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 $A(X)+B(X)Y$.


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?

  • $\begingroup$ 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.$ $\endgroup$ Feb 24 '16 at 17:26
  • $\begingroup$ Search the site also for "ring of trigonometric polynomials" $\endgroup$
    – rschwieb
    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.


The answer rschwieb has given is likely the best way to see what is happening. Nevertheless, I'd like to point out that one doesn't have to actually compute anything.

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.


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