Plane Curves, Local Intersection Multiplicity, and Bézout — Examples

Given: AC The Axiom of Choice, the algebraically closed field $k$, the plane projective curves $C=V(x_0)$ of degree one and $D=V(x_0{x}_1)$ of degree two, and an arbitrary point $p$ of the shared line. In a chart $x_i=1$ containing $p$ (where $i=1$ or $2$), write $f=x_0\mathrm{/}{x}_i$ and $g=(x_0\mathrm{/}{x}_i)(x_1\mathrm{/}{x}_i)$ for the local equations.

$x_0$ and $x_0{x}_1$ are nonconstant square-free forms, so $C$ and $D$ are plane projective curves; $V(x_0)$ is an irreducible component of both, so $C$ and $D$ share the component $V(x_0)$ Plane projective curves and their components.

At $p\in V(x_0)$ the local ideal is $(f,g)=(x_0\mathrm{/}{x}_i)$, whose nonunit irreducible generator is a local coordinate vanishing on the line; hence $O\mathrm{/}(f,g)$ is not of finite length and the definition records $I_p(C,D)=\mathrm{\infty }$ Local intersection multiplicity of two plane curves, Finite local length exactly when no common local branch, Prime ideals and maximal ideals in a commutative ring.

The Bezout theorem is stated under the no-common-component hypothesis; it computes a finite sum of finite multiplicities Bezout's theorem for plane projective curves. A module of infinite length is not a finite summand Composition series and length of a module.

Given: AC The Axiom of Choice, the imaginary conic $C=V(x_0^2+x_1^2+x_2^2)$ and the line $L=V(x_1)$, first over the ordered field $\mathbb{R}$ The reals form a totally ordered field and then over the algebraically closed field $\mathbb{C}$: the complex field construction $\mathbb{C}=\mathbb{R}[x]\mathrm{/}(x^2+1)$ is a field, every element is uniquely $a+bi$, and every nonzero element has inverse $(a-bi)\mathrm{/}(a^2+b^2)$ and the root theorem Fundamental theorem of algebra: every nonconstant complex polynomial has a complex root give the latter assertion.

The derivatives of the quadratic are $2x_0,2x_1,2x_2$. Any repeated irreducible factor would divide all three, impossible since these coordinates have no common nonconstant factor; thus the quadratic is square-free, as is the linear form $x_1$. Their degrees are two and one. Over $\mathbb{C}$ they define curves in the convention of Plane projective curves and their components. Over $\mathbb{R}$, $C(\mathbb{R})$ and $L(\mathbb{R})$ mean the rational zero sets of these homogeneous equations; the degrees refer to the equations (or their geometric curves after extension to $\mathbb{C}$), and no degree is assigned to the empty real point set.

A point $[a_0:a_1:a_2]\in L$ has $a_1=0$ and satisfies the conic equation exactly when $a_0^2+a_2^2=0$. In an ordered field a nonzero square is positive by trichotomy and closure of the positive cone Ordered field, so over $\mathbb{R}$ the sum of the two squares can be zero only if $a_0=a_2=0$, impossible for a projective point; over $\mathbb{C}$ the solutions are $[1:0:i]$ and $[1:0:-i]$ (or $[i:0:1]$, $[-i:0:1]$) Evaluation and roots of a polynomial in a commutative target ring, An algebraically closed field: every nonconstant polynomial has a root in the field.

Over the algebraically closed field, the Bezout identity gives ${\sum }_p{I}_p(C,L)=2\cdot 1=2$ Bezout's theorem for plane projective curves. At $p=[1:0:i]$ the gradients of $x_0^2+x_1^2+x_2^2$ and $x_1$ are $(2,0,2i)$ and $(0,1,0)$, both nonzero, so both curves are smooth at $p$ with tangent lines $x_0+i{x}_2=0$ and $x_1=0$, which are distinct; hence $I_p(C,L)=1$ Transversal smooth curves meet with multiplicity one, Multiplicity one characterises smooth points with a unique tangent, and symmetrically at the conjugate point Intersection multiplicity dominates the product of multiplicities, with equality for separated tangent cones, Local intersection multiplicity of two plane curves. The two points are conjugate, with non-real coordinates, and are not $\mathbb{R}$-rational: at either point the ratio $x_2\mathrm{/}{x}_0=\pm i$ is non-real, since a real square cannot equal $-1$ $\mathbb{C}=\mathbb{R}[x]\mathrm{/}(x^2+1)$ is a field, every element is uniquely $a+bi$, and every nonzero element has inverse $(a-bi)\mathrm{/}(a^2+b^2)$, Ordered field.

