Let $R=k[x,y]/(x^{2}+y^{2}1)$, and let $\mu=(x,y1)\subset R$. 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 don´t know how to prove that $\mu$ is not a free module of rank 1.
Thank you for your time.
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Locally free but no globally

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Re: Locally free but no globally
hI PJPu17. The ring \(\displaystyle{R}\) is commutative. We observe that
\(\displaystyle{x^2+y^21=x\cdot x+(y+1)\cdot (y1)\in\langle{x,y1\rangle}}\), so,
\(\displaystyle{\langle{x^2+y^21\rangle}\leq \langle{x,y1\rangle}\leq \mathbb{K}[x,y]}\)
and according to the 3rd Ring Isomorphism Theorem, we get
\(\displaystyle{\left(\mathbb{K}[x,y]/\langle{x^2+y^21\rangle}\right)/\langle{x,y1\rangle}/\langle{x^2+y^21\rangle}\cong \mathbb{K}[x,y]/\langle{x,y1\rangle}}\)
or equivalently,
\(\displaystyle{R/\mu\cong \mathbb{K}[x,y]/\langle{x,y1\rangle}}\) 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 \(\displaystyle{\rm{Ker}(\Phi)=\langle{x,y1\rangle}}\)
Then,
\(\displaystyle{R/\mu\cong \mathbb{K}[x,y]/\langle{x,y1\rangle}=\mathbb{K}[x,y]/\rm{Ker}(\Phi)\cong Im(\Phi)=\mathbb{K}}\)
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 \(\displaystyle{R}\).
\(\displaystyle{x^2+y^21=x\cdot x+(y+1)\cdot (y1)\in\langle{x,y1\rangle}}\), so,
\(\displaystyle{\langle{x^2+y^21\rangle}\leq \langle{x,y1\rangle}\leq \mathbb{K}[x,y]}\)
and according to the 3rd Ring Isomorphism Theorem, we get
\(\displaystyle{\left(\mathbb{K}[x,y]/\langle{x^2+y^21\rangle}\right)/\langle{x,y1\rangle}/\langle{x^2+y^21\rangle}\cong \mathbb{K}[x,y]/\langle{x,y1\rangle}}\)
or equivalently,
\(\displaystyle{R/\mu\cong \mathbb{K}[x,y]/\langle{x,y1\rangle}}\) 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 \(\displaystyle{\rm{Ker}(\Phi)=\langle{x,y1\rangle}}\)
Then,
\(\displaystyle{R/\mu\cong \mathbb{K}[x,y]/\langle{x,y1\rangle}=\mathbb{K}[x,y]/\rm{Ker}(\Phi)\cong Im(\Phi)=\mathbb{K}}\)
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 \(\displaystyle{R}\).