Show that an irreducible curve \(Y\) of degree \(d > 1\) in \(\PP^2\) cannot have a point of multiplicity \(\geq d\).
If \(Y\) is an irreducible curve of degree \(d > 1\) having a point of multiplicity \(d - 1\), show that \(Y\) is a rational curve.
Solution
After a projective linear change of coordinates, put the point under consideration at \[ P=[0:0:1]. \] Let \(F(x,y,z)\) be an irreducible homogeneous equation of degree \(d\) for \(Y\), and write \[ f(x,y)=F(x,y,1)=f_0+f_1+\cdots+f_d \] with each \(f_i\) homogeneous of degree \(i\) in \(x,y\).
1. The multiplicity of \(Y\) at \(P\) is at most \(d\).
By Exercise I.5.3, the multiplicity is the least index \(m\) for which \(f_m\ne0\). Since \(f\) has degree at most \(d\) and is nonzero, such an index satisfies \(m\le d\).
2. If the multiplicity at \(P\) were \(d\), then \(F\) would be reducible for \(d>1\).
Multiplicity \(d\) means \[ f_0=f_1=\cdots=f_{d-1}=0, \] so \(f=f_d\) is homogeneous of degree \(d\). Homogenizing back to degree \(d\) gives \[ F(x,y,z)=f_d(x,y), \] which is independent of \(z\). Over the algebraically closed field \(k\), the binary homogeneous form \(f_d(x,y)\) factors into \(d\) linear forms, counted with multiplicity. For \(d>1\) this contradicts irreducibility of \(F\).
Together with step 1, this proves \[ \boxed{\mu_P(Y)\le d-1} \] for every point of an irreducible plane curve of degree \(d>1\), proving part (a).
3. If \(\mu_P(Y)=d-1\), then the equation of \(Y\) has the form\[ F(x,y,z)=z f_{d-1}(x,y)+f_d(x,y), \]with \(f_{d-1}\ne0\).
The multiplicity hypothesis gives \[ f_0=\cdots=f_{d-2}=0, \qquad f_{d-1}\ne0. \] Since \(F\) is homogeneous of total degree \(d\), the degree-\((d-1)\) affine term acquires one factor of \(z\) on homogenization, while the degree-\(d\) term acquires none. This is exactly the displayed equation.
4. Projection from \(P\) gives a birational map\[ \pi:Y\dashrightarrow\PP^1, \qquad [x:y:z]\longmapsto[x:y]. \]
The map is defined away from \(P\). For a direction \([u:v]\in\PP^1\), the corresponding line through \(P\) consists of points with \([x:y]=[u:v]\). Substituting \(x=u\), \(y=v\) into step 3 gives \[ z f_{d-1}(u,v)+f_d(u,v)=0. \] On the nonempty open subset \[ W=\{[u:v]:f_{d-1}(u,v)\ne0\}\subseteq\PP^1, \] there is therefore exactly one residual point of \(Y\) on that line, namely \[ \rho([u:v]) = [u f_{d-1}(u,v):v f_{d-1}(u,v):-f_d(u,v)]. \] The three displayed coordinates are homogeneous of the same degree \(d\), and on \(W\) the first two do not vanish simultaneously, so \(\rho\) is a morphism there. Direct substitution gives \(F(\rho([u:v]))=0\). Moreover \[ \pi\rho([u:v])=[u:v] \] on \(W\).
Conversely, on the dense open subset of \(Y\) where \((x,y)\ne(0,0)\) and \(f_{d-1}(x,y)\ne0\), the equation \[ z f_{d-1}(x,y)+f_d(x,y)=0 \] forces the \(z\)-coordinate to be the one in the formula for \(\rho\) up to projective scaling. Thus \(\rho\pi\) is the identity there. Hence \(Y\) and \(\PP^1\) are birational.
5. The curve \(Y\) is rational.
By definition, an irreducible curve is rational precisely when it is birationally equivalent to \(\PP^1\). Step 4 supplies such a birational equivalence, proving part (b).
6. Q.E.D.
Steps <1>1–<1>2 prove part (a), and steps <1>3–<1>5 prove part (b).