• On Bezout's theorem Mc coy
• Let $k$ be algebraically closed.
• Let $R \equiv k[x, y, z]$ be ring.
• We wish to detect number of intersections betweeen $f, j \in k[x, y, z]$ counted upto multiplicity.
• For any point $a \in k$, denote $R_a$ to be the localization of $R$ at the multiplicative subset $D_a \equiv \{ f \in R: f(a) \neq \}$( $D$ for does not vanish).
• So $R_a \equiv D_a^{-1}(R)$, which concentrates attention around point $a$.

§ Intersection multiplicity $i[f \cap g](a)$

• Define the intersection multiplicyt of $f, g$ at $a$ by notation $i[f \cap g](a)$.
• Defined as $i[f \cap g](a) \equiv dim_k(R_a/(f, g)_a)$.
• That is, we localize the ring at $a$ and quotient by the ideal generated by $f, g$, and then count the dimension of this space as a $k$ vector space.

§ $f(a) \neq 0$ or $g(a) \neq 0$ implies $i[f \cap g](a) \equiv 0$

• WLOG, suppose $f(a) \neq 0$. Then localization at $a$ makes $f$ into a unit. The ideal $(f, g)_a \equiv R_a$ since the ideal explodes due to the presence of the local unit $f_a$. Thus, $R_a/(f, g)_a \equiv 0$.

§ $f(a) = 0$ and $g(a) = 0$ implies $i[f \cap g](a) \neq 0$.

• If both vanish, then $(f, g)_a$ is a real ideal of $R_a$.

§ Examples

• $x-0$ and $y-0$ at $(0, 0)$ have multiplicity $D_{(0, 0)}^{-1}(k[x, y]/(x, y))$ which is just $k$, which has dimension $1$. So they intersect with dimension $1$.
• $x-1$ and $y-1$ at $(0, 0)$ have multiplicity $D_{(0, 0)}^{-1}(k[x, y]/(x - 1, y - 1))$. The ideal $(x - 1, y - 1)$ blows up because $x - 1 \in D_{(0, 0)}$, and thus the quotient is $0$, making the dimension $0$.
• $x^2-y$ and $x^3-y$ at $(0, 0)$ gives quotient ring $k[x, y]/(x^2-y, x^3-y)$, which is the same as $k[x, y]/(x^2 - y, x^3 - y, 0)$, which is equal to $k[x,y]/(x^2, x^3, y)$, which ix $k[x]/(x^2)$. This is the subring of the form $\{ a + bx : a,b \in k \}$ which has dimension $2$ as a $k$vector space. So this machinery actually manages to captures the degree 2 intersection between $y=x^2$ and $y=x^3$ at $(0, 0)$.

§ Intersection cycle ( $f \cap g$)

• Define $f \cap g \equiv \sum_{a \in \texttt{space}} i[f \cap g](a) \cdot a.$
• It's a generating function with intersection multiplicity as coefficients hanging on the clothesline of points.

§ Intersection number $\#(f \cap g)$

• Given by $\#(f \cap g) \equiv \sum_{a \in \texttt{space}} i[f \cap g](a)$. This is the count of number of intersections.

§ Lemma: $f \cap g = g \cap f$

• Mental note: replace $f \cap g$ with "ideal $(f, g)$" and stuff makes sense.
• Follow immediately since $(f, g) = (g, f)$ and the definition of $i[f \cap g](a) = R_a/(f, g)_a$ which is equal to $R_a/(g, f)_a = i[g \cap f](a)$

§ Lemma: $f \cap (g + fh) = f \cap g$

• $(f, g + fh) \equiv (f, g)$.

§ $f \cap gh \equiv f \cap g + f \cap h$

• Heuristic: if $f(a)$ and $gh(a)$ vanish, then either $f(a), g(a)$ vanish or $f(a), h(a)$ vanish, which can be counted by $f \cap g + f \cap h$

§ Lemma: if $f, g$ are nonconstant and linear then $\#(f \cap g) = 1$.

