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Let $A$ be a ring, $\mathfrak{q}$ an prime ideal and $\mathfrak{p}$ a maximal ideal such that $\mathfrak{q}\subset \mathfrak{p}$. Consider now the localized ring

$$(A/\mathfrak{q})_{\mathfrak{p}}$$

and denote by $\mathfrak{m}$ its maximal ideal. I saw somewhere that we have

$$\mathfrak{m}=(\mathfrak{p}+\mathfrak{q})/\mathfrak{q}$$

and so

$$\mathfrak{m}/\mathfrak{m}^2= \mathfrak{p}/(\mathfrak{p}^2+\mathfrak{q}).$$

I have two questions:

  • I dont understand how to get the expression of $\mathfrak{m}$ from the expression of the maximal ideal of a localization. So far I got something like $\mathfrak{m}=\mathfrak{p}(A_{\mathfrak{p}}/\mathfrak{q}_{\mathfrak{p}})$.

  • Assuming the expression of $\mathfrak{m}$, I dont understand the computation of the vector space $\mathfrak{m}/\mathfrak{m}^2$.

For the sake of completeness this is part of the proof of the Jacobian criterion for singularity of varieties.

BinAcker
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    If $\mathfrak q \subset \mathfrak p$ then $\mathfrak p + \mathfrak q = \mathfrak p$. Are you sure you transcribed that section of the proof correctly? What book is that from? – Ayman Hourieh Aug 25 '19 at 09:43
  • @AymanHourieh I saw it here https://pdfs.semanticscholar.org/e951/27c8bfb9b38d3f41571619b780a3f85b421c.pdf not exactly like I wrote but if $\mathfrak{q}$ defines a variety $X$ and a point $x\in X$ corresponds to the maximal ideal $\mathfrak{p}$ then I we have $\mathfrak{q}\subset \mathfrak{p}$. But you are right that it looks wrong then – BinAcker Aug 25 '19 at 09:57
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    I think this post has the answers you're looking for: https://math.stackexchange.com/q/730680/4583 – Ayman Hourieh Aug 25 '19 at 10:26

0 Answers0