# 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.

## marked as duplicate by Dietrich Burde, user91500, A.D, drhab, Gabriel RomonJun 16 '15 at 13:48

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.
Since $\sin(t)^2 + \cos(t)^2 = 1$, you can think of $x$ and $y$ as representing the sine and cosine functions. Then just write sine and cosine in terms of $e^{it}$ to get your isomorphism.
• Dear @DanielMcLaury thank you for your hint. Yes, it is exactly what I tried, but...mmm...$(e^{it})^2 + (e^{-it})^2$ is not 1...am I missing some point? – user233650 Jun 4 '15 at 8:06
• $e^{it}$ isn't $\sin(t)$ or $\cos(t)$, it's $\cos(t) + i \sin(t)$. This lets you write $\sin(t)$ and $\cos(t)$ in terms of $e^{it}$ and $e^{-it}$. – Daniel McLaury Jun 4 '15 at 8:07