3Affine algebraic curves

We will define an affine plane algebraic curve over a field $k$ as the zero set $V(F)$ , where $F\in k[x, y]$ is an irreducible polynomial. So (at least in this chapter) we do not allow algebraic curves to have more than one component i.e., $V(x y)$ is not considered an algebraic curve.

Algebraic curves are, in a sense we will not have time to explain, algebraic geometric objects of dimension one. The next step on the ladder is algebraic surfaces. Here we could consider $V(f)$ , where $f\in k[x, y, z]$ . An example plot is given below, where

$f = (x^2 + \frac{9}{4} y^2 + z^2 - 1)^3 - x^2 z^3 - \frac{9}{200} y^2 z^3 \tag{3.1}$ and $k = \mathbb{R}$ . This surface could be called the Valentine's surface.

3.1 Intersection of two plane curves

Below you see an extract from Disquisitiones Arithmeticae of relevance in the proof of Lemma 3.2.

For an arbitrary integral domain $R$ , a greatest common divisor for $a, b\in R$ is an element $(a, b)\in R$ satisfying

$(a, b) \mid a$ and $(a, b) \mid b$
If $d\mid a$ and $d\mid b$ , then $d\mid (a, b)$ for $d\in R$ .

It follows from these two conditions that a greatest common divisor is uniquely determined up to multiplication by a unit. Furthermore, a greatest common divisor satisfies $(c a, c b) = c (a, b)$ and $((a, b), c) = (a, (b, c))$ for $a, b, c\in R$ . The latter condition brings meaning to a greatest common divisor $(a_0, a_1, \dots, a_n)$ of finitely many elements $a_0, a_1, \dots, a_n\in R$ . A polynomial

$p(x) = a_0 + a_1 x + \dots + a_n x^n\in R[x]$ is called primitive if $(a_0, a_1, \dots, a_n) = 1$ .

We assume now that $R$ is a UFD. This ensures the existence of $(a, b)$ for $a, b\in R$ . Let $K$ denote the fraction field of $R$ . Then any polynomial $f(x)\in K[x]$ can be written as

$f(x) = \frac{a}{s}\, p(x),$ where $p(x)\in R[x]$ is a primitive polynomial and $\frac{a}{s}\in K$ . The following result is (probably) hidden in the above excerpt from Disquisitiones.

Suppose that

$\frac{a_1}{s_1} p_1(x) = \frac{a_2}{s_2} p_2(x),$ where $p_1(x), p_2(x)\in R[x]$ are primitive polynomials. Then there exists a unit $u\in R^*$ , such that

$u \frac{a_1}{s_1} = \frac{a_2}{s_2}.$ One may also prove that a product of two primitive polynomials is a primitive polynomial. Now one can conclude that if $(f, g) = 1$ in $R[x]$ , then $(f, g) = 1$ in $K[x]$ .

If $f(x, y), g(x, y)\in k[x, y]$ are polynomials with no common prime factors, then there exists $\lambda(x, y), \mu(x, y)\in k[x, y]$ with

$\lambda(x, y) f(x, y) + \mu(x, y) g(x, y) = h(x)\in k[x]\setminus\{0\}.$

Let $R = k[x]$ and $K = k(x)$ . If $f, g\in k[x, y] = R[y]$ have no common factors in $R[y]$ , they also have no common factors in $k(x)[y] = K[y]$ , which is a PID. Therefore

$\lambda_1 f + \lambda_2 g = 1$ for suitable $\lambda_1, \lambda_2\in k(x)[y]$ . Multiplying by a suitable polynomial $h(x)\in R$ , we can clear the denominators on the left hand side and get the result.

Get inspired by the proof of Lemma 3.2 and explain how the extended euclidean algorithm applied to the polynomials

$y^3 - x^2\qquad\text{and}\qquad y^2 - x^3 + x$ in $k(x)[y]$ (which is a euclidean ring!) helps in finding a polynomial $h \in k[x]$ , such that

$h \in (y^3 - x^2, y^2 - x^3 + x).$ Can we be sure that $k[x] \cap (y^3 - x^2, y^2 - x^3 + x) = (h)$ ?

