How to describe the ring $\mathbb{C}[\cos{x},\sin{x}]$ as a quotient ring of $\mathbb{C}[X,Y]$? [duplicate]

Solution 1:

First of all, $\mathbb{C}[\sin,\cos]$ certainly cannot be a proper subring of $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ (with $X$ corresponding to $\cos$ and $Y$ corresponding to $\sin$). As a $\mathbb{C}$-algebra, $\mathbb{C}[X,Y]/(X^2+Y^2-1)$ is generated by $X$ and $Y$, so there is not any proper $\mathbb{C}$-subalgebra that contains both $X$ and $Y$. A priori, $\mathbb{C}[\sin,\cos]$ could be a proper quotient of $\mathbb{C}[X,Y]/(X^2+Y^2-1)$, since there could be more relations between $\sin$ and $\cos$ that are not generated by the single relation $X^2+Y^2-1$. However, this is not the case (and the proof is basically the same as the proof over $\mathbb{R}$ that you say you know), so $\mathbb{C}[\sin,\cos]$ really is isomorphic to $\mathbb{C}[X,Y]/(X^2+Y^2-1)$.

So what's going on? Well, the equation $\sin^2=(1+\cos)(1-\cos)$ simply does not imply that the ring is not a UFD. After all, $6^2=4\cdot 9$ but $\mathbb{Z}$ is still a UFD. To conclude that you have a failure of unique factorization, you would need to know something more about the factors $\sin,1-\cos,$ and $1+\cos$, such as that they are irreducible and not associate to each other.

In fact, none of these factors are irreducible. To see this, let us use the isomorphism $\mathbb{C}[X,Y]/(X^2+Y^2-1)\cong\mathbb{C}[T,T^{-1}]$ which is given by mapping $T$ to $X+iY$ (and $T^{-1}$ to $X-iY$). So in terms of $T$, $\sin$ (or $Y$) would be $\frac{T-T^{-1}}{2i}=\frac{T^2-1}{2iT}$. This is not irreducible, because $\frac{1}{2iT}$ is a unit (so can be ignored) and $T^2-1$ factors as $(T+1)(T-1)$ (and neither factor is a unit). Similarly, $1\pm\cos$ becomes $1\pm\frac{T+T^{-1}}{2}=\pm\frac{T^2\pm 2T+1}{2T}$ and the numerator factors as $(T\pm 1)^2$. So up to units, $\sin$ is $(T+1)(T-1)$ and $1+\cos$ and $1-\cos$ are $(T+1)^2$ and $(T-1)^2$, so the factorization $\sin^2=(1+\cos)(1-\cos)$ is just like the factorization $6^2=4\cdot 9$ of integers (with $T+1$ corresponding to $2$ and $T-1$ corresponding to $3$).