Calculating the local Milnor number of a polynomial I've seen this previous thread Difficulty with Milnor number
which was helpful, but doesn't quite seem to translate to the problem I'm looking at. I came across in an exercise the notion of a local Milnor number, defined the following way:
If $f$ is a polynomial in $K[x_{1}, \ldots x_{n}]$ defining an affine variety in $\mathbb{A}^{n}$, and $C(f)$ is the set of critical points of $f$ generated by the Jacobian ideal $J_{f}$, then $\mu_{0} := dim_{K}\mathcal{O}_{\mathbb{A}^{n},0}/(J_{f})$
Explicitly, I need to calculate this for the polynomial $f = x^{3} +y^{3} +z^{3} +3xy +2y^{2}$. From this definition and my searches I believe this $\mu_{0}(f)$ to be the local Milnor number at $0$ of $f$. 
So far, I have the following - the ring of regular functions of a variety at a point $\mathcal{O}_{Y,p}$ is isomorphic to the affine coordinate ring of the variety $Y$ localised at the maximal ideal at that point $p$. In this case, the coordinate ring of $\mathbb{A}^n$ is just the polynomial ring $K[x_{1}, \ldots x_{n}]$, since the generating ideal of $\mathbb{A}^n$ is the zero ideal. $\mathcal{O}_{\mathbb{A}^{n},0}/(J_{f})$ is therefore isomorphic to $A(Y)_{M_{0}}$ where $M_{0}$ is the maximal ideal at $0$ - i.e, the ideal containing all polynomials with no linear term. So elements of this ring are elements of the form $p/q$ where $q$ is any polynomial with a nonzero linear term.
I can easily take a monomial basis of the Affine coordinate ring of $J_{f}$ and determine which monomials are linearly independent. Explicitly, I calculated these as $1, x, z, x^{2}, xz, x^{3}, x^{2}z, x^{3}z$ and $x^{4}z$. But this does not take into account the localisation, as I could have elements of the form, say, $\frac{z}{1+x^{2}z}$, and proving this is linearly independent is proving challenging. Is there some neat way to shortcut checking this explicitly for however many monomials? I'm not really familiar with taking vector space bases for a polynomial fraction ring.
As a small aside, I notice in the wikipedia page the following - "It follows from Hilbert's Nullstellensatz that $\mu (f)$ is finite if and only if the origin is an isolated critical point of $f$; that is, there is a neighbourhood of 0 in $\mathbb {C} ^{n}$ such that the only critical point of $f$ inside that neighbourhood is at $0$." This statement seems non-obvious to me - and the only proofs I found were not using the Nullstellensatz but instead machinery that I'm not familiar with (fibrations?). Is there a simple way of showing this?
 A: Let's deal with the problem of the denominators first. The good news is that we do not need to worry about this, and the solution is not even too tricky. Every element $d$ in the denominator is a unit, so there exists some element $u$ with $ud=1$. Multiplying by $1=\frac uu$, we get that $\frac pd = \frac{up}{ud} = up$, which is an expression without any denominators. For an explicit application of this to your example of $\frac{z}{1+x^2z}$, we have $d=1+x^2z$, and we can find $u=1-x^2z$, as $ud=1+x^4z^2$, and $x^4z^2\in J$, so it is zero in the quotient. So $\frac{z}{1+x^2z}=z-x^2z^2$ and you can proceed from here.
(The trick to find $u$ explicitly here is that the identity $\frac{1}{1-q}=1+q+q^2+q^3+q^4+\cdots$ which holds in the ring of formal power series also holds in our ring because eventually $q^n=0$.)
As for your aside about why $\mu(f)$ is finite, suppose the origin is not an isolated critical point: this means that the ideal $\sqrt{J}\subset \mathcal{O}_{\Bbb A^n,0}$ is not maximal, so we can find $f\in \mathcal{O}_{\Bbb A^n,0}$ which is a nonzero nonunit and not in $\sqrt{J}$. This means that any polynomial in $f$ isn't in $J$, which means $\mu(f)$ cannot be finite. On the other hand, if the origin is an isolated critical point, then clearly $\sqrt{J}=\mathfrak{m}$ inside $\mathcal{O}_{\Bbb A^n,0}$ which implies that $\mu(f)<\infty$ by the definition of the radical of an ideal.
