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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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    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 $x\ge y$.2011-10-08
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    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.2011-10-08
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    Take $\mathfrak m^2\not=\mathfrak m=\mathfrak p_0, n=2$.2011-10-08
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    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.2011-10-08
  • 0
    Dear @Georges: Thank you very much! (By the way, you forgot the `@` sign...)2011-10-08
  • 0
    Dear @Pierre-Yves: True, I often forget that @ sign...2011-10-08
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    @Georges: That's the only imperfection I've been able to find so far in your personality...2011-10-08
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    Thank you @MichaelHardy for your comment.2011-10-09

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