Question about ideals and linear polynomials

As stated we do not have (i)$\implies$ (ii). For a counterexample take $$g=x_2, \qquad f_1=x_1,\,\,f_2=x_1+1.$$

Then $$g=x_2f_2-x_2f_1,$$ so $g\in\langle f_1,f_2\rangle$. However $g\notin {\rm Span}_k\{f_1,f_2\}$.

This example only works because the variety defined by the $f_i$ is empty. As long as this variety is non-empty your argument for (i)$\implies$ (ii) is valid:

If $g$ is not in the span of the $f_i$ and the variety defined by the $f_i$ is non-empty, then throwing $g$ into the system adds a pivot, which reduces the dimension of the variety (or makes it empty). This makes the new variety a proper subset of the original, which contradicts (iii).

It is a nice result - definitely worth mentioning to your students. In my example the $f_i$ are multiplied by linear polynomials, and the degree $2$ terms cancel yielding a linear polynomial not in the span. It is surprising to me that this can only happen in the restricted case that the variety is empty. Otherwise the linear polynomials obtained through this kind of cancellation will always happen to lie in the $k$-linear span of the $f_i$.