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Let us consider the map $\phi : \mathbb{C}[X,Y] \to \mathbb{C}[e^{iT}, e^{-iT}]$ defined by $\phi(X)=\frac{e^{iT}+e^{-iT}}{2}$ and $\phi(Y)=\frac{e^{iT}-e^{-iT}}{2i}$. (We can notice that $\phi$ is in particular surjective).

We want to show that $\ker(\phi)=(X^2+Y^2-1)$.

One inclusion is clear.

Now we want to prove the other. Let $P \in \ker(\phi)$ and let us use the euclidean division in $(\mathbb{C}[X])[Y]$: $P = Q(X^2+Y^2-1) +R$ with $\deg_Y(R)<2.$

We can deduce that $R$ is of the form : $R(X,Y)=A(X,Y) + B(X,Y).Y$ with $A,B\in \mathbb{C}[X,Y]$. But because of the hypothesis on the degree of $R$ we can write that $R(X,Y)=A(X,0)+B(X,0).Y$.

Moreover as $P\in \ker(\phi)$ we obtain that $P(\frac{e^{iT}+e^{-iT}}{2},\frac{e^{iT}-e^{-iT}}{2i})=0$ hence $R(\frac{e^{iT}+e^{-iT}}{2},\frac{e^{iT}-e^{-iT}}{2i})=0$. It gives $A(\frac{e^{iT}+e^{-iT}}{2},0)+B(\frac{e^{iT}+e^{-iT}}{2},0).(\frac{e^{iT}+e^{-iT}}{2i})=0.$ Then I am stuck to show that $R(X,Y)=0$.

Thanks in advance !

Maman
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    At a high level: the image is an integral domain, so the kernel is a prime ideal. $\mathbb{C}[X,Y]$ has dimension 2, the image has dimension $>0$, and $(X^2+Y^2-1)$ is a prime ideal of height $1$, so this forces the kernel to be exactly $(X^2+Y^2-1)$. – diracdeltafunk Aug 12 '22 at 02:32
  • Presumably this is not an acceptable proof for your course, but hopefully it helps to see that with enough theory building, one day this kind of problem becomes easier! – diracdeltafunk Aug 12 '22 at 02:33
  • Does this answer your question? Ring of trigonometric functions with real coefficients – Viktor Vaughn Aug 12 '22 at 03:03
  • @ViktorVaughn It is similar indeed (even if in your link it is in $\mathbb{R}$) but I wanted to complete the proof with the method in my question. – Maman Aug 12 '22 at 11:15
  • @diracdeltafunk thank you for the complements ! I have never heard of height of ideals before but it seems really interesting to dig ! – Maman Aug 12 '22 at 11:16
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    Should $\phi(Y) = \frac{e^{iT}-e^{-iT}}{2i}$? – cat Aug 12 '22 at 21:13
  • Also, can't you just assume that $A, B \in \mathbb C[X]$, i.e. $R(X, Y)$ is at most a degree 1 polynomial with coefficients in $\mathbb C[X]$? – cat Aug 12 '22 at 21:23
  • @cat Indeed sorry for the typo in the change of the indeterminate ! I guess you're right for the degree ! – Maman Aug 13 '22 at 11:48

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