1.1 A brief introduction to computational algebraic geometry  (Page 20/20)

If we replace $\mathbf{Z}$ with a polynomial ring $\mathbf{C}\left[x\right]$ in one variable, we can do more or less the same thing. Given a non-zero polynomial $f\in \mathbf{C}\left[x\right]$ of degree $n$ , we define

to again be the set of possible remainders upon division by $f$ . Again, for $g,h\in \mathbf{C}\left[x\right]/⟨f⟩$ we can define the sum or product of $g$ and $h$ to be the remainder upon division by $f$ of their sum or product in $\mathbf{C}\left[x\right]$ .

With the help of Gröbner bases, we can do the same thing with polynomials in several variables. Fix a monomial order on $\mathbf{C}[x_1,...,x_k]$ . Given an ideal $I\subseteq \mathbf{C}[x_1,...,x_k]$ we can find a Gröbner basis $G$ for $I$ and then define $\mathbf{C}[x_1,...,x_k]/I$ to be the set of possible possible remainders upon division by $G$ . But what are the possible remainders upon division by $I$ ?

The possible remainders are those polynomials none of whose terms is divisible by a leading term of a polynomial in $G$ , or in other words the finite sums ${\sum }_{\alpha }{a}_{\alpha }{x}^{\alpha }$ where each $x^{\alpha }\notin ⟨LT\left(I\right)⟩$ . It may then be the case that there are infinitely many monomials $x^{\alpha }$ which are not in $I$ : for example, we may take $I=⟨x^{2}y,x{y}^2⟩\subset \mathbf{C}[x,y]$ . Then the monomials that may appear in a remainder upon division by $I$ are $x^n$ and $y^n$ for all $n\ge 0$ , and the monomial $xy$ . The term order used to find the Gröbner basis for $I$ doesn't matter here because $I$ itself is a monomial ideal In this case $\mathbf{C}[x,y]/I$ is an infinite-dimensional vector space over $\mathbf{C}$ , with basis the infinite set $\{x^{\alpha }:x^{\alpha }\notin ⟨LT\left(I\right)⟩\}$ .

It is also possible that there are only finitely many monomials $x^{{\alpha }_1},...,x^{{\alpha }_n}$ not in $LT\left(I\right)$ , so that

is an $n$ dimensional vector space over $\mathbf{C}$ . For example, consider the ideal

We saw on the homework that in lex order, $G=\{x^2+2y^2-3,xy-y^2,y^3-y\}$ is a Gröbner basis for $I$ . There are thus exactly 4 monomials not in $⟨LT\left(I\right)⟩$ , namely 1, $x$ , $y$ , and $y^2$ .

If $X=V\left(I\right)$ is the set of four points where the polynomials of $I$ all vanish, then we can think of $\mathbf{C}[x,y]/I$ as being the polynomial functions on $X$ . In our construction of $\mathbf{C}[x,y]/I$ we're making functions $f$ and $g$ on the whole plane equivalent if they have the same remainder upon division by $I$ , i.e. if $f=g+h$ where $h$ is in $I$ and thus is identically zero on $X$ . Thus in this example, the space of functions on the four points of $X$ is four-dimensional, and so is $\mathbf{C}[x,y]/I$ .

The dimension of $\mathbf{C}[x,y]/I$ isn't always equal to the number of points in $X=V\left(I\right)$ though. For example, consider the ideal

Here, $\{8x+15y^2-16,225y^4-224y^2\}$ is a Gröbner basis for $J$ in lex order, and $Y=V\left(J\right)$ contains only 3 points, but there are 4 monomials not in $⟨LT\left(J\right)⟩$ , namely 1, $y$ , $y^2$ , and $y^3$ .

Nor is this a peculiarity of lex order: if we use graded lex instead, we find that $\{15y^2+8x-16,15x^2-32x+4\}$ is a Gröbner basis and that $\{1,x,y,xy\}$ is a vector space basis for $\mathbf{C}[x,y]/J$ , which still has dimension 4 as a vector space over $\mathbf{C}$ . More generally, the particular monomials not in $⟨LT\left(J\right)⟩$ may depend on the monomial order, but the dimension of $\mathbf{C}[x,y]/J$ as a $\mathbf{C}$ -vector space (i.e. the number of such monomials) does not. This is because $\mathbf{C}[x,y]/J$ can be defined abstractly in terms of congruence classes modulo $J$ , and this agrees with the construction of $\mathbf{C}[x,y]/J$ using each monomial order. This reflects the fact that while the curves $x^2+4y^2=4$ and $4x^2-8x+y^2=0$ only intersect in 3 points, their intersection at $(1,0)$ has “multiplicity 2” because the two curves are tangent there. We've only defined the intersection multiplicity of a curve and a line. For more information on the intersection multiplicity of two curves at a point in general, see section 8.7 of Cox, Little, and O'Shea, where it is defined using resultants.

