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.

Tags : modules


Answers 1


Denote $A=\mathbb {R}[x,y]/<x^2+y^2-1>$. 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 $A^r\oplus \langle x,y-1\rangle$.

Hans
Hans
August 14, 2019 15:12 PM

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