Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following is a quotation from the proof of Proposition 11.10 in "Introduction to Commutative Algebra" by Atiyah and MacDonald.

Also if ${\mathfrak m}'$ is the maximal ideal of $A'$,
$A'/{\mathfrak m}'^n$ is a homomorphic image of $A/{\mathfrak m}^n$, hence $l(A/{\mathfrak m}^n) \geq l(A'/{\mathfrak m}'^n)$.

In the above, $A$ is a Noetherian local ring, ${\mathfrak m}$ is its maximal ideal, and $A'=A/{\mathfrak p}_0$ where ${\mathfrak p}_0$ is a prime ideal in $A$. Also, $l(M)$ is the length of $M$.

It seems to me that
$A'/{\mathfrak m}'^n$ is an isomorphic image of $A/{\mathfrak m}^n$, hence $l(A/{\mathfrak m}^n) = l(A'/{\mathfrak m}'^n)$.

Am I wrong ?

share|cite|improve this question
Without looking at whether you're right to say that $A'/{\mathfrak m}'^n$ is an isomorphic image of $A/{\mathfrak m}^n$, or whether you're right to say that $l(A/{\mathfrak m}^n) = l(A'/{\mathfrak m}'^n)$, one can say that if those are right, then $A'/{\mathfrak m}'^n$ is a homomorphic image of $A/{\mathfrak m}^n$ and $l(A/{\mathfrak m}^n) \ge l(A'/{\mathfrak m}'^n)$, since every isomorphism is a homomorphism, and generally if $x=y$ then $x\ge y$. – Michael Hardy Oct 8 '11 at 14:38
That still leaves your question intact, except for the words "rather than" in the title. "Am I right in thinking this homomorphism is an isomorphism" might fit better. – Michael Hardy Oct 8 '11 at 14:38
Take $\mathfrak m^2\not=\mathfrak m=\mathfrak p_0, n=2$. – Pierre-Yves Gaillard Oct 8 '11 at 15:18
Dear Pierre-Yves, if you upgraded that nice but extremely concise suggestion to an answer (with an example, maybe) , I'd be glad to upvote you and I hope other users would do the same. – Georges Elencwajg Oct 8 '11 at 15:49
Dear @Georges: Thank you very much! (By the way, you forgot the @ sign...) – Pierre-Yves Gaillard Oct 8 '11 at 16:18

The natural epimorphism from $A/\mathfrak m^n$ to $A'/\mathfrak m'^n$ is not injective in general.

Indeed, if you put $$ A:=\mathbb Z/4\mathbb Z,\quad\mathfrak p_0=\mathfrak m=(2),\quad n=2, $$ you get $$ A'=\mathbb Z/2\mathbb Z,\quad\mathfrak m'^n=\mathfrak m'^2=\mathfrak m'=0,\quad A'/\mathfrak m'^n=\mathbb Z/2\mathbb Z, $$ $$ \mathfrak m^n=\mathfrak m^2=0,\quad A/\mathfrak m^n=A=\mathbb Z/4\mathbb Z. $$ Thus $A/\mathfrak m^n$ has four elements, whereas $A'/\mathfrak m'^n$ has only two elements.

share|cite|improve this answer
Thank you for answering my question. If ${\mathfrak p}_0={\mathfrak m}$, we have $A'/{\mathfrak m}'^n=A'=A/{\mathfrak m}$. So, the kernel of the mapping $A \rightarrow A'/{\mathfrak m}'^n$ is ${\mathfrak m}$ not ${\mathfrak m}^n$. Do you think that this is the only case when the mapping has the kernel other than ${\mathfrak m}^n$ ? – Aki Oct 9 '11 at 8:02
Dear @Aki: You're welcome. The kernel is $$\frac{\mathfrak p+\mathfrak m^n}{\mathfrak m^n}\simeq\frac{\mathfrak p}{\mathfrak p\cap\mathfrak m^n}\quad.$$ If you take $\mathfrak p=\mathfrak m^2$, $n=3$, you get $\mathfrak m^2/\mathfrak m^3$, which is nonzero in general. The map is injective iff $\mathfrak p\subset \mathfrak m^n$. – Pierre-Yves Gaillard Oct 9 '11 at 9:23
Thank you @Pierre-Yves. This time I fully understand. – Aki Oct 10 '11 at 7:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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