Macaulay2 code - viewer discretion advised

Use Lemma 3.2 to show that $V(f)\cap V(g)$ is a finite set, when $f$ and $g$ have no common factors.

We consider two polynomials $f, g\in k[x, y]$ with no common prime factors. As we saw in Lemma 3.2 this implies that $V(f, g)\in k^2$ is finite. It could be empty. The distant goal is to prove that in the right setting $V(f)$ and $V(g)$ intersect in preciselt $\deg(f) \deg(g)$ points.

Let $f, g\in k[x, y]$ be polynomials with no common factors, $m_1 = \deg(f)$ and $m_2 = \deg(g)$ . Then

$\dim_k k[x, y]/(f, g) \leq m_1 m_2.$

Let $R = k[x, y]$ and $I = (f, g)$ . We let $R_{\leq d}$ denote the finite dimensional vector space of polynomials of degree $\leq d$ . in $R$ . Check that

$\dim(R_{\leq d}) = \frac{(d+1)(d+2)}{2}.$ Define for $d\geq m_1, m_2$ ,

$V_d = R_{\leq d-m_1} f + R_{\leq d-m_2} g \subset I\cap R_{\leq d}. \tag{3.2}$ Notice that

$R_{\leq d - m_1} f \cap R_{\leq d-m_2} g = R_{\leq d-m_1 - m_2} f g$ for $d \geq m_1 + m_2$ . This is because $f$ and $g$ are relatively prime (no common factors). Now consider the subspaces $U = R_{\leq d-m_1} f$ and $V = R_{\leq d-m_2} g$ and use the isomorphism $U/(U\cap V) \cong (U+V)/V$ :

to conclude that $\dim_k (R_{\leq d}/V_d) = \dim_k(R_{\leq d}) - \dim_k V_d$ is equal to $(d+1)(d+2)/2$ minus the dimension of $U + V$ or $\dim(U) + \dim(V) - \dim(U\cap V)$ , which is

$\frac{(d-m_1+1)(d-m_1+2)}{2} + \frac{(d-m_2+1)(d-m_2+2)}{2} - \frac{(d-m_1-m_2 + 1)(d - m_1 - m_2 + 2)}{2}.$ Running this through a computer algebra system you magically end up with $m_1 m_2$ .

Since we have a surjective homomorphism

$R_{\leq d} /V_d \rightarrow R_{\leq d} / I\cap R_{\leq d},$ it follows that $\dim_k R/I \leq m_1 m_2$ .

In fact, suppose that $f_1, \dots, f_m\in R$ are polynomials so that $[f_1], \dots, [f_m]$ are linearly independent in $R/I$ . Choosing $d \geq \max\left(\deg(f_1), \dots, \deg(f_m)\right)$ , the equivalence classes $[f_1], \dots, [f_m]$ now in $R_{\leq d}/I\cap R_{\leq d}$ must be linearly independent, because we cannot find $\lambda_1, \dots, \lambda_m\in K$

$\lambda_1 f_1 + \cdots + \lambda_m f_m \in I\cap R_{\leq d} \subset I$ unless $\lambda_1 = \cdots = \lambda_m = 0$ .

Show that the inclusion in (3.2) in the proof of Proposition 3.5 can be strict.

Consider $f = x-1$ and $g = y-x^2$ .

Below we generate two random cubics in two variables over $\mathbb{Q}$ and output the dimenson (or rather degree of $(f, g)$ ) of $\mathbb{Q}[x, y]/(f, g)$ .

The example given above are random cubics. Find examples of $f$ and $g$ in the Macaulay2 window above so that the degree becomes strictly less than $\deg(f) \deg(g)$ .