• Recall that we are stating this within the context of $k[x, y, z]$.
• So $f, g$ are homogeneous linear polynomials $f(x, y, z) = ax + by$, $g(x, y, z) = cx + dy$.
• Sketch: if they have a real solution, then they will meet at unique intersection by linear algebra.
• if they do not have a unique solution, then they are parallel, and will meet at point at infinity which exists because we have access to projective solutions.

§ Lemma: homogeneous polynomial $g \in k[p, q]$ factorizes as $\alpha_0 p^t \prod_{i=1}{n-t}(p - \alpha_i q)$: $\alpha_0 \neq 0$ and $t > 0$

• Key idea: see that if it were $g \in k[p]$, then it would factorize as $p^t \prod_i (p - \alpha_i)$
• To live in $k[p, q]$, convert from $g(p, q) \in k[p, q]$ to $g(p/q, q/q) \in k[(p/q)]$, which is the same as $g(t, 1) \in k[t]$.
• Since we are homogeneous, we know that $g(\lambda p, \lambda q) = \lambda^{deg(g)} g(p, q)$. This lets us make the above transform:

• $g(p/q, q/q) = g(p/q, 1) = (p/q)^k \prod_{i : i +k = n} (p/q - \alpha_i)$.
• $g(p/q, q/q) = g(p/q, 1) = (p/q)^k \prod_{i : i + k = n} (p - \alpha_i q)/q$.
• $g(p/q, q/q) = g(p/q, 1) = p^k/q^k \cdot (1/q^{n-k}) \cdot \prod_{i : i + k = n} (p - \alpha_i q)$.
• $g(p/q, q/q) = g(p/q, 1) = p^k / q^n \prod_{i : i + k = n} (p - \alpha_i q)$.
• $g(p, q) = q^n \cdot g(p/q, 1) = q^n \cdot p^k / q^n \prod_{i : i + k = n} (p - \alpha_i q)$.
• $g(p, q) = q^n \cdot g(p/q, 1) = p^k \prod_{i : i + k = n} (p - \alpha_i q)$.
• This proves the decomposition that $g(p, q) = q^k \prod_i (p - \alpha_i q)$.

§ Lemma: homogeneous polynomial $g \in k[p, q]$ factorizes as $\alpha_0 q^t \prod_{i=1}{n-t}(p - \alpha_i q)$ with $t > 0$.

• This is different from the previous step, since we are pulling out a factor of $q^t$ this time!
• We cannot argue "by symmetry" since the other terms are $(p - \alpha_i q)$. If it really were symmetry, then we should have $(q - \alpha_i p)$ which we don't.
• So this new lemma is in fact DIFFERENT from the old lemma!

• Key idea: see that if it were $g \in k[p]$, then it would factorize as $p^t \prod_i (p - \alpha_i)$
• To live in $k[p, q]$, convert from $g(p, q) \in k[p, q]$ to $g(p/q, q/q) \in k[(p/q)]$, which is the same as $g(t, 1) \in k[t]$.
• Since we are homogeneous, we know that $g(\lambda p, \lambda q) = \lambda^{deg(g)} g(p, q)$. This lets us make the above transform:

• $g(p/q, q/q) = g(p/q, 1) = (p/q)^k \prod_{i : i +k = n} (p/q - \alpha_i)$.
• $g(p/q, q/q) = g(p/q, 1) = (p/q)^k \prod_{i : i + k = n} (p - \alpha_i q)/q$.
• $g(p/q, q/q) = g(p/q, 1) = p^k/q^k \cdot (1/q^{n-k}) \cdot \prod_{i : i + k = n} (p - \alpha_i q)$.
• $g(p/q, q/q) = g(p/q, 1) = p^k / q^n \prod_{i : i + k = n} (p - \alpha_i q)$.
• $g(p, q) = q^n \cdot g(p/q, 1) = q^n \cdot p^k / q^n \prod_{i : i + k = n} (p - \alpha_i q)$.
• $g(p, q) = q^n \cdot g(p/q, 1) = p^k \prod_{i : i + k = n} (p - \alpha_i q)$.
• This proves the decomposition that $g(p, q) = q^k \prod_i (p - \alpha_i q)$.

