Help proving $I^2$ is a principal ideal and $I$ isn't with $I=(x,y-1)$ in the ring $R := \mathbb{R}[x,y]/(x^2+y^2-1)$

Question

So we have this ring $R := \mathbb{R}[x,y]/(x^2+y^2-1)$ with an ideal $I=(x,y-1)$ generated by the elements $x$ and $y-1$. I'm having trouble proving that $I^2=(x^2,x(y-1),(y-1)^2)$ is a principal ideal, because the gcd of those elements is 1. But the ideal $I^2=(1)$ is just $R$, so I dont know if that's a principal ideal. I'm also having trouble proving that $I$ is not a principal ideal, because their gcd is also $1$. So if I say that it isn't a principal ideal because their gcd is $1$ then i'm contradicting my first "proof". Is there any chance my ideal $I^2$ is wrong, from what I understood the square of an ideal generated by 2 elements $a,b$ is (a^2,ab,b^2), but I may be wrong. Does anybody have any tips?

Answer 1

Actually $I^2 \neq (1)$ (indeed the gcd of $x$ and $y-1$ is $1$ so if your argument were correct we'd have $I = R$, which is not the case because $R/I \cong \mathbb{R} \neq 0$.). This idea only works in Euclidean domains, where you can use the division algorithm to prove that $(a,b) = (\gcd(a,b))$. There is no division algorithm in $R$, so that doesn't hold here!

To clarify a couple points from your post:

• $R$ certainly is a principal ideal of $R$. Remember that a principal ideal is just any ideal of the form $(r) = \{rx : x \in R\}$ for some $r \in R$. Since $R = (1)$, this is principal.

• You're correct that $(a,b)^2 = (a^2, ab, b^2)$. Indeed, generally we have that $(x_1, \dots, x_n)^2$ is the ideal generated by all products $x_i x_j$ where $i,j \in \{1, \dots, n\}$. You should try to prove this! Hint: show both sets are subsets of one another.

Anyway, to show that $I^2$ is principal, you should try to find some element $p \in R$ such that the image of $I$ is $0$ under the quotient $R \to R/(p)$ – this will tell you that $I^2 \subseteq (p)$, and you will only need to prove $p \in I^2$ to conclude that $I^2 = (p)$. To show that $I$ is not principal, you could try to argue by contradiction: e.g. it's possible to prove that $R/(p) \not\cong \mathbb{R} \cong R/I$ for any $p \in R$ (first of all, if $R/(p) \cong \mathbb{R}$ then $(p)$ is maximal...).

If you know anything about algebraic geometry, you can gain a lot of intuition here by noting that $R$ is the coordinate ring of the circle.