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Let $k$ be a field, $R=k\times k$ and $M=k\times\{0\}$ endowed with a structure of $R$-module. Let $\mathfrak{P}=\{0\}\times k$, it is a maximal ideal of $R$ such that $R_{\mathfrak{P}}\simeq k$. I want to show that $M_{\mathfrak{P}}$ is not a free $R_{\mathfrak{P}}$-module, but I found that $M_{\mathfrak{P}}\simeq k$ which is free so I must be mistaken. Any help is appreciated.

Here is how I proceeded : $M_{\mathfrak{P}}=M\times(R\setminus\mathfrak{P})/\sim$ where $\sim$ is such that for $(m_1,x_1,y_1),(m_2,x_2,y_2)\in M\times (R\setminus\mathfrak{P})=k\times k^{*}\times k$, we have $$ \begin{aligned} (m_1,x_1,y_1)\sim (m_2,x_2,y_2) &\iff \exists (u,v)\in k^*\times k,(u,v)\cdot((x_1,y_1)\cdot m_2-(x_2,y_2)\cdot m_1)=0 \\ &\iff \exists u\in k^*, u(x_1m_2-x_2m_1)=0 \\ &\iff x_1m_2=x_2m_1 \end{aligned} $$ Therefore, the morphism $M_{\mathfrak{P}}\rightarrow k$ defined by $\frac{m}{(x,y)}\mapsto mx^{-1}$ is well defined and it is an isomorphism so that the $R_{\mathfrak{P}}$-module $M_{\mathfrak{P}}$ is isomorphic to the $k$-vector space $k$.

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    $\begingroup$ I'm not entirely sure what happens to the zero divisors, i.e. what if $x=0$? Then we don't have $x^{-1}$. On the other hand, $k$ is not a free module over $R$. By the way, what $R$-action do you obtain on $k$ via this isomorphism? $\endgroup$
    – Berci
    Dec 12, 2021 at 16:54
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    $\begingroup$ For the first question : $(x,y)\in k^*\times k$ so $x\neq 0$. Moreover, $k$ is not a free module over $R$ indeed, but I have to show that $M_{\mathfrak{P}}$ is not free over $R_{\mathfrak{P}}\simeq k$. As for the isomorphism, I obtain a structure of $k$-vector space on $k$, not a $R$-action. $\endgroup$
    – Tuvasbien
    Dec 12, 2021 at 17:09
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    $\begingroup$ Ok, thanks, I misunderstood. Indeed, you took $R\setminus\mathfrak P$. $\endgroup$
    – Berci
    Dec 12, 2021 at 17:18
  • $\begingroup$ Yes absolutely. $\endgroup$
    – Tuvasbien
    Dec 12, 2021 at 17:24
  • $\begingroup$ what you have written looks completely correct to me. are you sure you are supposed to show that $M_\mathfrak{P}$ is not a free $R_\mathfrak{P}$-module, and not that it is one? $\endgroup$ Dec 14, 2021 at 18:30

1 Answer 1

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What you have written is absolutely correct. Here is a more general way of arguing it; let $R$ be a (commutative, unital) ring and $S\subset R$ a multiplicatively closed set, and let $R_S$ denote the localization of $R$ at $S$, with localization map $\lambda:R\to R_S$. Recall that there is a natural map $\mu:\operatorname{Ideals}(R)\to\operatorname{Ideals}(R_S)$ induced by localization; it takes an ideal $J\vartriangleleft R$ to the ideal of $R_S$ generated by $\lambda(J)\subset R_S$.

Fact: For any ideal $J\vartriangleleft R$, we have $\mu(J)\cong J_S$ as $R_S$-modules. $\blacksquare$

Now suppose the ideal $J$ contains an element $s\in S$; then necessarily $\mu(J)=R_S$, since $\mu(J)$ is an ideal containing the unit $\lambda(s)\in R_S^\times$. In particular, by the highlighted fact, $J_S$ is then a free $R_S$-module of rank $1$.

In your case, taking $R=k\times k$, $S=R\setminus\mathfrak{P}$, $M=k\times 0\vartriangleleft R$, and $s=(1,0)$ gives exactly the desired result.

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  • $\begingroup$ Thank you for your answer, it is indeed more clever. $\endgroup$
    – Tuvasbien
    Dec 14, 2021 at 19:37
  • $\begingroup$ @Tuvasbien my pleasure, happy it helped! :) $\endgroup$ Dec 14, 2021 at 19:41

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