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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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    Thank you @MichaelHardy for your comment.2011-10-09

1 Answers 1

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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.

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    Thank you @Pierre-Yves. This time I fully understand.2011-10-10