# Localization of the $k\times k$-module $k\times\{0\}$ by $\{0\}\times k$

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 $$k$$.

• 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? Dec 12, 2021 at 16:54
• 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. Dec 12, 2021 at 17:09
• Ok, thanks, I misunderstood. Indeed, you took $R\setminus\mathfrak P$. Dec 12, 2021 at 17:18
• Yes absolutely. Dec 12, 2021 at 17:24
• 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? Dec 14, 2021 at 18:30

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.