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How to compute localization of module $\mathbb Z/n\mathbb Z$ at the prime ideal $(p)$?

For example, let $M = \mathbb Z/12\mathbb Z$. Then how to compute $M_{(2)}$?

I'm also wondering if this can be generalized into polynomial case. For example, let $M=k[x_1,\dots,x_n]/I$. Then could we compute $M_{\mathfrak p}$ for prime ideal $\mathfrak p$?

Thank you!

Smiley1000
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Hydrogen
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    What have you tried? What does the definition of localization force you to invert? – Qiaochu Yuan Sep 25 '20 at 00:21
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    @QiaochuYuan Here is my attempt. $(\mathbb Z/12\mathbb Z){(2)} = \mathbb Z{(2)}/12\mathbb Z_{(2)} = \mathbb Z_{(2)}/4\mathbb Z_{(2)}$. But I don't know what to do next. – Hydrogen Sep 25 '20 at 16:12
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    You’re almost done. What are the cosets? – Qiaochu Yuan Sep 25 '20 at 16:18
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    Oh, It seems to be $\mathbb Z/4\mathbb Z$. – Hydrogen Sep 25 '20 at 23:47
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    Yes, that's right! Localizing at $(2)$ means you invert all other primes so the $3$ disappears. Once you get to $\mathbb{Z}/4$ all the other primes are invertible so there's nothing left to invert. – Qiaochu Yuan Sep 25 '20 at 23:48
  • @QiaochuYuan I think I actually prove $(\mathbb Z/4\mathbb Z)\otimes_{\mathbb Z}\mathbb Z_{(2)} = \mathbb Z/4\mathbb Z$. I'm wondering if there is any generalization of this identity. When will we have $M/nM = \mathbb Z/n\mathbb Z$ in general? – Hydrogen Sep 25 '20 at 23:49
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    Tensoring by $\mathbb{Z}_{(p)}$ fixes any abelian $p$-group. – Qiaochu Yuan Sep 25 '20 at 23:52

1 Answers1

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Consider $\newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} R = \quotient{\mathbb{Z}}{\langle 12 \rangle}$, $\mathfrak{p} = \quotient{\langle 2 \rangle}{\langle 12 \rangle} = \{0, 2, 4, 6, 8, 10\}$ and $U = R \setminus \mathfrak{p} = \{1, 3, 5, 7, 9, 11\}$. We want to compute the localization $U^{-1} R$ of $R$ at $U$.

According to the following questions, we can also construct $U^{-1} R$ as $\quotient{R[x_s \ : \ s \ \in \ U]}{\langle s x_s - 1 \ : \ s \ \in \ U\rangle}$.

This means that we have the following isomorphisms: $$ \begin{array}{cl} & \{1, 3, 5, 7, 9, 11\}^{-1} \left(\quotient{\mathbb{Z}}{\langle 12 \rangle}\right) \\ \cong & \quotient{\left(\quotient{\mathbb{Z}}{\langle 12 \rangle}\right)[x_1, x_3, x_5, x_7, x_9, x_{11}]}{\langle x_1 - 1, 3x_3 - 1, 5x_5 - 1, 7x_7 - 1, 9x_9 - 1, 11x_{11} - 1 \rangle} \\ \cong & \quotient{\mathbb{Z}[x_1, x_3, x_5, x_7, x_9, x_{11}]}{\langle 12, x_1 - 1, 3x_3 - 1, 5x_5 - 1, 7x_7 - 1, 9x_9 - 1, 11x_{11} - 1 \rangle} \\ \cong & \quotient{\mathbb{Z}[x_3, x_5, x_7, x_9, x_{11}]}{\langle 12, 3x_3 - 1, 5x_5 - 1, 7x_7 - 1, 9x_9 - 1, 11x_{11} - 1 \rangle} \\ \end{array} $$ But now, consider the ideals $$ I = \langle 12, 3x_3 - 1, 5x_5 - 1, 7x_7 - 1, 9x_9 - 1, 11x_{11} - 1 \rangle $$ and $$ J = \langle 4, x_3 - 3, x_5 - 1, x_7 - 3, x_9 - 1, x_{11} - 3 \rangle $$ of $\mathbb{Z}[x_3, x_5, x_7, x_9, x_{11}]$. We claim that $I = J$.

Indeed, we have:

  • $4 = 4 - 12x_3 + 12x_3 = 4(1 - 3x_3) + 12x_3 \in I$
  • $x_3 - 3 = 9x_3 - 3 - 8x_3 = 3(3x_3 - 1) - 4(2x_3) \in I$
  • $x_5 - 1 = 5x_5 - 1 - 4x_5 \in I$
  • $x_7 - 3 = 21x_7 - 3 - 20x_7 = 3(7x_7 - 1) - 4(5x_7) \in I$
  • $x_9 - 1 = 9x_9 - 1 - 8x_9 = 9x_9 - 1 - 4(2x_9) \in I$
  • $x_{11} - 3 = 33x_{11} - 3 - 32x_{11} = 3(11x_{11} - 1) - 4(8x_{11}) \in I$

This shows that $J \subseteq I$.

On the other hand, we also have:

  • $12 = 3 \cdot 4 \in J$
  • $3x_3 - 1 = 3x_3 - 9 + 8 = 3(x_3 - 3) + 4 \cdot 2 \in J$
  • $5x_5 - 1 = 5x_5 - 5 + 4 = 5(x_5 - 1) + 4 \in J$
  • $7x_7 - 1 = 7x_7 - 21 + 20 = 7(x_7 - 3) + 4 \cdot 5 \in J$
  • $9x_9 - 1 = 9x_9 - 9 + 8 = 9(x_9 - 1) + 4 \cdot 2 \in J$
  • $11x_{11} - 1 = 11x_{11} - 33 + 32 = 11(x_{11} - 3) + 4 \cdot 8 \in J$

This shows that $I \subseteq J$. Hence $I = J$ and we may continue our calculation: $$ \begin{array}{cl} & \quotient{\mathbb{Z}[x_3, x_5, x_7, x_9, x_{11}]}{\langle 12, 3x_3 - 1, 5x_5 - 1, 7x_7 - 1, 9x_9 - 1, 11x_{11} - 1 \rangle} \\ \cong & \quotient{\mathbb{Z}[x_3, x_5, x_7, x_9, x_{11}]}{\langle 4, x_3 - 3, x_5 - 1, x_7 - 3, x_9 - 1, x_{11} - 3 \rangle} \\ \cong & \quotient{\mathbb{Z}}{\langle 4 \rangle} \\ \end{array} $$ Putting everything together, we obtain: $$ \{1, 3, 5, 7, 9, 11\}^{-1} \left(\quotient{\mathbb{Z}}{\langle 12 \rangle}\right) \cong \quotient{\mathbb{Z}}{\langle 4 \rangle} $$ The following questions are similar:

Smiley1000
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