The Tjurina number of a plane curve singularity

Suppose $C=V\left(f\right)$ , where $f\in \mathbf{C}[x,y]$ is a plane curve with a unique singular point $p$ . We consider the Tjurina ideal

and set $T_f=\mathbf{C}[x,y]/I_f$ and define the Tjurina number $\tau \left(f\right)$ to be the dimension of $T_f$ as a vector space over $\mathbf{C}$ , or in other words, the number of monomials not in $⟨LT\left(I_f\right)⟩$ for a fixed monomial order on $\mathbf{C}[x,y]$ . Since $p$ is the only singular point of $C$ , it is also the only common zero of $f$ , $\frac{\partial f}{\partial x}$ , and $\frac{\partial f}{\partial y}$ .

It turns out that, like the multiplicity of the singularity, the Tjurina number $\tau \left(f\right)$ is an invariant (e.g. if $f$ is affine equivalent to $g$ , then $\tau \left(f\right)=\tau \left(g\right)$ ). Also, it is a new invariant: the Tjurina number is not simply a function of the multiplicity.

Exercises

1. Calculate the Tjurina numbers for the following singular curves (each has a single singularity at the origin). The first few are some of the curves from the beginning of the course.
   1. $f(x,y)=y^2-x^3-x^2$
   2. $f(x,y)=y^2-x^3$
   3. $f(x,y)=x^4+y^4-x^2$
   4. $f(x,y)=x^{2}y+x{y}^2-x^4-y^4$
   5. $f(x,y)=x^{3}y-x{y}^2$
   6. $f(x,y)=y^3-x^4$
   7. $f(x,y)=x^3+y^3-x^{2}y$
   8. $f(x,y)=y^3-x^7+x^{5}y$
2. Calculate the Tjurina number of $f(x,y)=x^p+y^q$ for positive integers $p$ and $q$ .
3. Is there any relation between the Tjurina number and the multiplicity? (e.g. is it possible for the Tjurina number to be large and the multiplicity to be small or vice versa?) Explain.
4. We mentioned an alternative definition for the resultant: given two monic polynomials $f\left(t\right)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$ and $g\left(t\right)=t^m+b_{m-1}{t}^{m-1}+\cdots +b_0$ , by the fundamental theorem of algebra, we can factor them over the complex numbers as
   $$\begin{array}{cc}\hfill f\left(t\right)& =(t-{\alpha }_1)(t-{\alpha }_2)(t-{\alpha }_3)\cdots (t-{\alpha }_{n-1})(t-{\alpha }_n),\hfill \\ \hfill g\left(t\right)& =(t-{\beta }_1)(t-{\beta }_2)\cdots (t-{\beta }_m).\hfill \end{array}$$
   Then if we form the product
   $$R(f,g,t)=\prod _{j=1}^n\prod _{k=1}^m({\alpha }_j-{\beta }_k)$$
   which is clearly a degree $nm$ polynomial in the ${\alpha }_j,{\beta }_k$ . On the other hand, our original definition of the resultant $\text{Res}(f,g,t)$ is a polynomial in the coefficients $a_j,b_k$ , but we can express the coefficients in terms of the roots by
   $$\begin{array}{cc}\hfill a_{n-r}& ={(-1)}^r\sum _{i_1<i_2<\cdots <i_r}{\alpha }_{i_1}{\alpha }_{i_2}\cdots {\alpha }_{i_r}\hfill \\ \hfill b_{m-s}& ={(-1)}^s\sum _{i_1<i_2<\cdots <i_s}{\beta }_{i_1}{\beta }_{i_2}\cdots {\beta }_{i_s}\hfill \end{array}$$
   which allows us the express the original resultant $\text{Res}(f,g,t)$ as a polynomial in the ${\alpha }_j,{\beta }_k$ . Show that $\text{Res}(f,g,t)$ also has degree $nm$ as a polynomial in the ${\alpha }_j,{\beta }_k$ . After this problem, we've shown that $\text{Res}(f,g,t)$ and $R(f,g,t)$ are polynomials of the same degree with the same zero locus. This doesn't quite show that they're equal though, even up to a constant: for example, consider the polynomials $x^{2}y$ and $x{y}^2$ . If, however, we could show that either $\text{Res}(f,g,t)$ or $R(f,g,t)$ is irreducible, and check that $R(f,g,t)=\text{Res}(f,g,t)\ne 0$ for some $f$ and $g$ , that would be enough to show that they are equal.