At the end of the proof of Artin's approximation theorem, and using all his notation, he reduces to finding a solution $y\in A$ such that $$y\equiv\overline y\mod \mathfrak m^c$$ $$\tag{*}f(y)\equiv0\mod\delta^2(X,y^\circ). \mathfrak m^c.$$ To do this he proves a lemma which finds elements $y\in A$ such that $$f_i(X,y)\equiv0\mod g(X,y)$$ $$y\equiv\overline y\mod\mathfrak m^d,$$ where $g=\delta^2$ and $d$ is any positive integer. But a priori the ideal $\delta^2(X,y^\circ). \mathfrak m^c$ may be smaller than $(g(X,y))=(\delta^2(X,y))$.

To produce the desired statement $(\ast)$, I reason as follows. We may assume $g(X,\overline y)$ not invertible (and the $f_i$ not zero, otherwise there is no question); in that case suppose the initial form of $g(X,\overline y)$ lies in $\mathfrak m^i/\mathfrak m^{i+1}, i>0$. If we take $d$ large enough, $g(X,y)$ will have the same initial form. Now take $d$ large enough, and also larger than $i+c$. Then as $\overline y$ is a solution of $f$ (i.e. all the $f_i$), $f_i(X,y)$ will agree with $f_i(X,\overline y)$ up to at least degree $d$, hence its initial form will be in $\mathfrak m^j/\mathfrak m^{j+1}$ for $j\geq i+c$. And we know the associated graded of $A$ is a polynomial algebra over the residue field since $A$ is the henselization of a regular local ring hence a regular local ring. So indeed $(\ast)$ holds.

I would like to know if there is an easier way to see how to apply the lemma to get what we want. When I first saw it I thought perhaps there was a typo.

Secondarily, I would like to know why we can choose the coordinates $y_1,\ldots,y_N$ at the end of the Néron approximation section so that $l(s')=l(\tilde{s}')$. Artin addresses this in a parenthetical statement but I don't see why it is always possible.

Thanks!