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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.

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  • $\begingroup$ I don't know if this gets to the heart of your question but the usual statement of Artin's theorem is about finitely many functions $f_i$ of finitely many variables $x_j$ and $y_k$, so you are only talking about a special case. Perhaps a more helpful remark was that I thought that the point was that the theorem said something even when there was branching 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
    Commented Feb 24, 2016 at 19:46
  • $\begingroup$ If $f(x,y)=x^2-y^2$ then there is branching and all partial derivatives vanish at the origin but there is a formal solution $y(x)=x$. So perhaps I don't understand your assertion "..and this means there is no formal solution in powers of x..." $\endgroup$
    – wrigley
    Commented Feb 24, 2016 at 19:48
  • $\begingroup$ @wrigley thanks for your comment. I overlooked this case, however I think that my question is still valid. I think that it is not correct to say that $f(x,y)=x^2-y^2$ has branching, at least not in the sense I meant. $f(x,y)$ has 2 solutions that cross, hence the implicit function theorem fails, but each one is an entire function. Also I calling $y(x)=x$ a formal solution seems a bit odd. So I should add an assumption that I exclude multiple solutions, since we can consider each one separately. $\endgroup$
    – tst
    Commented Feb 24, 2016 at 20:11
  • $\begingroup$ Compare this with $f(x,y)=x^2-y^3$, which has only one solution that branches at $0$, in which case no formal power series can be a solution. $\endgroup$
    – tst
    Commented Feb 24, 2016 at 20:12
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    $\begingroup$ @tst: The power of Artin's theorem is that in any (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
    Commented Feb 25, 2016 at 12:22

1 Answer 1

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In the case when y is a single variable the set of formal solutions is a finite set and Artin's Theorem implies that all these solutions are convergent. But this is a very particular case and not the most interesting application of this theorem.

In the case when y is a set of several variables then the formal solutions are not always convergent. For example if you consider the equation $y_1^2-y_2^3=0$ as an equation with coefficients in $\mathbb{C}\{x\}$, the formal solutions are couples $(y_1(x),y_2(x))=(z(x)^3,z(x)^2)$ where $z(x)$ is a formal power series. But if $z(x)$ is not convergent then the corresponding solution is not convergent. On the other hand if you replace $z(x)$ by one of its truncations (you remove in $z(x)$ all the monomials of degree higher than a given number $c$), then you obtain a polynomial $z'(x)$ and the couple $(z'(x)^3,z'(x)^2)$ is a solution of the equation which is close to the given formal solution. And its components are convergent power series (here in fact they are even polynomials).

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    $\begingroup$ To fit the hypotheses of Artin's theorem, you would have to set $\mathbb{k} = \mathbb{C}\{x\}$. Artin's power series would then be with respect to either $y_1$ or $y_2$, but not $x$. So your example, while related, falls outside the scope of the theorem as stated by the OP. $\endgroup$
    – Igor Khavkine
    Commented May 17, 2016 at 23:31
  • $\begingroup$ @Igor No. I set $\Bbbk=\mathbf{C}$ and my power series is $f(x,y)=y_1^2-y_2^3$ with $y=(y_1,y_2)$. Even if the variable $x$ does not appear in the expression of $f$ I can consider $f$ as a power series in the variables $x$, $y_1$ and $y_2$. So this is exactly Artin's theorem. $\endgroup$
    – hello
    Commented Jan 26, 2017 at 13:27
  • $\begingroup$ If you prefer you can set $f(x,y)=x(y_1^2-y_2^3)$. It does not change the conclusion. $\endgroup$
    – hello
    Commented Jan 26, 2017 at 13:30
  • $\begingroup$ Wow, what a blast from the past! :-) But having read over your original answer and your comments, I think I see what you meant. Thanks. $\endgroup$
    – Igor Khavkine
    Commented Jan 26, 2017 at 19:11

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