Algebraic structure of Riemann Sphere Does Riemann sphere have any algebraic structure (Field, Ring, Algebra)?
It appears that 'as a set', construction of Riemann Sphere is similar to the construction of the ring of polynomials over a field. At least the first step is same. 
Consider the set $ \mathbb{C} $. Insert a symbol $ \infty $ in that set. That is $ \hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\} $
(Then define the topology suitably)
On the other hand.
Consider any field $ F $. Insert a symbol $ x $ in that set. 
(Then insert a whole lot more (all 'powers' of x, and formal sums of F-coefficient 'powers' of x etc.)
Clearly, in Riemann Square, the extra symbol inserted in the set does not produce linearly independent elements $ \infty, \infty^2  ... $ 
But I am having a hard time understanding how the 'algebra' of the set $ \mathbb{C} $ is disrupted by this new symbol. 
 A: No, the Riemann sphere in its entirety is not a ring, algebra or field in any interpretation I know.
I think its most natural algebraic interpretation the Riemann sphere has is its identification with the complex projective line, so that it is a set upon which the Möbius group acts upon. So, not a field, ring or algebra, but the underlying space of a group action.
I think probably (but I don't know, personally) that one can probably say something more about it being a topological group action on the complex projective line.
A: The traditional algebraic structures of rings or fields do not allow you to include a multiplicative inverse of $0$. The existence of such an element would violate the ring axioms, except in the most degenerate cases. 
However there is another algebraic structure that axiomatizes and allows to reason about division by zero, which is called a wheel. Any commutative ring may be embedded in a wheel. The algebraic structures of the real and complex projective line are examples of wheels (as long as you also include an element for $0/0$). 
