# Finitely generated Projective Modules over Real Circle

by Dibya Banerjee   Last Updated August 14, 2019 15:20 PM

I need to find the isomorphism classes of finitely generated projective modules over $$\mathbb{R}[x,y] / \langle x^2 + y^2 - 1 \rangle$$.

Now what I tried was since $$\mathbb{R}$$ is a field $$\mathbb{R}[x]$$ is a PID which implies $$\mathbb{R}[x][y]$$ is noetherian. Thus the quotient is noetherian. After this I am stuck.

I was thinking if it is a local ring or not then projective modules over this could be free. Is it local? For any hint or help I am grateful, thanks.

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#### Answers 1

Denote $$A=\mathbb {R}[x,y]/$$. We first observe that $$A$$ is a Dedekind domain since the corresponding plane curve $$C$$ is smooth. The characterization of finitely generated modules over a Dedekind domain says that every such $$P$$ is isomorphic to $$I\oplus A^r$$ where $$I\subset A$$ is an ideal. We thus have to classify ideals of $$A$$ up to isomorphism of $$A$$-modules, i.e., we have to compute the class group of $$A$$. Every non-zero prime ideal of $$A$$ is a maximal ideal and thus is the set of all polynomials that vanish on a real point of $$C$$ or on a pair of complex conjugate points on $$C$$. In the latter case the ideal is a principal ideal: It is generated by the linear polynomial defining the line spanned by the pair of complex conjugate points. On the other hand, the vanishing ideal of a single real point is not a principle ideal: If it was generated by a polynomial $$f$$, then the zero set of $$f$$ would intersect the real part of the curve $$C$$ transversally in exactly one point which is not possible. However, the vanishing ideal of the union of any to real points is again a principle ideal (take again the line spanned by both points). Thus the class group of $$A$$ is the cyclic group of order two generated by the vanishing ideal of a single real point, say $$\langle x,y-1\rangle$$.

Therefore, every finitely generated projective module over $$A$$ is either isomorphic to $$A^r$$ or to $$A^r\oplus \langle x,y-1\rangle$$.