Welcome to mathimatikoi.org forum; Enjoy your visit here.

Locally free but no globally

Linear Algebra, Algebraic structures (Groups, Rings, Modules, etc), Galois theory, Homological Algebra
Post Reply
Articles: 0
Posts: 10
Joined: Mon Oct 17, 2016 9:33 pm

Locally free but no globally


Post by PJPu17 » Sat Oct 07, 2017 3:25 pm

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 don´t know how to prove that $\mu$ is not a free module of rank 1.

Thank you for your time.
Papapetros Vaggelis
Community Team
Community Team
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Locally free but no globally


Post by Papapetros Vaggelis » Sat Oct 07, 2017 8:17 pm

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 \(\displaystyle{\rm{Ker}(\Phi)=\langle{x,y-1\rangle}}\)


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

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 \(\displaystyle{R}\).
Post Reply