Given: AC The Axiom of Choice, an algebraically closed field $k$, the curve $C=V(y^{2}z-x^3)$ with the affine chart $z=1$, affine equation $f=y^2-x^3$, the cusp $p=(0,0)$, and $O=k[x,y{]}_{(x,y)}$.

The homogeneous equation $y^{2}z-x^3$ is primitive and linear in $z$ over $k[x,y]$, so it is 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. Its dehomogenisation is square-free, so $C$ is a plane projective curve; in the chart $z=1$ the point $p$ has $m_p(C)=2$ and the lowest-degree part of $f$ is $y^2$, so the tangent cone is the double line $V(y)$ Plane projective curves and their components, Multiplicity of a plane curve at a point, Tangent cone and tangent lines at a point.

The local intersection multiplicity is the length of $O\mathrm{/}(f,g)$ for a local equation $g$ of the second curve, and it is computed by $I_p(C,D)={\mathrm{\ell }}_O(O\mathrm{/}(f,g))$ Local intersection multiplicity of two plane curves. Simplicity: $O\mathrm{/}(y,x^3)$ has $k$-basis the classes of $1,x,x^2$, and $O\mathrm{/}(x,y^2)$ has $k$-basis the classes of $1,y$ Each quotient is respectively $k[x{]}_{(x)}\mathrm{/}(x^3)$ or $k[y{]}_{(y)}\mathrm{/}(y^2)$; the descending-power flag has simple residue-$k$ factors, hence lengths three and two Composition series and length of a module, Finite local length exactly when no common local branch.

The point is a double point, so the product bound gives $I_p(C,L)\ge 2$ for every line $L$ through $p$, with equality exactly when the tangent cones are separated; the tangent line $V(y)$ shares its (double) tangent direction with the cusp, so there the value is at least $3$ Intersection multiplicity dominates the product of multiplicities, with equality for separated tangent cones.

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.

Given: AC The Axiom of Choice, affine coordinates $(x,y)$ on ${\mathbf{A}}^2\subset {\mathbf{P}}^2$ with $x=x_1\mathrm{/}{x}_0$, $y=x_2\mathrm{/}{x}_0$, the affine lines $V(x)$ and $V(x-1)$, and their projective closures $C_1=V(x_1)$, $C_2=V(x_1-x_0)$ in ${\mathbf{P}}^2$.

$x_1$ and $x_1-x_0$ are square-free linear forms, so $C_1$ and $C_2$ are plane projective curves of degree one, with no common component; each is a line Plane projective curves and their components.

The affine parts of $C_1$ and $C_2$ are the parallel lines $x=0$ and $x=1$, which are disjoint in ${\mathbf{A}}^2$; hence the count of affine intersection points counted with multiplicity is $0$: $x=0$ and $x=1$ cannot hold simultaneously since $0\mathrm{\ne }1$.

The projective closures meet in the point $[0:0:1]$: solving $x_1=0$ and $x_1-x_0=0$ gives $x_0=x_1=0$ with $x_2\mathrm{\ne }0$, i.e. $[0:0:1]$, a point at infinity of the affine chart $x_0=1$; in the chart $x_2=1$ the local ideal is $(x_1,x_1-x_0)=(x_0,x_1)$, so its quotient is the residue field $k$, of length one projective space points, Local intersection multiplicity of two plane curves.

Bezout for the two projective lines gives ${\sum }_p{I}_p(C_1,C_2)=1\cdot 1=1$, realised at the single point at infinity Bezout's theorem for plane projective curves.

