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 ?

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    $\begingroup$ 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 $x\ge y$. $\endgroup$ – Michael Hardy Oct 8 '11 at 14:38
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    $\begingroup$ 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. $\endgroup$ – Michael Hardy Oct 8 '11 at 14:38
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    $\begingroup$ Take $\mathfrak m^2\not=\mathfrak m=\mathfrak p_0, n=2$. $\endgroup$ – Pierre-Yves Gaillard Oct 8 '11 at 15:18
  • $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Oct 8 '11 at 15:49
  • $\begingroup$ Dear @Georges: Thank you very much! (By the way, you forgot the @ sign...) $\endgroup$ – 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.

  • $\begingroup$ 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 ${\mathfrak m}^n$ ? $\endgroup$ – Aki Oct 9 '11 at 8:02
  • $\begingroup$ 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 $\mathfrak p\subset \mathfrak m^n$. $\endgroup$ – Pierre-Yves Gaillard Oct 9 '11 at 9:23
  • $\begingroup$ Thank you @Pierre-Yves. This time I fully understand. $\endgroup$ – Aki Oct 10 '11 at 7:57

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