Given a polynomial $p\in k[x, y]$ , there is a very natural way of associating a homogeneous polynomial $p^*\in k[x, y, z]$ by

$p^* = z^{\deg(f)} p\left(\frac{x}{z}, \frac{y}{z}\right).$

Suppose that $p(x, y) = y^2 - x^3 + x$ , then

$p^*(x, y, z) = z^3 \left(\left(\frac{y}{z}\right)^2 - \left(\frac{x}{z}\right)^3 + \left(\frac{x}{z}\right)\right) = y^2 z - x^3 + x z^2.$

Similarly having a homogeneous polynomial $P\in k[x,y , z]$ we can dehomogenize by

$P_*(x, y) = P(x, y, 1).$

For $P(x, y, z) = z^3 + x z^2 + y^2 z$ ,

$P_*(x, y) = 1 + x + y^2.$ Notice here that

$P_*(x, y)^* = z^2 + x z + y^2 \neq P(x, y, z).$ However,

$P = z (P_*)^*.$

Similar to the situation in $k[x, y]$ , we can also homogenize a polynomial $f\in k[x_1, x_2, x_3]$ and obtain a homogeneous polynomial $f^*\in k[x_0, x_1, x_2, x_3]$ . Explain how this is done.

For an ideal $I\subset k[x_1, x_2, x_3]$ , we define

$I^* = \left(f^* \mid f\in I\right)$ i.e., as the ideal in $k[x_0, x_1, x_2, x_3]$ generated by $f^*$ for $f\in I$ .

Is it true that

$(x_2 - x_1^2, x_3 - x_1^3)^* = (x_0 x_2 - x_1^2, x_0^2 x_3 - x_1^3)?$

Suppose that $f, g\in k[x, y]$ have no common factors and that $V(f)$ and $V(g)$ do not intersect at infinity i.e.

$V(f^*) \cap V(g^*) \cap V(z) = \emptyset \quad\text{in } \mathbb{P}^2.$ Then

$\dim_k k[x, y]/(f, g) = m_1 m_2,$ where $m_1 = \deg(f)$ and $m_2 = \deg(g)$ . Here we assume $k = \overline{k}$ .

It is not too difficult to compute the points at infinity for a polynomial. Write the highest degree terms in $f$ as $f^h$ . Here $f^h$ is a homogeneous polynomial of degree $m = \deg(f)$ and $f^h$ survives in $f^*$ without any multiplication of a power of $z$ . As $k$ is algebraically closed, we may write

$f^h = (a_1 x + b_1 y) \cdots (a_m x + b_m y),$ for $a_1, b_1, \dots, a_m, b_m\in k$ , since

$\left(\frac{1}{y}\right)^m f^h(x, y) = f^h\left(\frac{x}{y}, 1\right) = \left(\frac{x}{y} - \alpha_1\right) \cdots \left(\frac{x}{y} - \alpha_d\right),$ for $d\leq m$ and $\alpha_1, \dots, \alpha_d\in k$ because $k$ is algebraically closed.

The points at infinity of $f$ are

$(-b_1, a_1, 0), \dots, (-b_m, a_m, 0).$ If $f$ and $g$ do not have any common points at infinity, then $f^h$ and $g^h$ are relatively prime polynomials.

This implies that

$R_{d-m_1} f + R_{d - m_2} g = I \cap R_d \tag{3.3}$ and therefore that

$\dim R/I = \dim R_{\leq d} / V_d = m_1 m_2$ in the notation used in the proof of Proposition 3.5.