Given: AC The Axiom of Choice, an algebraically closed field $k$ of characteristic not two, the conic $C=V(x_0^2-x_1{x}_2)$, the line $L=V(x_1-x_2)$, and the parametrisation $(s:t)\mapsto [s:t:t]$ of $L$.

Viewed in $k[x_0,x_1][x_2]$, the polynomial $x_0^2-x_1{x}_2$ is primitive (its two nonzero coefficients are coprime) and linear, hence irreducible over $k(x_0,x_1)$ and over $k[x_0,x_1]$ 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. Thus it is a square-free quadratic form, so $C$ is a plane projective curve of degree two with no linear component; $L$ is a line and $L\subseteq ̸C$ Plane projective curves and their components.

Substituting the parametrisation into the defining form gives the binary quadratic $s^2-t^2=(s-t)(s+t)$; its roots are $[s:t]=[1:1]$ and $[-1:1]$, corresponding to $[1:1:1]$ and $[-1:1:1]$, and both roots are simple Intersection with a line is the order of vanishing of the restricted equation.

At each of the two points the gradients of $x_0^2-x_1{x}_2$ and of $x_1-x_2$ are nonzero with distinct tangent directions, so the curves meet transversally and $I_p(C,L)=1$; the line-intersection count confirms the total $2$ Transversal smooth curves meet with multiplicity one, A line meets a degree-d curve in d points counted with multiplicity, Local intersection multiplicity of two plane curves.

Given: AC The Axiom of Choice, an algebraically closed field $k$ of characteristic not three, the Fermat cubic $C=V(x_0^3+x_1^3+x_2^3)$ and the triangle $D=V(x_0{x}_1{x}_2)$.

A repeated irreducible factor of the Fermat form would divide all three derivatives $3x_0^2,3x_1^2,3x_2^2$, which have no common nonconstant factor because $3\mathrm{\ne }0$. Thus it is square-free; the triangle is a product of three distinct prime linear factors. None of those factors divides the Fermat form (setting each coordinate to zero leaves a nonzero binary cubic), so the two degree-three forms have no common factor, so $C,D$ are plane projective curves of degree three with no common component Plane projective curves and their components. In particular their intersection is nonempty and finite Two plane projective curves meet, Curves without a common component meet finitely often, and Bezout gives ${\sum }_p{I}_p=9$ Bezout's theorem for plane projective curves.

A point lies on $D$ exactly when one of its coordinates vanishes. If, say, $x_0=0$, then the cubic equation reads $x_1^3+x_2^3=0$ with $x_1{x}_2\mathrm{\ne }0$, so $[0:a:b]$ with $a^3+b^3=0$; over the algebraically closed field and in characteristic not three the ratio $b\mathrm{/}a$ solves $1+t^3=0$, which splits over $k$ A field is algebraically closed exactly when every nonconstant polynomial splits, equivalently when it has no nontrivial finite extension. No root is repeated: a repeated root would annul the derivative $3t^2$, whereas every root is nonzero and $3\mathrm{\ne }0$. Thus the ratio takes three distinct values, so this coordinate line contributes three points, and the same holds for the other two coordinate lines; their intersection subsets with $C$ are disjoint, because the pairwise intersections of the coordinate lines are the coordinate vertices and no such vertex satisfies the cubic equation, giving exactly nine points Evaluation and roots of a polynomial in a commutative target ring.

At each intersection point $C$ is smooth: not all of $3x_0^2,3x_1^2,3x_2^2$ vanish for a nonzero point, and the characteristic is not three. At a point with exactly one vanishing coordinate $D$ is also smooth, with tangent line the corresponding coordinate line, and the tangent line of $C$ there is not that coordinate line. Hence the two curves meet transversally at each of the nine points and every local multiplicity equals one Transversal smooth curves meet with multiplicity one, Local intersection multiplicity of two plane curves.

