Reading this questions Ring of trigonometric functions with real coefficients we can see that $\mathbb{C}[x, y] / (x^2 + y^2 - 1)$ is a UFD because it's isomorphic to Laurent polynomial ring which is UFD. On the one hand we have the following equality $$ x^2 = x * x = (1-y)(1+y) $$ So if $\mathbb{C}[x, y] / (x^2 + y^2 - 1)$ is a UFD, it must be that $$ x = u(1+y) $$ where $u$ is a unit. Also i have found that $x+iy$ and $x-iy$ are units in this ring, but i can't find $u$. It must be $u = c(x+iy)^n(x+iy)^m$ for some $n, m \in \mathbb{N}$ and $c \in \mathbb{C^{*}}$ but it seem to me that such $u$ doesn't exist.
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1I would first transform everything to the Laurent polynomial ring parameters. – Trevor Gunn Oct 06 '22 at 12:31
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See the proofs given at this duplicate. They are also helpful. – Dietrich Burde Oct 06 '22 at 12:39
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1I think the comments on this post are related, especially my comment here. – Viktor Vaughn Oct 06 '22 at 14:39
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@ViktorVaughn yes, they are good – NeoFanatic Oct 06 '22 at 14:42
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There is an isomorphism $\mathbb{C}[X,Y]/(X^2+Y^2-1)\cong\mathbb{C}[T,T^{-1}]$ given by $T^{\pm1}=X\pm iY$. With this identification in mind, we see $X=(T+T^{-1})/2=\frac{1}{2}T^{-1}(T+i)(T-i)$ is not irreducible because the factors $T\pm i$ are not units (though $\frac{1}{2}T^{-1}$ is), and $1\pm Y=\pm\frac{1}{2i}T^{-1}(T\pm i)^2$ (which we find by substituting $T$ into $X$ and $Y$, combining into one fraction, and factoring numerators.)
Thus, $X^2=1-Y^2$ does not tell us $X$ is associate to either of $1\pm Y$ (which it isn't, as we can see by simplifying the quotient $X/(1\pm Y)$ as a rational function of $T$).
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