# algebraic multivariate power series over a field

Is there a "simple" proof that any power series in $\mathbb Q[[X,Y]]$ algebraic over $\mathbb Q(X,Y)$ is in the Henselization of $\mathbb Q[X,Y]$ localised in $(X,Y)$?

• What is the meaning of a "simple" proof? Do you mean a proof which avoids using certain kind of notions, theories or constructions? – user42090 Nov 15 '13 at 16:02
• It is a direct consequence of Artin's approximation theorem. – Andrew Stout Nov 15 '13 at 16:51
• @AndrewStout To get the ball rolling: would you consider explaining your comment in an answer? – Todd Trimble Nov 16 '13 at 22:30

## 2 Answers

I am not sure if this is what the OP really wants, but this is the proof I had in mind.

Theorem. Let $(A,m)$ be a local ring which is obtained by localizing a finitely generated $k$-algebra where $k$ is a field. Let $\hat A$ be the $m$-adic completion of $A$. Let $\mathcal{E}$ be collection of elements of $A[x_1,\ldots,x_n]$ where $x_i$ are variables. If $\mathcal{E}$ has a solution $(a_1,\ldots,a_n)$ with $a_i \in \hat A$, then it has a solution $(b_1,\ldots,b_n)$ with $b_i\in A^h$ where $A^h$ is the henselization of $A$. Moreover, given any $t$, we can choose the solution in $A^h$ such that the residues of $a_i$ and $b_i$ in $\hat A/m^t\hat A$ are equal for all $i$.

This is almost a verbatim restatement of Theorem 8.4.5 of the great book Cohen-Macaulay Rings by Bruns and Herzog, which is more or less a restatement of Artin's Approximation Theorem.

So, if $A = \mathbb{Q}[X,Y]_{(X,Y)}$, then $\hat A = \mathbb{Q}[[X,Y]]$.Then choosing n=1, $\mathcal{E}=\{P(x_1)\}$ for some arbitrary monic polynomial $P(x_1)$, and t=1 gives the result provided one accepts the following trick: we may apply affine transformations to the coefficients (possibly, defined over $\bar{\mathbb{Q}}$) of a monic polynomial $P(x_1)$ in $\mathbb{Q}(X,Y)[x_1]$ so that each coefficient of $P(x_1)$ lies in $A$.

• Sure, my pleasure. I can think of another proof in the univariant case. I believe it should follow from cell-decomposition for definable subassignments in the language of Denef-Pas over $\mathbb{Q}[[X]]$, but this model-theoretic. Even though this is neither here nor there, I would like to know how to prove the multivariant case model-theoretically. – Andrew Stout Nov 19 '13 at 4:10

This is also a direct consequence of a result of M. Nagata (see Theorem 44.1 of his book "Local Rings"). And this result of Nagata is much more easier to prove than Artin's approximation theorem.