Ring of trigonometric functions with real coefficients Let $R$ be the ring of functions that are polynomials in $\cos t$ and $\sin t$ with real coefficients. 


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*Prove that $R$ is isomorphic to $\mathbb R[x,y]/(x^2+y^2-1)$. 

*Prove that $R$ is not a unique factorization domain.

*Prove that $S=\mathbb C[x,y]/(x^2+y^2-1)$ is a principal ideal domain and hence a unique factorization domain. 

*Determine the units of the rings $S$ and $R$. (Hint: Show that $S$ is isomorphic to the Laurent polynomial ring $\mathbb C[u,u^{-1}]$.)

(1) Send $X$ to $\cos$ and $Y$ to $\sin$. The kernel of this homomorphism consists from the polynomials $f\in\mathbb{R}[X,Y]$ with the property $f(\cos t,\sin t)=0$ for any $t\in\mathbb{R}$. Now prove that this implies $f$ divisible by $X^2+Y^2-1$: consider $f$ as a polynomial in $X$ with coefficients in $\mathbb{R}[Y]$ and write $f=(X^2+Y^2-1)g+r$, where $\deg_Xr\le 1$. Then $r=a+bX$, where $a,b\in\mathbb{R}[Y]$. From $f(\cos t,\sin t)=0$ for any $t\in\mathbb{R}$ we get that $r(\cos t,\sin t)=0$ for any $t\in\mathbb{R}$, that is, $a(\sin t)+b(\sin t)\cos t=0$ for any $t\in\mathbb{R}$. This gives us that $a^2(\sin t)=b^2(\sin t)(1-\sin^2t)$ for any $t\in\mathbb{R}$. But $a,b$ are polynomials and the last relation implies that $a^2(Y)=b^2(Y)(1-Y^2)$ and this is enough to deduce $a=b=0$ (why?).
(2) We show that $x$, the residue class of $X$ in $\mathbb{R}[X,Y]/(X^2+Y^2-1)$ is irreducible and does not divide $1+y$ and $1-y$, the residue classes of $1+Y$ and $1-Y$. (Note that in $\mathbb{R}[x,y]$ we have $x^2=(1+y)(1-y)$.)
Edit. If $x=z_1z_2$, then, by using (4) (see below) we get $N(x)=N(z_1)N(z_2)$ $\Leftrightarrow$ $Y^2-1=N(z_1)N(z_2)$. Now we have the following cases:
(i) $\deg N(z_1)=0\Leftrightarrow z_1\in\mathbb R^*$,
(ii) $\deg N(z_1)=2$ $\Leftrightarrow$ $\deg N(z_2)=0$ $\Leftrightarrow$ $z_2\in\mathbb R^*$,
(iii) $\deg N(z_1)=\deg N(z_2)=1$. If $N(z_1)=Y-1$, then $a_1^2(Y)+b_1^2(Y)(Y^2-1)=Y-1\Rightarrow Y-1\mid a_1(Y)\Rightarrow\exists a_2\in\mathbb R[Y]$ such that $a_1=(Y-1)a_2$ and pluggin this in the foregoing equation we get $a_2^2(Y)(Y-1)+b_1^2(Y)(Y+1)=1$. Looking now at the leading coefficients of $a_2$ and $b_1$ we find that one of these is zero (false!) or the sum of their square is zero (false!).
Assume that $x\mid 1-y$. Then $N(x)\mid N(1-y)\Leftrightarrow Y^2-1\mid (Y-1)^2$, a contradiction.
(3) I've proved here all you need for this part.
(4) As one can see from $(1)$ the elements of $\mathbb{R}[x,y]$ can be uniquely written as $a(y)+b(y)x$. Define a "norm" $N:\mathbb{R}[x,y]\to\mathbb{R}[Y]$ by $N(a(y)+b(y)x)=a^2(Y)+b^2(Y)(Y^2-1)$. $N$ is multiplicative and using this we get that the units of $\mathbb{R}[x,y]$ are the non-zero elements of $\mathbb{R}$.
For $\mathbb{C}[x,y]$ we can see, via the isomorphism from part $(3)$ that the invertible elements are the non-zero elements of $\mathbb{C}$, the powers of $x+iy$ and $x-iy$, and products of the non-zero elements of $\mathbb{C}$ with the powers of $x+iy$ and $x-iy$.