§ Lemma: $f \in k[x, y, z]$ and $g \in [y, z]$ homogeneous have $def(f) deg(g)$ number of solutions

• This is the base case for an induction on the degree of $x$ in $g$. here, the degree of $x$ in $g$ is zero.
• to compute $i[f(x, y, z) \cap g(y, z)]$, we write it as $i[f(x, y, z) \cap z^k \prod_{i : i + k = n} (y - \alpha_i z)]$
• This becomes $i[f(x, y, z) \cap y^k] + \sum_i i[f(x, y, y) \cap (y - \alpha_i z)]$.
• Intersecting with $y^k$ gives us $k$ times the intersection of $y$ with $f(x, y, z)$, so we have the eqn $i[f(x, y, z) \cap z^k] = k i[f(x, y, z) \cap z]$.
• The full eqn becomes $k i[f(x, y, z) \cap z] + \sum_i i[f(x, y, z) \cap (y - \alpha_i z)]$.

§ Solving for $i[f(x, y, z) \cap z]$

• Let's deal with the first part.
• See that $i[f(x, y, z) \cap z]$ equals $i[f(x, y, 0) \cap z]$, because we want a common intersection, thus can impose $z = 0$ on $f(x, y, z)$.
• We now write $f(x, y, 0) = \mu y^t \prod_j (x - \beta_j y)$.

§ Solving for $i[f(x, y, z) \cap (y - \alpha_i z)]$

• Here, we must impose the equation $y = \alpha_i z$.
• Thus we are solving for $f(x, z, \alpha_i z)$. Once again, we have an equation of two variables, $x$ and $z$.
• Expand $f(x, z, \alpha_i z) = \eta_i z^{l_i} \prod_{j=1}{m - l_i}(x - \gamma_{ij} z)$
• This makes the cycles to be $l_i (z \cap (y - \alpha_i z)) + \sum_j (x - \gamma_{ij} z) \cap (y - \alpha_i z)$.
• The cycle $(z \cap (y - \alpha_i z))$ corresponds to setting $z = 0, y - \alpha_i z = 0$, which sets $y=z=0$. So this is the point $[x:0:0]$.
• The other cycle is $(x - \gamma_{ij} z) \cap (y - \alpha_i z)$, which is solved by $(\gamma_{ij} z : \alpha_i z : z)$.
• In total, we see that we have a solution for every cycle.

§ Inductive step

• Let $deg(f)$ denote total degree of $f$, $deg_x(f)$ denote $x$ degree.
• Let $\deg_x(f) \geq deg_x(g)$.
• We treat $f, g$ as polynomials in a single variable $x$, ie, elements $(k[y, z])[x]$.
• We want to factorize $f$ as $f = Qg + R$. But to do this, we need to enlarge the coefficient ring $k[y, z]$into the coefficient field $k(y, z)$ so the euclidean algorithm can work.
• So we perform long division to get polynomials $Q, R \in (k(y, z)[x]$ such that $f = Qg + R$.
• Since $f, g$ are coprime, we must have $R$ nonzero. Now these $Q, R$ are rational functions since they live in $k(y, z)$.
• Take common denominator of $Q, R$ and call this $h \in k[y, z]$ (ie, it is the polynomial denominator).
• Then $hf = (hQ)g + (hR)$ which is $hf = qg + r$ where $q \equiv hQ \in k[y, z]$ and $r \equiv hR \in k[y, z]$. So we have managed to create polynomials $q, r$ such that $hf = qg + r$.
• Let $c = gcd(g, r)$. $c$ divides $g$ and $r$, thus it divides $qg + r$, ie, it divides $hf$.
• Dividing through by $c$, we get $h'f = qg' + r'$, where $h = h'c$, $g = g'c$, $r = r'c$.
• We assume (can be shown) that these are all homogeneous.
• Furthermore, we started with $gcd(g, f) = 1$. Since $g'$ divides $g$, we have $gcd(g', f) = 1$.
• $c$ cannot divide $f$, since $c = gcd(g, r)$, and $g, f$ cannot share nontrivial common divisors. Thus, $gcd(c, f) = 1$.
• We have some more GCDs to check, at the end of which we write the intersection equation:

• Borcherds Video
• On Bezout's Theorem: Dathan Ault-McCoy