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I just was thinking about mobius strip, and asked myself simple questions, which I could not convince myself of answers and hope someone can give my a concrete answer.

I know Mobius strip is a line bundle over the circle as real manifolds, but how this transfer to algebraic geometry.

Q1- What algebro-geometric properties of analogous Mobius strip in algebraic geometry? i.e. is it a variety, or some sort of a general scheme?

in case the above answer to Q1 is yes please see Q2 , otherwise thanks.

2-how it's constructed in this respect? i,e. if it's any of the above, any informain about its structure sheaf if it's a scheme, or coordinate ring in case it's a variety?

UPDATE: After the discussion bellow with kenny Wong, I realized that the rightwo question is this: 1- Can we have a mobuis strip analogy as a complex manifold, I mean will it look the same band that twisted around the edges and glued?

I appreciate your help here, thank you.

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    Is your Mobius strip a $[0,1]$ bundle over $S^1$, with the obvious twist in the $[0,1]$ as you go around the circle? This most certainly isn't an algebraic variety, as far as I can tell! Neither is $[0,1]$ or $S^1$, for that matter! – 2017-02-26
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    If I understood you right, yes my bundle is the twist around the circle, but why did you say $S^1$ is not algebric, I think it's $\spec \C[x,y]/$ isn't it? – 2017-02-26
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    ${\rm Spec \ }\mathbb C [x,y]/(x^2 + y^2 - 1)$ is defined over the complex numbers. $S^1$ is defined over the reals. – 2017-02-26
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    Thank you so much. so I see now. Can mobuis strip be defined over the circle $${\rm Spec \ }\mathbb C [x,y]/(x^2 + y^2 - 1)$ ? is there an analogy here?. I appreciate your help. – 2017-02-26
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    You might try defining a complex line bundle over this variety. Though I have a feeling that any such line bundle might be trivial in the complex algebraic category - I'm not sure. – 2017-02-26
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    In fact, unless I'm terribly mistaken, any line bundle on this variety is trivial because the variety is smooth and any divisor on this variety is linearly equivalent to the zero divisor. – 2017-02-26
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    So let's assume it's true that it is trivial, which I will try to prove, what would be the picture of this mobuis strip, I might be naively having a wrong picture, but would the mobuis strip be just a sort of a a cylinder? like ${\rm Spec \ }\mathbb C [x,y]/(x^2 + y^2 - 1)\times \math C$ ? I mean we don't have a mobuis strip anymore if we go to the category of complex, am I right? – 2017-02-26
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    Dear user93893 and commenters: the Möbius bundle is a perfectly legitimate real algebraic, non trivial line bundle on the real circle. There is no need to mention $\mathbb C$ in this question, which only confuses the issue. See my answer. – 2017-02-26
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    No offense to anyone, but why this question is closed as unclear? – 2018-05-07

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Yes, the Möbius strip is a perfectly valid algebraic line bundle bundle on the real circle , which is the perfectly valid real algebraic variety $$S=\operatorname {Proj}\mathbb R[X,Y]=\mathbb P^1_{\mathbb R}$$ We can look at the circle also as the subscheme $S^1=V(X^2+Y^2-1)\subset \mathbb A^2_\mathbb R$, i.e. $$S^1=\operatorname {Spec}\frac {\mathbb R[X,Y]}{\langle X^2+Y^2-1\rangle}=:\operatorname {Spec}\mathbb R[x,y]$$ Hence the real circle is both an affine and a projective variety, which is of course only possible because $\mathbb R$ is not algebraically closed!

Over $S$
The Möbius bundle is the total space of the tautological line bundle $\mathcal O_{\mathbb P^1_{\mathbb R}}(-1)$.

Over $S^1$
In the second incarnation $S^1$ of the circle, the Möbius bundle is the line bundle associated to the ideal $\langle y,x-1\rangle\subset \mathbb R[x,y]$, which is a non free projective module of rank one over the ring $\mathbb R[x,y]$.
In other words the Möbius bundle is the line bundle associated to the divisor $1.P$ of the circle, where $P$ is the closed point with coordinates $(1,0)$.

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    Thank you so much. very elegant and clear answer. – 2017-02-26
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    Thank you for the kind words, dear user93893. – 2017-02-26