I asked this on math stackexchange but I had no luck, so I am posting my question also here.

I am not an algebraist so my question might be stupid. I am doing mainly complex analysis and recently I was informed about the existence of Artin's theorem and it sounded like it could be of interest to me. I have found a survey on the subject and I started reading it. Here's the link.

So to the actual theorem (section 1.1)

Let $\mathbb{k}$ be a field of characteristic 0 and let $f(x,y)$ be a vector of convergent power series in two variables $x$ and $y$. Assume given a formal power series $\hat{y}(x)$ vanishing at 0, $$f(x,\hat{y}(x))=0.$$ Then for any $c\in\mathbb{N}$, there exists a convergent power series solution $\tilde{y}(x)$ $$f(x,\tilde{y}(x))=0$$ which coincides with $\hat{y}(x)$ up to degree $c$, $$\hat{y}(x)\equiv \tilde{y}(x) \mbox{ mod }x^c. $$

I really care only for the case where $k=\mathbb{C}$. Using the implicit function theorem for some analytic $f$ we get the existence of an analytic solution as long as the Jacobian has full rank at 0. If on the other hand the Jacobian does not have full rank then we generically get some king of branching and this means that there is no formal solution in powers of $x$ that solves the equation. So in that sense I don't see how Artin's theorem is stronger than the implicit function theorem in the analytic setting.

Is this true or do I miss something? By the way I don't know what happens when we consider other fields and I don't imply that the theorem is trivial or useless.

EDIT: As was pointed out correctly by wrigley the implicit function theorem can fail at points where there are multiple solutions. For example when $f(x,y)=x^2-y^2$. In this case the workaround is to consider one solution at a time, i.e. factorize $f$ and look at one factor each time.

wasbranching and that this was in some sense the power of the method. But I only learnt this stuff myself recently and on the algebraic side of things. Can you clarify what you mean by "Jacobian is invertible" in your context? A 2x1 matrix can't be... $\endgroup$ – wrigley Feb 24 '16 at 19:46any(finite) number of variables, if there is a formal solution then there is a convergent one (the approximation aspect is much simpler than existence). The geometry of $f=0$ can be very nasty when not smooth at the origin, so one cannot see this by a simple method with the implicit function theorem. In 2 variables there are at most finitely many formal solutions, and the content of Artin's theorem is that they're convergent (can you really prove that directly?); in $> 2$ variables there are generally many solutions which are only formal. $\endgroup$ – nfdc23 Feb 25 '16 at 12:22