Fundamental group of the complement to a plane curve with unramified normalization

Suppose that $$C\subset\mathbb P^2$$ is an irreducible projective curve over $$\mathbb C$$ such that the normalization morphism $$\bar C\to C$$ is unramified (i.e., the induced morphism $$\bar C\to\mathbb P^2$$ is an immersion). Could you help me with examples of curves with this property for which $$\pi_1(\mathbb P^2-C)$$ is not commutative?

• Oh. One should properly wake up before writing anything on the Internet. (...Cusps are unibranch, so in my simple topologist mind they have unramified normalisation, because "homeomorphisms cannot be ramified"). Am I right that you are asking for an example of immersed plane curve $C: \,q(x, y, z)$ such that its Milnor fiber $F:\, q(x, y, z) = 1$ has nontrivial $\pi_1$? For irreducible curve $\pi_1(F)$ should be equal exactly to $\pi_1(P^2 \setminus C)'$, unless I'm making some obvious mistake again. (Numerous articles by Degtyarev, Shimada and Dimca should contain multifarious examples..) Commented Sep 24, 2022 at 8:18
• @JasonStarr It is, yes, but it does not tell us much about global fundamental groups. For example, if $C$ is a quartic with one simple triple point (that is formally isomorphic to the union of three concurrent lines), then $\pi_1$ of its complement is abelian. Commented Sep 24, 2022 at 13:21