I would like to find reference for the following statement. I need it only in the particular case when $A=\mathcal{O}_{(\mathbb{C}^n, 0)}$ is the local algebra of holomorphic germs $(\mathbb{C}^n, 0) \to (\mathbb{C}, 0)$, but I'm also interested in the general case. (The $n=2$ case - intersection multiplicity of plane curves - can be found e.g. in Fulton.)

Let $A$ be a commutative algebra (over $\mathbb{R}$ or $\mathbb{C}$), $I$ is the ideal generated by the elements $(a_i)_{i=1}^n$. Assume that $A$ is UFD. The irreducible decomposition of $a_i$ is $$ a_i =\prod_{j=1}^{m_i} p_{ij},$$ where $p_{ij}$ are different irreducible elements in $A$. Then $$ \dim \frac{A}{I}=\sum_{\mbox{all choice functions } f,} \dim \frac{A}{I(p_{1, f(1)}, p_{2, f(2)}, \dots, p_{n, f(n)})} .$$

Example: $A=\mathcal{O}_{(\mathbb{C}^2, 0)}$, $n=2$. In this case $\dim \frac{A}{I(a_1, a_2)}$ is the intersection multiplicity of the plane curves $C_i=\{a_i=0\}$.

  • $\begingroup$ Please use the Contact Us form to have your accounts merged, to regain full control over your posts. $\endgroup$
    – Glorfindel
    Oct 3, 2022 at 12:46

2 Answers 2


This has a lot to do with Atiyah--Macdonald Proposition 1.10 (the Chinese Remainder Theorem).

  • 1
    $\begingroup$ Welcome on MathOverflow! -- Can you perhaps elaborate, to turn this into a complete answer? $\endgroup$
    – Stefan Kohl
    Apr 3, 2022 at 15:30
  • $\begingroup$ Well, let us think. Under what conditions can we claim that $A/(pq,r)\simeq A/(p,r)\oplus A/(q,r)$, or, equivalently, $\bar A/(\bar p\bar q)\simeq \bar A/(\bar p)\oplus \bar A/(\bar q)$, where $\bar p$ and $\bar q$ are the images of $p$ and $q$ in $\bar A=A/(r)$ ? It is sufficient that $(p,q,r)=(1)$. $\endgroup$ Apr 3, 2022 at 16:13
  • $\begingroup$ ...but $(p,q,r)=(1)$ does not necessarily hold in our case. Luckily, we are not aiming for direct sum, only additivity of dimension. Thus $((\bar p)\cap(\bar q))/(\bar p\bar q)\simeq \bar A/(\bar p, \bar q)$ would suffice. $\endgroup$ Apr 4, 2022 at 4:17
  • $\begingroup$ Thank you for your answer! I also associated to Chinese reminder theorem, but I was unsure how can we use it. I think we have $(p, q, r)=1$, since I assumed that the decompositions contains different irreducible elemnts. Hence, do I have a direct sum at each step? $\endgroup$ Apr 4, 2022 at 15:11
  • $\begingroup$ No, consider $p=x$, $q=y$, $r=x^2+y^2+x$ in $A=\mathbb C[x,y]$. $\endgroup$ Apr 4, 2022 at 18:03

I provide here a proof for the case $A=\mathcal{O}_{\mathbb{C}^n, 0}$ is the local algebra of holomorphic germs of functions. Surprisingly I haven't found the general statement in the classical books, except the $n=2$ case (intersection multiplicity of plane curves), which can be found e.g. in pg. 38. Fulton. For the proof below thanks to A. Némethi, A. Sándor, P. Frenkel. I'm curious in which generality is it true for other algebras.

Of course we have to assume that the dimensions are finite. The statement follows from the following lemma by induction:

Lemma: Let $a_1=bc$ a nontrivial decomposition, assume that $c$ is irreducible. Then we have an exact sequence $$ 0 \to \frac{A}{(b, a_2, \dots, a_n)} \to \frac{A}{(bc, a_2, \dots, a_n)} \to \frac{A}{(c, a_2, \dots, a_n)} \to 0. $$

Proof of lemma: Of course we have an exact sequence

$$ 0 \to \frac{(c, a_2, \dots, a_n)}{(bc, a_2, \dots, a_n)} \to \frac{A}{(bc, a_2, \dots, a_n)} \to \frac{A}{(c, a_2, \dots, a_n)} \to 0. $$

We will be ready if we show that the multiplication by $c$ induces an isomorphism $$ \mu: \frac{A}{(b, a_2, a_3 \dots, a_n)} \to \frac{(c, a_2, a_3 \dots, a_n)}{(bc, a_2, a_3 \dots, a_n)}. $$ Clearly $\mu$ is well defined and surjective. For the injectivity take $x \in A$ such that $\mu[x]=0$, we have to show that $[x]=0$. That is, $cx \in (bc, a_2, a_3, \dots, a_n)$ implies $x \in (b, a_2, a_3, \dots, a_n)$. That is, we assume that $$cx=l_1 bc+ l_2 a_2 + l_3 a_3 + \dots + l_n a_n$$ holds with some $l_i \in A$, and we want to conclude that there are $l_i' \in A$ such that $$ x=l'_1 b+l_2' a_2 +l_3' a_3 + \dots + l_n' a_n.$$ From the condition we obtain $$ c(x-l_1 b)=l_2 a_2 + l_3 a_3 + \dots + l_n a_n.$$ Here we see why the $n=2$ case is special: since $a_2$ is not divisible by $c$, $l_2$ must be divisible, hence we can divide both sides by $c$ and obtain the conclusion with $l_1'=l_1$ and $l_2'=l_2/c$.

For arbitrary $n$ we need some singularity theory. Namely, $$ \dim \frac{\mathcal{O}_{(\mathbb{C}^n, 0)}}{(a_1, a_2, a_3, \dots , a_n)} $$ is finite if and only is the common zero set of the germs $a_i$ contains only one ponint, the origin. In this case the ideal $I=I(a_i)_{i=n}^n$ is called complete intersection. Another equivalent algebraic characterization is that $a_i$ form a regular sequence, i.e. for all $i$ $$[a_i] \in \frac{\mathcal{O}_{(\mathbb{C}^n, 0)}}{(a_1, a_2, a_3, \dots , a_{i-1})} $$ is not a zero divisor. I guess it can be found in many books e.g. Looijenga, Mond-Ballesteros.

Therefore $[a_1]$ is not a zero divisor in $$ \frac{\mathcal{O}_{(\mathbb{C}^n, 0)}}{( a_2, a_3, \dots , a_n)} ,$$ hence $[c]$ is not a zero divisor. That is, $$ c(x-l_1 b)=l_2 a_2 + l_3 a_3 + \dots + l_n a_n.$$ implies $$ x-l_1 b=l'_2 a_2 + l'_3 a_3 + \dots + l'_n a_n,$$ what we wanted to prove.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.