Let us explain (3.3). Suppose that

$F \in \left(I\cap R_{\leq d}\right) \setminus \left( R_{d-m_1} f + R_{d-m_2} g\right).$ Then $F = \lambda_1 f + \lambda_2 g$ with $\deg(F)\leq d$ . Let us assume that $\lambda_1, \lambda_2\in k[x, y]$ are picked so that $\deg(\lambda_1)$ is minimal. A moment of reflection will convince you that

$\lambda_1^h f^h + \lambda_2^h g^h = 0,$ since we must have $\lambda_1\not\in R_{\leq d-m_1}$ and $\lambda_2\not\in R_{\leq d - m_2}$ . As $f^h$ and $g^h$ are relatively prime, $g^h \mid \lambda_1^h$ . Therefore there exists a polynomial $q\in k[x, y]$ , such that

$\deg(\lambda_1 - q g) < \deg(\lambda_1).$ Now focus on

$F = \lambda_1 f + \lambda_2 g = (\lambda_1 - q g) f + (\lambda_2 + q f) g.$

3.2 Rational functions and local rings

The quotient ring

$\Gamma(F) := R/(F)$ is an integral domain and we denote its field of fractions by $k(F)$ . The elements in $k(F)$ are called rational functions on $V(F)$ .

Notice that

$\frac{[f]}{[g]} = \frac{[r]}{[s]} \iff [f] [s] = [g] [r] \iff f s - g r \in (F) \iff F \mid f s - g r. \tag{3.4}$

A rational function $r\in k(F)$ is defined at $P\in V(F)$ if there exists $f, g\in k[x, y]$ , such that

$r = \frac{[f]}{[g]},$ and $g(P)\neq 0$ . The pole set of $r$ is the set of points on $V(F)$ , where $r$ is not defined.

The pole set of $r\in k(F)$ is the zero set of the ideal $J_r$ , where

$J_r = \{G\in k[x, y] \mid [G] r\in \Gamma(F)\}.$ The pole set is finite. Also, if $r$ is defined at $P$ , then its value is well defined by

$r(P) = \frac{f(P)}{g(P)},$ where $r = [f]/[g]$ .

Observe that $[g]$ is a denominator for $r$ if and only if $[g]r \in \Gamma(F)$ .

If $v\in V(J_r)$ , then $v$ has to be a pole for $r$ . If not, then $r = [f]/[g]$ with $g(v)\neq 0$ , but $g\in J_r$ . If $v\not\in V(J_r)$ , we may find $f, g\in k[x, y]$ , such that $[g] r = [f]$ and $g(v)\neq 0$ . Therefore $v$ is not a pole of $r$ . We are assuming that $F$ is an irreducible polynomial. If $r = [f]/[g]$ and $[g]\neq 0$ , then $F\nmid g$ . Therefore $V(F)\cap V(g)$ is finite by Exercise 3.4.

If $r = [f]/[g_1] = [f_2]/[g_2]$ with $g_1(P)\neq 0$ and $g_2(P)\neq 0$ , then $f(P) g_2(P) = g_1(P) f_2(P),$ since $F(P)=0$ . Therefore

$\frac{f(P)}{g_1(P)} = \frac{f_2(P)}{g_2(P)}.$

Consider $F = x^2 + y^2 - 1\in \mathbb{R}[x]$ from Example 1.9. In $\mathbb{R}(F)$ we have

$\frac{[y-1]}{[x]} = -\frac{[x]}{[y+1]}.$ This follows from (3.4).

Let $F = y^2 - x^3 + x$ . What is the value of the rational function

$r = \frac{[x]}{[y]}$ evaluated at $(0, 0)\in V(F)$ ? What about $\displaystyle\frac{[y]}{[x]}$ ?

Suppose that $R$ is an arbitrary commutative ring with ideals $I, J\subseteq R$ . Then the colon ideal is defined by

$I : J = \{r\in R \mid r J \subseteq I\}.$

Suppose that you have a rational function $r\in K(F)$ . Then its pole set is the zero set of the ideal $J_r$ from Proposition 3.13. If $r$ is represented by $[f]/[g]$ , then

$J_r = (F, g):(f).$ Here are some examples in Macaulay2 related to the computation of $J_r$ using computation of the socalled colon ideal.

3.2.1 The local ring at a point

A local ring is a commutative ring with precisely one maximal ideal.

