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Let $k$ be an algebraically closed field. In the discussion of Bezout's theorem, the following definiton of multiplicity has been given:

Definition. Let $R$ be a noetherian graded $k$-algebra and $M$ a finitely generated $R$-module. The multiplicity $\mu_{\mathfrak{p}}$ of $M$ at a prime ideal $\mathfrak{p} \subset R$ is defined as the length $l$ of $M_{\mathfrak{p}} = M \otimes_R R_\mathfrak{p}$ as $R_{\mathfrak{p}}$-module.

I have trouble understanding the definition:

  1. Why do we want our ring and module to be graded? I do not see why this is essential (or 'needed').

  2. I tried to work out the following special case: If $(R, \mathfrak{m})$ is local and noetherian with residue field $K = R/\mathfrak{m}$, then $R \cong R_\mathfrak{p}$ and $M \cong M_\mathfrak{p}$. So, $\mu_{\mathfrak{p}}(M)$ is finite if and only if $\mathfrak{m}^d M = 0$ for some $d$. Indeed, $$M \supset \mathfrak{m}M \supset \mathfrak{m}^2M \supset \cdots \supset \mathfrak{m}^s M$$ gives a composition series and, $M$ has finite length if and only if this composition series is finite, that is $\mathfrak{m}^s M = 0$ for some $s \in \mathbb{N}$. What confuses me is the following: In this case, the lectures notes define $$\mu_\mathfrak{m}(M) := \dim_K(M \otimes_R K),$$ but $M \otimes_R K \cong M/\mathfrak{m}M$, and this is a $K$-vector space of dimension one. So, the multiplicity of $M$ over a local ring is always one? This seems rather unlikely to me. Where is the reasoning error? (See Edit below)

  3. How does this capture the geometric intuition of multiplicity? I am trying to understand how one comes up with this definition, and usually going from geometry backwards seems helpful.

Here is my attempt: For simplicity, let $X = \mathbb{A}^2$ be the affine plane and $R = A(\mathbb{A}^2) \cong k[x,y]$ the coordinate ring. Suppose $p \in X$, corresponding to the maximal ideal $\mathfrak{m}_p \subset k[x,y]$. Then, $R_{\mathfrak{m}_p} \cong \mathcal{O}_{X,p}$. Now, assume we have $f,g \in \mathcal{O}_{X,p}$ which cut out two curves $C_1 = Z(f)$ and $C_2 = Z(g)$ in the plane and $f(p) = g(p) = 0$. That is, $f,g$ belong to the unique maximal ideal $\mathfrak{m}_{X,p}$ of $\mathcal{O}_{X,p}$. At this point, I am not completely sure how to continue and what the '$M$' should be. Does it make sense to set $M: = \mathcal{O}_{X,p}/I$ for $I = (f,g)$? This is a finitely generated $R$-module, and in fact, it has vector space structure over $\mathcal{O}_{X,p}/\mathfrak{m}_{X,p}$ if we assumed that $I$ is $\mathfrak{m}_{X,p}$-primary. But what do we gain by setting $f$ and $g$ to zero? If this is the approach to pursue, why does the dimension of this vector space correspond to the multiplicity in the sense of order of vanishing? This still remains in the dark, from my point of view.

Note: Please assume no scheme theory knowledge in your answer.

Edit: I realized that $\dim_K(M/\mathfrak{m}M)$ is not necessarily one-dimensional, but finite, since $M$ is a finite module. But then the asserted chain of submodules is not a composition series. So, one should possibly argue follows: $M/\mathfrak{m}M$ has finite length and from $$ l_R(M/\mathfrak{m}M) + l_R(\mathfrak{m}M) = l_R(M), $$ we get that $M$ has finite length if and only if $\mathfrak{m}M$ has finite length.

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  1. Because we're working in projective space, whose coordinate ring is (rather importantly) graded.

  2. $M/\mathfrak{m}M$ need not be of dimension one. For instance, if $M$ is such that $M/\mathfrak{m}M$ is a one-dimensional $K$-vector space, then $M^{\oplus 2}$ has $M^{\oplus 2}/\mathfrak{m}M^{\oplus 2}$ a two-dimensional $K$-vector space. (Let me also point out that usually local rings are not graded, at least in the specific algebro-geometric context you're coming from.)

  3. When you want to think about intersections of varieties, we need a way to capture "multiple intersections". Think about the parabola $y=x^2$ and a line $y=a$: if $a\neq 0$, we get two points of intersection. But if $a=0$, we only have one, but it should "count twice", whatever that means - if you think about what happens as $a\to 0$, you can see the two points coming together. It turns out that this notion of multiplicity is the right way to "count twice" here and count multiple times later for more interesting intersections.

As far as your attempt, yes, you're on the right track. You do indeed want to pick $M=\mathcal{O}_{X,p}/I$ here, and setting $f$ and $g$ to zero is how you get the appropriate intersection of $V(f)$ and $V(g)$. It turns out that if you come up with a list of geometric axioms that intersection multiplicity should satisfy, there is a unique answer to the problem of defining intersection multiplicity and the procedure you discuss in your post will give the right answer. I'd recommend trying to prove the equivalence yourself (or at least walking through each of the axioms and checking why it corresponds to something you'd expect geometrically).

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