Quotient Rings of Polynomial Rings + Isomorphisms =?

Question

I'm taking a course in commutative rings (at least, that's what it's supposed to be) but the instructor has never constructed a single isomorphism. Point being, he assigned some problems and I don't even know how to think about them, moreover explicitly find an isomorphism.

For instance, how would you go about this problem?

Let C denote the field of complex numbers and <f> the principal ideal generated by f.

Show that

C[x, y] / <x^2 + y^2 - 1> = C[x, x^-1]

( = denotes an isomorphism )

Low Key Lyesmith · Accepted Answer

Let's rewrite the right side as C[t, t^-1] to avoid collision of notation.

The equation x^2 + y^2 - 1 = 0 is equivalent to
(x + iy)(x - iy) = 1, which suggests that we should map 
x + iy to t and x - iy to t^-1.
Solving for x and y gives
x = (t + t^-1)/2 and y = (t - t^-1)/(2i).

So our candidate homomorphism is defined by
f(x) = (t + t^-1)/2 and f(y) = (t - t^-1)/(2i).

I will leave it to you to check that this is a well-defined map and that it is an isomorphism of commutative rings.
--
Another point of view: C[x,y]/<x^2+y^2-1> is isomorphic to the ring of polynomials in cosθ and sinθ, owing to the identity (cosθ)^2 + (sinθ)^2 - 1 = 0. But any polynomial in cosθ and sinθ can be written as a Laurent polynomial in e^(iθ).

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Quotient Rings of Polynomial Rings + Isomorphisms =?

1 Answer