Define

$\mathbb{C}[z]_0 = \left\{\frac{f(z)}{g(z)}\, \middle|\, f(z), g(z)\in \mathbb{C}[z], \,z \nmid g(z)\right\}$ and the ring of power series

$\mathbb{C}\llbracket z\rrbracket = \left\{a_0 + a_1 z + a_2 z^2 + \cdots\, \middle|\, a_i\in \mathbb{C}\right\}$ over $\mathbb{C}$ .

Prove that $\mathbb{C}[z]_0$ is a subring of $\mathbb{C}\llbracket z \rrbracket$ and that they both are local rings. Show that
$e^z = 1 + z + \frac{z^2}{2!} + \cdots \in \mathbb{C}\llbracket z \rrbracket \setminus \mathbb{C}[z]_0.$
Show that the field of fractions $\mathbb{C}((z))$ of $\mathbb{C}\llbracket z\rrbracket$ is the Laurent series ring
$\left\{a_m z^m + a_{m+1} z^{m+1} + \cdots\, \middle|\, a_i\in \mathbb{C}, \, m\in \mathbb{Z}\right\}$ over $\mathbb{C}$ .
Show that $\mathbb{C}((z))$ is not an algebraically closed field.
One can prove that
$\mathbb{C}\langle\langle z\rangle\rangle = \bigcup_{n\geq 1} \mathbb{C}((z^{1/n}))$ is an algebraically closed field (Puiseux series) containing $\mathbb{C}((z))$ . Is $K\langle\langle x \rangle\rangle$ algebraically closed if $K$ is an algebraically closed field of characteristic $p>0$ ?

Let $P\in V(F)$ . Then we let $\mathcal{O}_P(F)$ denote the subset of $k(F)$ given by the rational functions defined at $P$ . This is a subring of $k(F)$ and contains $\Gamma(F)$ as a subring:

$\Gamma(F)\subset \mathcal{O}_P(F) \subset k(F).$

Prove that $\mathcal{O}_P(F)$ is a local ring and that its unique maximal ideal is formed by the rational functions $r\in \mathcal{O}_P(F)$ with $r(P) = 0$ .

Suppose that $P=(0,0)$ . Show that the maximal ideal in $\mathcal{O}_P(F)$ is given by $([x], [y]) \mathcal{O}_P(F)$ i.e. it is the ideal

$\left(\frac{[x]}{[1]}, \frac{[y]}{[1]}\right) \subset \mathcal{O}_P(F).$

Prove that

$\Gamma(F) = \bigcap_{P\in V(F)} \mathcal{O}_P(F).$

3.3 Dimension in local rings

Consider $I = (f, g) \subset R = K[x, y]$ , where $K$ is any field and $f$ and $g$ have no common factors in $R$ . Then we know by Proposition 3.5 that

$\dim_K R/I \leq m_1 m_2,$ where $m_1 = \deg(f)$ and $m_2 = \deg(g)$ .

We now wish to zoom in one of the finitely many points in $P\in V(I)$ . We do this by focusing on the local ring

$R_P = \left\{\frac{f}{g}\, \middle|\, g(P) \neq 0\right\} \subset K(x, y).$

Define the ideal

$I_P = R_P\, f + R_P\, g \subset R_P.$

To get grounded, consider first the one variable case $R = \mathbb{C}[x]$ and for example $I = (x^2 - 5 x + 6)\subset R$ along with the point $P = 3\in \mathbb{C}$ . Then

$R_P = \left\{\frac{f(x)}{g(x)} \middle| g(3) \neq 0\right\}.$ Show that $I_P = (x-3)_P$ and thatCheck that

$\frac{\lambda}{g} \equiv \frac{\lambda}{g(3)} \pmod{x-3},$ where $\lambda\in \mathbb{C}$ . $\dim_\mathbb{C} R_P/I_P = 1$ , whereas $\dim_\mathbb{C} R/I = 2$ .

Prove that $I_P = R_P$ if $P\not\in V(I)$ .

