Proving uniqueness of intersection multiplicities I'm trying to understand a proof from my Algebraic curves course. Additional context is provided Geometric Interpretation of Intersection Multiplicities, but I'll copy the relevant details here. Partial answers (e.g. answering only one of the questions) are very welcome, and I'll upvote them! This looks like a long question, but only because I have provided unnecessary details for the sake of completeness.
(Axioms of intersection multiplicity)

*

*Symmetry: $\mathbf{I}(p, P, Q)=\mathbf{I}(p, Q, P)$

*Detects intersection points: $\mathbf{I}(p, P, Q) \neq 0$ if and only if $P(p)=Q(p)=0$.

*Detects common components: $\mathbf{I}(p, P, Q)=\infty$ if and only if $P$ and $Q$ have a common irreducible factor that vanishes at $p$.

*Transversality: suppose that $P$ and $Q$ both have degree one and that $Q$ is not of the form $\lambda P$ for some $\lambda \in \mathbb{C}^{*} .$ If $p$ is the unique point in $\mathbb{P}_{\mathbb{C}}^{2}$ such that $P(p)=Q(p)=0,$ then $\mathbf{I}(p, P, Q)=1$

*Additivity: if $R$ is another non-zero homogeneous polynomial in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right],$ then $\mathbf{I}(p, P, Q R)=$ $\mathbf{I}(p, P, Q)+\mathbf{I}(p, P, R)$

*Deformation: assume that $\operatorname{deg}(Q) \geq \operatorname{deg}(P),$ and let $R$ be a homogeneous polynomial in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right]$ of degree $\operatorname{deg}(Q)-\operatorname{deg}(P)$ such that $Q+P R \neq 0 .$ Then $\mathbf{I}(p, P, Q)=\mathbf{I}(p, P, Q+P R)$
Moreover, if $P$ and $Q$ are non-constant and have no repeated factors, and we denote by $C$ and $D$ their zero sets in $\mathbb{P}_{\mathbb{C}}^{2},$ then $\mathbf{I}(p, P, Q)=\mathbf{I}(p, C, D)$
Proposition:

There is at most one invariant $\mathbf{I}(p, P, Q)$ that satisfies all the axioms above.

