Show that $\mathbb{C}[x,y]/(x^2+y^2-1)$ is a UFD. [duplicate]

I am trying to prove that the ring $\mathbb{C}[x,y]/(x^2+y^2-1)$ is a UFD. I have an hint, that suggests to find an isomorphism between $\mathbb{C}[x,y]/(x^2+y^2-1)$ and $\mathbb{C}[e^{it},e^{-it}]$, but to be honest I don't see it. Please, could you give me a hand?

I would to solve my problem completely, any kind of suggestion is fully appreciated.

Update: user Daniel McLaury helped me in showing the isomorphism - so now it remains only to see that this ring is actually an UFD.

Thanks in advance. Cheers

Solution 1:

The Euler formula $e^{it} = \cos(t) + i \sin(t)$ let you rewrite $x = \frac{e^{it} + e^{-it}}{2}$ and $y = \frac{e^{it} - e^{-it}}{2i}$, and it's more-or-less straightforward to see that the induced morphism of rings $\mathbb{C}[x,y]/(x^2+y^2-1) \to \mathbb{C}[e^{it}, e^{-it}]$ is an isomorphism.

Now you can see $\mathbb{C}[e^{it}, e^{-it}]$ as the ring of Laurent polynomials in a variable $z$, say $\mathbb{C}[e^{it}, e^{-it}] \cong \mathbb{C}[z, z^{-1}]$ (where $z = e^{it}$). In other words, it's the localization of the polynomial ring $\mathbb{C}[z]$ at the ideal $(z)$. The ring $\mathbb{C}[t]$ is a UFD (this is well-known), and the localization of a UFD is a UFD (as long as you don't invert $0$, which is the case here). Thus $\mathbb{C}[z, z^{-1}]$ is a UFD.