3.4 The crucial isomorphism

Suppose that $V(I) = \{P_1, \dots, P_r\}$ with $I = (f, g) \subset K[x, y]$ , where $K$ is algebraically closed. For each $P_i$ we have a well defined canonical homomorphism

$\varphi_i: R/I\rightarrow R_{P_i}/I_{P_i}$ given by $\varphi([f]) = [f/1]$ . Our objective is to show that

$\varphi = (\varphi_1, \dots, \varphi_r): R/I\rightarrow \prod_{j=1}^r R_{P_j}/I_{P_j} \tag{3.5}$ is an isomorphism $K$ -algebras and therefore that

$\dim_K R/I = \sum_{j=1}^r \dim_K R_{P_j}/I_{P_j}.$

3.4.1 Constructing orthogonal idempotents in $R/I$

We are proving that $\varphi$ is an isomorphism by constructing socalled orthogonal idempotents in $R/I$ i.e., elements $e_1, \dots, e_r$ satisfying

$e_i^2 = e_i,\qquad e_i e_j = 0\qquad\text{and}\qquad 1 = e_1 + \cdots + e_r \tag{3.6}$ for $i = 1, \dots, r$ and $j\neq i$ . These correspond to the natural basis elements in the product ring on the right hand side of (3.5). A first approach is to work with polynomials $F_1, \dots, F_r$ , such that $F_i(P_j) = \delta_{ij}$ . We know these polynomials from the proof of Corollary 2.39.

Does it work to put $e_i = [F_i]$ ? Certainly $F_i^2 - F_i$ vanishes on $\{P_1, \dots, P_r\}$ , but this does not necessarily imply that $F_i^2 - F_i\in I$ . We only have $F_i^2 - F_i \in \operatorname{rad}(I)$ . We need to adjust $F_i$ to make it work. This is done using Theorem 2.38 (the Hilbert Nullstellensatz).

Notice that $I(\{P_1, \dots, P_r\}) = M_1 \cap \cdots \cap M_r$ , where $M_i$ denotes the (maximal) ideal of polynomials vanishing on $P_i$ . For example, $M_i = (x - a_i, y- b_i)$ if $P_i = (a_i, b_i)$ . By Theorem 2.38,

$(M_1\cap \cdots \cap M_r)^d \subset I \tag{3.7}$

for sufficiently big $d$ . We keep this $d$ fixed for the rest of the proof.

The elements that make (3.6) work are $e_i = [E_i]$ , where

$E_i = 1 - (1 - F_i^d)^d.$ Notice that $E_i = F_i^d D_i$ for a suitable polynomial $D_i$ by the binomial expansion of $(1 - F_i^d)^d$ . Therefore $E_i \in M_j^d$ for $j\neq i$ and

$1 - (E_1 + \cdots + E_n) = (1 - E_j) - (E_1 + \cdots + \xcancel{E_j} + \cdots + E_n) \in M_i^d$ for every $i = 1, \dots, r$ . This implies that $e_1 + \cdots + e_r = 1$ by Lemma 3.25. We continue to use this lemma below.

Also if $i\neq j$ , then $E_i E_j\in M_h^d$ for every $h = 1, \dots, r$ and finally

$E_i^2 - E_i = 1 - 2(1-F_i^d)^d + (1 - F_i^d)^{2d} - (1 - (1 - F_i^d)^d) = (1 - F_i^d)^{2d} - (1 - F_i^d)^d$ So we see that $E_i^2 - E_i \in M_i^d$ and since

$(1 - F_i^d)^{2d} - (1 - F_i^d)^d = (F_i^d D_i)^2 - F_i^d D_i$ it also belongs to $M_j^d$ for $j\neq i$ .

3.4.2 Comaximal ideals

$M_1^d \cap \cdots \cap M_r^d \subset I.$

Is it true in general that

$(I_1 \cap \cdots \cap I_r)^d = I_1^d \cap \cdots \cap I_r^d$ for ideals $I_1, \dots I_r$ in a commutative ring $R$ ?

Two ideals $I, J\subset R$ in a commutative ring $R$ are called comaximal if $I + J = R$ .

