Lines through a node and its two branches

Given: AC The Axiom of Choice, an algebraically closed field $k$ of characteristic not two, the curve $C=V(y^{2}z-x^2(x+z))$, the chart $z=1$ with affine equation $f=y^2-x^2(x+1)=y^2-x^2-x^3$, the node $p=(0,0)$, and $O=k[x,y{]}_{(x,y)}$.

The homogeneous equation $(y^2-x^2)z-x^3$ is primitive and linear in $z$ over $k[x,y]$, since $y^2-x^2$ and $x^3$ are coprime. It is therefore irreducible by Gauss lemma over a UFD, Every finite-variable polynomial ring over a field is a UFD, with prime irreducibles and principal height-one primes, and $f$ is square-free, so $C$ is a plane projective curve of degree three; at $p$ the lowest part is $y^2-x^2=(y-x)(y+x)$, so $m_p(C)=2$ and the tangent lines are $V(y-x)$ and $V(y+x)$ Plane projective curves and their components, Multiplicity of a plane curve at a point, Tangent cone and tangent lines at a point.

Local intersection multiplicities are lengths of local quotients, $I_p(C,L)={\mathrm{\ell }}_O(O\mathrm{/}(f,\mathrm{\ell }))$ Local intersection multiplicity of two plane curves, Finite local length exactly when no common local branch, and the product bound gives $I_p\ge 2$ on every line through $p$, with equality exactly for separated tangent cones Intersection multiplicity dominates the product of multiplicities, with equality for separated tangent cones.

$O\mathrm{/}(y-x,x^3)$ has $k$-basis $1,x,x^2$, and $O\mathrm{/}(y,x^2+x^3)=O\mathrm{/}(y,x^2(1+x))$ has $k$-basis $1,x$ because $1+x$ is a unit of $O$; The descending-power flags in the quotients $k[x{]}_{(x)}\mathrm{/}(x^3)$ and $k[x{]}_{(x)}\mathrm{/}(x^2)$ have simple residue-$k$ factors, so lengths equal the number of these basis monomials Composition series and length of a module.

For a line $L$ with $L\subseteq ̸C$, Bezout applied to the cubic $C$ and the line of degree one gives ${\sum }_{q\in L\cap C}{I}_q(C,L)=3$ Bezout's theorem for plane projective curves.

The two branches here are formal graph branches. In $k\mathrm{⟦}x\mathrm{⟧}$ Formal power series over a commutative ring and the coefficient-extraction functional $[x^n]$ the coefficient operations form a ring Cauchy multiplication makes $R\mathrm{⟦}x\mathrm{⟧}$ a commutative ring containing $R[x]$ as the finitely supported subring, which is a domain For a field $K$, $K\mathrm{⟦}x\mathrm{⟧}$ is a domain and its nonunits form the unique maximal ideal $xK\mathrm{⟦}x\mathrm{⟧}$. Construct $s(x)=1+{\sum }_{r\ge 1}{s}_r{x}^r$ with $s(x{)}^2=1+x$: the coefficient of degree $r$ is $2s_r+{\sum }_{i=1}^{r-1}{s}_i{s}_{r-i}$, so it uniquely determines $s_r$ because $2$ is invertible. Then $f=(y-xs(x))(y+xs(x))$ in $k\mathrm{⟦}x\mathrm{⟧}[y]$. Substitution gives the two graph branches $y=\pm xs(x)$, with distinct linear terms $\pm x$; each factor gives the domain $k\mathrm{⟦}x\mathrm{⟧}$ as quotient, and no other factor remains. This is a formal splitting, not a factorisation of the irreducible curve in the algebraic local ring.