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 Post subject: Locally free but no globallyPosted: Sat Oct 07, 2017 3:25 pm

Joined: Mon Oct 17, 2016 9:33 pm
Posts: 6
Let $R=k[x,y]/(x^{2}+y^{2}-1)$, and let $\mu=(x,y-1)\subset R$. I want to prove that $\mu$ is locally free (i.e that the localization in the multiplicative system defined by each prime $\mathfrak{p}$ is a free $R_{\mathfrak{p}}$-module). I have just proved that $\mu$ is locally free of rank 1, but I donĀ“t know how to prove that $\mu$ is not a free module of rank 1.

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 Post subject: Re: Locally free but no globallyPosted: Sat Oct 07, 2017 8:17 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 424
hI PJPu17. The ring $\displaystyle{R}$ is commutative. We observe that

$\displaystyle{x^2+y^2-1=x\cdot x+(y+1)\cdot (y-1)\in\langle{x,y-1\rangle}}$, so,

$\displaystyle{\langle{x^2+y^2-1\rangle}\leq \langle{x,y-1\rangle}\leq \mathbb{K}[x,y]}$

and according to the 3rd Ring Isomorphism Theorem, we get

$\displaystyle{\left(\mathbb{K}[x,y]/\langle{x^2+y^2-1\rangle}\right)/\langle{x,y-1\rangle}/\langle{x^2+y^2-1\rangle}\cong \mathbb{K}[x,y]/\langle{x,y-1\rangle}}$

or equivalently,

$\displaystyle{R/\mu\cong \mathbb{K}[x,y]/\langle{x,y-1\rangle}}$ as rings.

Now, the map $\displaystyle{\Phi:\mathbb{K}[x,y]\to \mathbb{K}\,,f(x,y)\mapsto f(0,1)}$

is a ring homomorphism, which is onto $\displaystyle{\mathbb{K}}$ and satisfies $\displaystyle{\rm{Ker}(\Phi)=\langle{x,y-1\rangle}}$

Then,

$\displaystyle{R/\mu\cong \mathbb{K}[x,y]/\langle{x,y-1\rangle}=\mathbb{K}[x,y]/\rm{Ker}(\Phi)\cong Im(\Phi)=\mathbb{K}}$

that is, the ideal $\displaystyle{\mu}$ is maximal. If $\displaystyle{\mu}$ is a free $\displaystyle{R}$ -module

of rank $\displaystyle{1}$, then $\displaystyle{R\cong \mu}$ as $\displaystyle{R}$- modules

and more specifically, $\displaystyle{|R|=|\mu|\,\,\land\,\,\mu\subset R\implies \mu=R}$,

a contradiction, since $\displaystyle{\mu}$ is a maximal ideal of $\displaystyle{R}$.

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