Given a plane algebraic curve $$ y^n + a_1(x)y^{n-1} + \dots +a_{n-1}(x) + a_n(x)y = 0, $$ with a branch point $P_0=(0, y_0)$ of order $n$. Can we prove that this curve is irreducible? What if the point $P_0$ is singular and unibranch?
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2$\begingroup$ How about $(y-\alpha _1(x))\ldots (y-\alpha _n(x))=0$, with $\alpha _i(0)=0$ for all $i$? $\endgroup$– abxCommented Aug 11, 2022 at 8:33
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2$\begingroup$ If by 'unibranch', you mean that the $x$-pre-image in the curve of any sufficiently small punctured disk $D^*_\epsilon = \{ x\ |\ 0<|x|<\epsilon\ \}$ is connected and the $x$-projection is $n$-to-$1$ onto $D^*_\epsilon$, then, yes, the curve is irreducible. $\endgroup$– Robert BryantCommented Aug 11, 2022 at 11:29
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1$\begingroup$ (Ir)reducibility of the curve depends only on the polynomial (not on a point): it is irreducible iff the polynomial does not factor into a product of non-constant polynomials. $\endgroup$– Alexandre EremenkoCommented Aug 11, 2022 at 12:26
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$\begingroup$ @abx: The example you gave is singular at $(0, 0)$, but wether it is unibranch? $\endgroup$– minxin jiaCommented Aug 12, 2022 at 1:05
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$\begingroup$ @RobertBryant: Thank you very much! I wonder if a unibranch point is a smooth branch point? In David Mumford's book Algebraic Geometry I, Complex Projective Varieties, on page 43 he defines what is a topologically unibranch: Let $X$ be an affine variety over $\mathbb{C}$ and $x\in X$. Then $X$ is topologically unibranch at $x$ if for all closed algebraic subsets $Y\subsetneq X$, $x$ has fundamental system neighborhoods $U_n$ in classical topology of $X$, s.t. $U_n - (U_n\cup Y)$ is connected. And he claims that smooth points are topologically unibranch. $\endgroup$– minxin jiaCommented Aug 12, 2022 at 1:19
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