I was looking at the Wikipedia page of the Milnor number found here, and specifically tried to work through proving for myself one of its stated properties. It mentions the Milnor number is finite if and only if the origin is an isolated critical point.
Now it seems someone gave the proof for this as part of an answer here, but it is going completely over my head. For one, I do not understand the seemingly "obvious" relation how the origin being an isolated critical point implies the radical of the Jacobian ideal $J_f$ to be maximal, and vice versa. Furthermore I also don't quite get why the definition of a radical ideal leads right into $\mu(f) < \infty$. I understand as intersection of prime ideals this must mean there aren't any more prime ideals between $J_f$ and $\mathfrak{m}$ but then what?
Recall the order-reversing bijection between ideals and closed subschemes in an affine scheme given by taking $V(-)$ and $I(-)$. If we have two ideals $I_1,I_2$, then the topological spaces underlying $V(I_1)$ and $V(I_2)$ are the same iff $\sqrt{I_1}=\sqrt{I_2}$. If the subschemes associated to the Jacobian ideal $J_f$ and the maximal ideal $\mathfrak{m}_0$ inside $\mathcal{O}_{\Bbb A^n,0}$ have the same underlying topological space (the unique closed point of that spectrum), we see that they must have the same radical, and vice-versa. As the maximal ideal is radical, this means that $\sqrt{J_f}=\mathfrak{m}_0$ is equivalent to the underlying topological spaces being the same, or the Jacobian ideal having an isolated zero at the origin.
To demonstrate that the closed subschemes associated to the maximal ideal and $J_f$ really do determine the same underlying topological space in $\operatorname{Spec}\mathcal{O}_{\Bbb A^n,0}$, let's remember what the points of that space are. The points of $\operatorname{Spec} \mathcal{O}_{\Bbb A^n,0}$ are in bijection with reduced irreducible closed subschemes of $\Bbb A^n$ passing through $0$. $J_f$ having a non-isolated zero at the origin would exactly mean that it vanishes on some positive-dimensional closed subscheme through the origin, which is equivalent to $V(J_f)\subset \operatorname{Spec} \mathcal{O}_{\Bbb A^n, 0}$ containing a point other than the unique closed point of $\operatorname{Spec}\mathcal{O}_{\Bbb A^n,0}$.
Alternatively, from the comments: Suppose $\sqrt{J_f}\neq \mathfrak{m}$. Then by the characterization of the radical of $J_f$ as the intersection of all prime ideals containing $J_f$, we can find a prime ideal $I$ between $J_f$ and $\mathfrak{m}$, that is $J_f\subset I \subsetneq \mathfrak{m}$. Taking varieties of this, we get that $V(J_f)\supset V(I)\supsetneq V(\mathfrak{m})$ where $V(I)$ is some irreducible subvariety properly containing $V(\mathfrak{m})$, the origin. This means that there's some curve through the origin on which $J_f$ vanishes, so the origin is not an isolated zero of $V(J_f)$.
As for why $\sqrt{J_f}=\mathfrak{m}_0$ implies that $\mu_f < \infty$, consider the standard generating set for $\mathfrak{m}_0\subset \mathcal{O}_{\Bbb A^n,0}$ given by $x_1,\cdots,x_n$. Then by the definition of the radical of an ideal, we have that for each $i$, we have there exists an positive integer $e_i$ with $x_i^{e_i}\in J_f$. So the monomials $\prod x_i^{d_i}$ with $0\leq d_i < e_i$ form a spanning set of $\mathcal{O}_{\Bbb A^n,0}/J_f$ as a vector space, which implies it's finite dimensional.
Alternatively, one may use the proof from the comments: $\mathcal{O}_{\Bbb A^n,0}/J_f$ is a $0$-dimensional local ring, which is Artinian. As an Artinian ring over a field is finite-dimensional, this also proves the claim.