Hello everyone I would like to know if
$R$:= $\Bbb{R}[X,Y]/(X^2+Y^2)$ is UFD or Noetherian.
I'm not really confortable in seeing how $\Bbb{R}[X,Y]/(X^2+Y^2)$ looks like. From what i've understood this is the set of all the equivalence classes
$$[p]\quad s.t \quad [p] = p + (X^2+Y^2)$$
Where $p \in \Bbb{R}[X,Y]$ is a polynomial in two variables and $(X^2+Y^2)$ is the ideal generated by the polynomial $X^2+Y^2$ that is $\{r\,(X^2+Y^2)\,|\,r\in R \}$. Is this correct? What are the element $r \in R$ that helps generating the ideal? Are they maybe elements of $\Bbb{R}[X,Y]$? Sorry if my questions sound stupid, but i'm really confused and it will be great if you help me figure this out.
I also tried to show that $R$ is not a UFD. I've thought that if it was, then every element should have had a factorization into irreducible elements of $\Bbb{R}[X,Y]/(X^2+Y^2)$ but if I take the equivalence class $[0]=(X^2+Y^2)$ i see that it cannot be found any factorization of $X^2+Y^2$ since this polynomial is irreducible in $\Bbb{R}[X,Y]$. Is it correct?
Also to show that $R$ is Noetherian or not i was thinking to show that in $R$ every ideal is finitely generated or that there is an ideal which is not finitely generated, but i don't know how the ideals of $R$ look like, could please help me? Thank you in advance
One way of thinking about this is to say that factoring out by the ideal is equivalent to setting $X^2+Y^2=0$ in the original ring. Hence $Y^2=-X^2$ and one can find a form of each element (coset representative) where $Y$ does not appear to any power higher than $1$, which is the same as saying the elements of the factor ring can be taken to be of the form $p(X)+q(X)Y$ where $p, q$ are arbitrary polynomials with real coefficients.
Can you see here a factorisation which is not unique?
If you have a polynomial like $X^2+3X+2+5XY^5+2X^3Y^2+20Y$ you have $5XY^3(X^2+Y^2)=0$ in the factor ring, so that $5XY^5=-5X^3Y^3$, and we have reduced the exponent of $Y$ by $2$. We can do the same trick again to give us $5XY^5=-5X^3Y^3=5X^5Y$.
Term by term we can reduce the exponent of $Y$ to $0$ or $1$.
Look carefully at $Y^2=-X^2$.