Given: AC The Axiom of Choice, an algebraically closed field $k$ of characteristic not two, the conic $C=V(x_0^2-x_1{x}_2)$, the line $L=V(x_1)$, and the point $p=[0:0:1]$.

The quadratic is primitive and linear in $x_2$ over $k[x_0,x_1]$, so Gauss lemma makes it irreducible and square-free Gauss lemma over a UFD, Every finite-variable polynomial ring over a field is a UFD, with prime irreducibles and principal height-one primes. Therefore $C$ is a plane projective curve of degree two and $L$ a line with $L\subseteq ̸C$; $p\in L\cap C$ because $x_1(p)=0$ and $x_0(p{)}^2-x_1(p)x_2(p)=0$ Plane projective curves and their components.

The gradient of $x_0^2-x_1{x}_2$ at $p$ is $(2x_0,-x_2,-x_1)=(0,-1,0)$, nonzero, and the tangent line it defines is $x_1=0$, i.e. $T_{p}C=L$; so $L$ is the tangent line of the conic at $p$ Multiplicity one characterises smooth points with a unique tangent, Tangent cone and tangent lines at a point.

Restricting the defining form to $L$: every point of $L$ has $x_1=0$, and the restriction is the binary form $x_0^2$ in the coordinates $(x_0,x_2)$, with a double root at $[x_0:x_2]=[0:1]$, the point $p$. By the order-of-vanishing formula $I_p(C,L)$ equals that root multiplicity Intersection with a line is the order of vanishing of the restricted equation, Local intersection multiplicity of two plane curves.

The line-intersection count for the degree-two conic and the degree-one line gives ${\sum }_{q\in L\cap C}{I}_q(C,L)=2$ A line meets a degree-d curve in d points counted with multiplicity.

Given: AC The Axiom of Choice, the conic $C=V(x_0^2-x_1{x}_2)$ and its tangent line $L=V(x_1)$ at $p=[0:0:1]$, both over an algebraically closed field of characteristic not two.

$L\cap C=\{p\}$ as a set, and $I_p(C,L)=2$: the restriction of the conic equation to $L$ is $x_0^2$, a double root at $p$ A tangent line meets a conic with multiplicity two at one point.

Bezout for $\mathrm{deg}C=2$, $\mathrm{deg}L=1$ gives ${\sum }_{q\in C\cap L}{I}_q(C,L)=2$ over an algebraically closed field Bezout's theorem for plane projective curves, and every local multiplicity at a point of the intersection is a positive integer Local intersection multiplicity of two plane curves.

Given: AC The Axiom of Choice, an algebraically closed field $k$ of characteristic not three, the Fermat cubic $C=V(F)$ with $F=x_0^3+x_1^3+x_2^3$, the point $p=[1:-1:0]$, and the line $T=V(x_0+x_1)$.

If $F$ had a repeated irreducible factor, it would divide all of $3x_0^2,3x_1^2,3x_2^2$, impossible since $3\mathrm{\ne }0$ and these polynomials have no common nonconstant divisor. Thus $F$ is square-free of degree three, and $p\in C$ because $1^3+(-1{)}^3+0^3=0$; the gradient of $F$ at $p$ is $(3x_0^2,3x_1^2,3x_2^2)=(3,3,0)$, nonzero since the characteristic is not three, so $p$ is a smooth point Plane projective curves and their components, Multiplicity one characterises smooth points with a unique tangent.

The tangent line at the smooth point $p$ is computed from the gradient: $3x_0+3x_1+0\cdot x_2=0$, i.e. $x_0+x_1=0$, so $T=T_{p}C$ Tangent cone and tangent lines at a point, Multiplicity one characterises smooth points with a unique tangent.

For a smooth point $p$ whose tangent line $T$ is not a component of $C$, $I_p(C,T)={\mathrm{ord}}_p(F{\mathrm{\mid }}_T)$, the order of vanishing of the nonzero restricted form, and $p$ is a flex exactly when that order is at least three; an ordinary flex is the case of order exactly three Flexes are contacts of order at least three with the tangent line, Intersection with a smooth curve is a vanishing order.