This is probably a well-known issue, but I could not find a clear discussion in the literature, and I think others could find it useful.


Consider a real-analytic function germ $f:(\mathbb R^2,0) \rightarrow \mathbb R$, it is represented by a convergent power series $\sum\_{i,j}a\_{ij}x^iy^j \in \mathbb R\{x,y\}$. Suppose $f$ has a singularity at $0$, i.e. $df(0)=0$. Define its *Jacobian ideal* as the ideal generated by its partial derivatives: $I\_{df} = \mathbb R\{x,y\}\langle f\_x,f\_y\rangle$, and its *real local algebra* as:
$$Q\_f := \frac{\mathbb R\{x,y\}}{I\_{df}}.$$


Let $\mu\_{\mathbb R} := \dim\_{\mathbb R} Q\_f$. The singularity is called *algebraically isolated* if $\mu\_{\mathbb R}< \infty$. It is called *isolated* if, in some small neighborhood of $0$, $df(x,y)=0$ only at $(x,y)=0$ (i.e., the usual meaning).


Now take the complexification of $f$ by simply considering the series as a complex one: $\sum\_{i,j}a\_{ij}z^iw^j$. It is convergent in some polydisc and defines an holomorphic germ $f^{\mathbb C}:(\mathbb C^2,0) \rightarrow \mathbb C$. Take its (complex) Jacobian ideal, that is: $I\_{df^{\mathbb C}} = \mathbb C\{x,y\}\langle f^{\mathbb C}\_z,f^{\mathbb C}\_w\rangle$, and the *complex local algebra*
$$Q^{\mathbb C}\_f := \frac{\mathbb C\{x,y\}}{I\_{df^{\mathbb C}}}.$$


The *Milnor number* of the complexified singularity is defined by: $\mu\_{\mathbb C} := \dim\_{\mathbb C} Q\_f^{\mathbb C}.$
A complex singularity is called *isolated* if $\mu\_{\mathbb C} < \infty$. (In the holomorphic case it can be proven that a singularity is isolated if and only if, in some small neighborhood of $0$, $df(z,w)=0$ only at $(z,w)=0$, so we don't need to specify ''algebraically'' here.)


**Questions**: 


1. What is the relationship between the real-analytic singularity and its complexification? In particular: between the two local algebras, and between $\mu\_{\mathbb R}$ and $\mu\_{\mathbb C}$? Is there a class of functions where things work better?
2. Is true (or false) that a real-analytic singularity is algebraically isolated if and only if its complexification is isolated?
3. In the real setting: is ''algebraically isolated'' stronger then simply ''isolated''?
4. Any reference!


Thank you.



Complexifying a real-analytic singularity
As David Speyer has already suggested, we have $Q\_f^\mathbb{C}\simeq Q\_f\otimes\mathbb{C}$, so that answers the first question (and the second question).


For the third question, consider $f = (x^2+y^2)^2$. The singularity at $(x,y)=0$ is clearly isolated, but it is not algebraically isolated, since $\mu\_\mathbb{R} = \infty$. Of course, over the complex numbers, $(z^2+w^2)^2$ has a non-isolated singularity at $(z,w)= (0,0)$; the two lines of singularities $z = \pm i\,w$ intersect at $(0,0)$.


The various confusions will be cleared up by noting that the ring of germs of analytic functions in two variables is a UFD, and, the singularity will be algebraically isolated as long as $f$ does not have any repeated irreducible factors. 'Isolated' is weaker over the reals simply because all the factors of $f$ might have no real zeros other than at the origin.


