $\Lambda.$ It is often claimed that the rank of $\Lambda$ and the rational rank, $\dim_\mathbb Q \Lambda\otimes \mathbb Q$, (which are also called the rank and the rational rank of a valuation) are bounded from above by the transcendence degree of $K/k,$ eg. http://www.cmls.polytechnique.fr/perso/favre/GT/2011-12/otal-survey.pdf (In fact, Abhyankar inequality provides an even stronger statement.)
But $\Lambda=\mathbb Z$ embeds into an ordered $\mathbb Z^N$ for any $N$ making any $v: K/k\to \mathbb Z$ into a rank $N$ valuation $v: K/k\to \mathbb Z^N$ for any $N> tr.deg(K/k)$. Am I wrong here, or (a) some people are sloppy defining the rank of a valuation or (b) some people are sloppy stating the above inequality?
You're right, you can embed $\Bbb Z$ in an order-preserving way in to any ordered $\Bbb Z^N$, but it is extremely common to assume that $v$ is surjective (which one can always do by restricting to the image of $v$ in $\Lambda$). I guess this falls under your option a).