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!