Suppose that $I$ and $J$ are comaximal ideals. Prove that $I J = I\cap J$ and that $I^m$ and $J^n$ are comaximal for every $m, n\in \mathbb{N}$ .

We know that $(M_1\cap \cdots \cap M_r)^d \subset I$ from (3.7). How do we get the much needed result in Lemma 3.25?

3.4.3 The image of the orthogonal idempotents

We have constructed orthogonal idempotents $e_1, \dots, e_r\in R/I$ . They will do their magic when $\varphi$ is applied to them. First, notice that $\varphi_i(e_i)$ is a unit in $R_{P_i}$ , since $E_i(P_i)\neq 0$ . This gives

$0 = \varphi_i(e_i e_j) = \varphi_i(e_i)\varphi_i(e_j) \implies \varphi_i(e_j) = 0$ for $i\neq j$ . Finally this implies

$1 = \varphi_i(e_1 + \cdots + e_r) = \varphi_i(e_i).$

We realize the canonical basis vectors of the right hand side of (3.5) as images of $e_1, \dots, e_r$ . But we need more.

3.4.4 Fractions are polynomials!

If $G\in k[x, y]$ and $G(P_i)\neq 0$ , then there exists $t\in R/I$ , such that $t g = e_i$ , where $g = [G]$ .

Assume that $G(P_i) = 1$ and put $H = 1-G$ . Then

$(1-H)(E_i + H E_i + \cdots + H^{d-1} E_i) = E_i - H^d E_i,$ where $H\in M_i$ , so that $H^d E_i\in I$ . Therefore $g ( e_i + h e_i + \cdots + h^{d-1} e_i) = e_i$ with $h = [H]$ . Notice that we have once again used Lemma 3.25. The geometric series rules. It rules!

Notice that Lemma 3.30 actually says that

$t = g^{-1}$ in $R_{P_i}/I_{P_i}$ (since $\varphi_i(e_i) = 1$ ).

3.4.5 Injectivity

Suppose that $\varphi(f) = 0$ and $f = [F]$ . This means that $F\in I_{P_j}$ for every $j = 1, \dots, r$ . Therefore there exists $G_j$ with $G_j(P_j)\neq 0$ and $F G_j\in I$ for $j = 1, \dots, r$ . Now use the notation $g_j = [G_j]$ to conclude

$f = f(e_1 + \cdots + e_r) = f (g_ 1 t_1 + \cdots + g_r t_r) = f g_1 t_1 + \cdots + f g_r t_r = 0,$ since $F G_j\in I$ for $j = 1, \dots, r$ . We have used Lemma 3.30 $r$ times.

3.4.6 Surjectivity

For a fraction $a/s\in R_{P_i}/I_{P_i}$ we have (using Lemma 3.30) with $t s = e_i$ for $t\in R/I$ , that

$\varphi_i(t) \varphi(s) = 1$ and therefore that $t = s^{-1}$ in $R_{P_i}/I_{P_i}$ . We get $\varphi_i(a t) = a/s$ . If presented with

$q = \left(\frac{a_1}{s_1}, \dots, \frac{a_r}{s_r}\right)\in \prod_{j=1}^r R_{P_j}/I_{P_j},$ we see that

$\varphi(a_1 t_1e_1 + \cdots + a_r t_r e_r) = q,$ where $t_i s_i = e_i$ for $i = 1, \dots, r$ .

Consider the two plane curves $V(f)$ and $V(g)$ in $\mathbb{C}^2$ , where

$\begin{aligned} f &= y^2 - x^3 + x\\ g &= y^3 - x^2. \end{aligned}$ Show that $f$ and $g$ do not intersect at $\infty$ . Now zoom in on $P = (0, 0)\in V(f, g)$ . What is

$\dim_\mathbb{C} R_P/I_P,$ where $I = (f, g)\subset R = \mathbb{C}[x, y]$ ? Compare the dimension with the output of