# Ring of polynomial functions modulo the circle

I need to answer the following question:

Let $$R$$ be the ring of real polynomial functions on the circle:

$$R=\mathbb R[X,Y]/I$$, with $$I=(X^2+Y^2-1)$$

Let $$x:=X\pmod I$$ and $$y:= Y\pmod I$$

a) Prove that every residue class mod $$I$$ has a unique representative of the form $$A(X)+B(X)Y$$.

(etc)

Now I don't really know how to get started. I'm not really understanding what is happening here and how I can see or simplify this ring of twovalued polynomials modulo this circle-ideal. Could anybody help me?

• What is asked is to show for polynomial $p(x,y)$ there are unique polynomials $A(x), B(x)$ such that $p(x,y)-A(x)-B(x)y$ belongs to $I,$ which means there is a poly'l $C(x,y)$ with $p(x,y)-A(x)-B(x)y=(1-x^2-y^2)C(x,y).$ Start by treating $x$ as a constant and use synthetic division to divide $(1-x^2-y^2)$ into $p.$ Feb 24 '16 at 17:26
• Search the site also for "ring of trigonometric polynomials" Feb 24 '16 at 19:52

Starting with an arbitrary polynomial in$R[X,Y]$ you replace each $Y^2$ with $1-X^2$ since $Y^2\equiv 1-X^2\pmod I$.
You are left with just powers of $X$ and lone $Y$'s. Now just factor all the Y's to one location and argue uniqueness.
As is often the case in these kind of problems, it is helpful to use the isomorphism $\mathbb{R}[X, Y] \cong (\mathbb{R}[X])[Y]$. Then, thinking of $Y^{2}+X^{2}-1$ as a polynomial in $Y$ with coefficients in $\mathbb{R}[X]$, we see that $Y^{2}+X^{2}-1$ is a monic degree $2$ polynomial, so for any $f \in (\mathbb{R}[X])[Y]$, we can perform Euclidean division by $Y^{2}+X^{2}-1$ to write $f$ as
$$f(Y) = q(Y)(Y^{2}+X^{2}-1) + r(Y)$$
for a unique $r(Y) \in (\mathbb{R}[X])[Y]$ which has degree at most $1$ in $Y$, i.e. $r(Y)$ is of the form $A(X)+B(X)Y$ for some $A(X), B(X) \in \mathbb{R}[X]$. Reducing mod $I = \langle Y^{2}+X^{2}-1 \rangle$, this is exactly the condition given.