Proof. Since the axioms are coordinate-independent, we may assume that $p=[1,0,0] .$ Indeed, if we could find two distinct invariants $\mathbf{I}(q, P, Q)$ and $\mathbf{I}^{\prime}(q, P, Q)$ at another point $q$ in $\mathbb{P}_{\mathbb{C}}^{2}$ satisfying all the axioms, then we could pick an invertible rank 3 matrix $A$ over $\mathbb{C}$ such that the projective coordinate transformation $\Phi_{A}$ maps $q$ to $p=[0,0,1],$ and produce two distinct invariants $\mathbf{I}(p, P, Q)=\mathbf{I}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$
and $\mathbf{I}^{\prime}(p, P, Q)=\mathbf{I}^{\prime}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$ satisfying all the axioms with respect to the point $p$.
Question 1: We know that $\mathbf{I}^{\prime}(q, P, Q)$ satisfy the axioms, but surely this doesn't imply that $\mathbf{I}^{\prime}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$ satisfies the axioms as well? (Because the polynomials with respect to which the intersection multiplicity is being measured has changed?) Then, we cannot claim that $\mathbf{I}^{\prime}(p, P, Q)$ satisfies the axioms? I suppose $\mathbf{I}(p, P, Q)=\mathbf{I}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$
and $\mathbf{I}^{\prime}(p, P, Q)=\mathbf{I}^{\prime}\left(q, P\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right), Q\left(\left(x_{0}, x_{1}, x_{2}\right) A^{t}\right)\right)$ holds because intersection multiplicities do not change under projective transformations.
So, suppose that we have two invariants $\mathbf{I}(p, \cdot, \cdot)$ and $\mathbf{I}^{\prime}(p, \cdot, \cdot)$ that satisfy all the axioms. We will show that $\mathbf{I}(p, P, Q)=\mathbf{I}^{\prime}(p, P, Q)$ for all non-zero homogeneous polynomials $P$ and $Q$ in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right] .$ Denote by $r$ and $s$ the degrees of the polynomials $P(1,0, x)$ and $Q(1,0, x)$ in $\mathbb{C}[x] .$ By symmetry, we may assume that $r \leq s .$ We will prove the result by induction on $r .$
First, suppose that $r \leq 0$. By the detection of intersection points, the equality $\mathbf{I}(p, P, Q)=\mathbf{I}^{\prime}(p, P, Q)$ is satisfied when $P$ or $Q$ does not vanish at $p$. Thus, by additivity and symmetry, we can reduce to the case where $P(p)=Q(p)=0$ and $P$ is irreducible. By the detection of common components, we can exclude the case where $Q$ is divisible by $P$.
Question 2: If $P$ has repeated factors, i.e. $P_1^2 \mid P$, then $\mathbf{I}(p,P,Q) = \mathbf{I}(p,P_1,Q) + \mathbf{I}(p,P_1,Q) + \dots$ so I don't think we can reduce to the case $\mathbf{I}(p,P,Q)$ where $P$ is irreducible? (As there may be more than one such irreducible $P_1$ in the factorization of the original $P$, in the sense that $P_1^2 \mid P$)
(The rest of the proof is provided for completeness, but is probably irrelevant to the above two questions)
Now $P(1,0, x)$ is a constant polynomial that vanishes at $x=0,$ so that $P(1,0, x)=0 .$ This implies that $P\left(x_{0}, x_{1}, x_{2}\right)$ is divisible by $x_{1} .$ Since we are assuming that $P$ is irreducible, it follows that $P=a x_{1}$ for some $a$ in $\mathbb{C}^{*}$. Now we can use the deformation axiom to remove all the terms in $Q$ that are divisible by $x_{1} ;$ this brings us to the case where $Q$ is a non-zero homogeneous polynomial in $\mathbb{C}\left[x_{0}, x_{2}\right] .$ Factoring $Q$ into linear homogeneous polynomials, we deduce from the additivity and transversality axioms that $\mathbf{I}(p, P, Q)=$ $\mathbf{I}^{\prime}(p, P, Q)=\operatorname{mult}_{0} Q(1,0, x):$ this is the number of linear factors in $Q$ that vanish at the point $p$.
Therefore, we may assume that $s \geq r \geq 1,$ and that $\mathbf{I}\left(p, P^{\prime}, Q^{\prime}\right)=\mathbf{I}^{\prime}\left(p, P^{\prime}, Q^{\prime}\right)$ for all non-zero homogeneous polynomials $P^{\prime}$ and $Q^{\prime}$ in $\mathbb{C}\left[x_{0}, x_{1}, x_{2}\right]$ such that the degree of $P^{\prime}(1,0, x)$ or $Q^{\prime}(1,0, x)$ is strictly less than $r$. We may again assume that $P(p)=Q(p)=0$ and that $P$ is irreducible and does not divide $Q$. Dividing $P$ and $Q$ by suitable elements in $\mathbb{C}^{*},$ we can reduce to the case where $P(1,0, x)$ and $Q(1,0, x)$ are monic. We set
$$
R\left(x_{0}, x_{1}, x_{2}\right)=x_{0}^{\operatorname{deg}(P)+s-r} Q\left(x_{0}, x_{1}, x_{2}\right)-x_{0}^{\operatorname{deg}(Q)} x_{2}^{s-r} P\left(x_{0}, x_{1}, x_{2}\right)
$$
This is a homogeneous polynomial such that the degree of $R(1,0, x)=Q(1,0, x)-x^{s-r} P(1,0, x)$ is at most $s-1 .$ Moreover, $R$ is non-zero because $P$ is irreducible and does not divide $x_{0}$ or $Q .$ By detection of intersection points, additivity and deformation, we have
$$
\mathbf{I}(p, P, Q)=\mathbf{I}\left(p, P, x_{0}^{\operatorname{deg}(P)+s-r} Q\left(x_{0}, x_{1}, x_{2}\right)\right)=\mathbf{I}(p, P, R)
$$
and the same equalities hold for $\mathbf{I}^{\prime}$. Replacing $Q$ by $R$ and repeating the argument, we arrive at a situation where the degree of $Q$ becomes strictly less than $r,$ where we can invoke the induction hypothesis.
 A: Q1: projective transformations do indeed preserve intersection multiplicity. On one hand, this should be geometrically clear: the axioms are coordinate-independent, so why should our specific choice of coordinates on $\Bbb P^2$ matter? If you want to prove it explicitly for yourself, the big idea of the proof is that we can build up any $I(q,P,Q)$ for any $P,Q$ via our axioms by starting with $I(q,\lambda_1,\lambda_2)$ for $\lambda_1,\lambda_2$ linear forms where we know the value of $I$ via axioms 2, 3, and 4 by combining them via the additivity and deformation axioms.
Q2: I don't understand your objection. Additivity says that if $P$ factors as $P_1P_2\ldots P_n$, then $I(q,P,Q)=I(q,P_1,Q)+I(q,P_2,Q)+\ldots+I(q,P_n,Q)$, and it places no restriction on the factorization - there is no requirement that any two $P_i$ be distinct, coprime, or anything. As $k[x,y,z]$ is a UFD, we can always break $P$ down uniquely (up to rearranging and multiplication by constants) in to a product of irreducible $P_i$, apply this result, and then we're dealing with a sum of $I(q,P_i,Q)$ where every $P_i$ is irreducible.
