$(\bar x , \bar y-1)$ is not a principal ideal of $A=\mathbb R[x,y]/(x^2+y^2-1)$

Question

Let $A=\mathbb R[x,y]/(x^2+y^2-1)$ ; then $A$ is an integral domain . Consider the maximal ideal $P=(\bar x , \bar y-1)$ of $A$ . Then how to show that $P$ is not a principal ideal ?

Answer 1

we can consider $S=R[x,y]/<x,x^2+y^2-1>$is isomorphic to $R[y]/<y^2-1>$.remark $S$ is a principal ideal ring.if the ideal you give is principal.denote $<f^->=<x^-,(y-1)^->$.in S,they also equal.So we can get $f=k(y-1)+gx+h(x^2+y^2-1)$,where $g,h\in R[x,y],k\in R$.the same argument,we can get $f=k(y-1)+lx+h(x^2+y^2-1)$,where $k,l\in R$,this is a contradiction.

Answer 2

We reason by contradiction, considering a generator $g$ of the principal ideal $(\overline{x},\overline{y} - 1)$. Let $B = \mathbb{R}[t, \frac{1}{1 + t^2}]$, that is, the localization $S^{-1} \mathbb{R}[t]$ where $S \subset \mathbb{R}[t]$ is the multiplicative set generated by $1 + t^2$. The map $(x, y) \mapsto (\frac{1 - t^2}{1 + t^2}, \frac{2t}{1 + t^2})$ induces a ring homomorphism $\phi: A \rightarrow B$. (This homomorphism can be checked to be injective, but we won't use that fact.) The ring $B$ is Euclidean, hence a UFD and it easily follows that $\phi(g)$ is of the form $\lambda (1 - t)(1 + t^2)^m$ for some $\lambda \in \mathbb{R} \setminus \{0\}$ and some $m \in \mathbb{Z}$. Let $C$ be the ring of functions from $\mathbb{R}$ to $\mathbb{C}$ of the form $\theta \mapsto \sum_{n = -N}^N \lambda_n e^{in\theta}$ with $\lambda_n \in \mathbb{C}$ and $N \ge 0$.Let $Q$ be the fraction field of $C$. It is immediate to see that $P(X) \mapsto (\theta \mapsto P(e^{i\theta}))$ induces a ring isomorphism from $\mathbb{C}[X^{\pm 1}]$ (resp. $\mathbb{C}(X)$) onto $C$ (resp. $Q$). The map $t \mapsto \tan(\frac{\theta}{2}) \Doteq \frac{e^{i\theta} - 1}{i(e^{i\theta} + 1)}$ induces a ring homomorphism $\psi: B \rightarrow Q$ such that $\psi \circ \phi(A) \subset C$ (check what are the images of $\overline{x}$ and $\overline{y}$ in $Q$). Since $\psi \circ \phi(g) = 2^m\lambda \frac{e^{i m \theta}((1 + i)e^{i\theta} + 1 - i)}{(e^{2i\theta} + 1)^m(e^{i\theta} + 1)} \in C$, we have $2^m\lambda \frac{X^m ((1 + i)X + 1 - i)}{(X^2 + 1)^m (X + 1)} \in \mathbb{C}[X^{\pm 1}]$. This is impossible because $-1$ is not a root of the numerator of the latter fraction.

Answer 3

TLDR Outline: If $g$ were a generator of the ideal, then over $\mathbb{C}[x,y] / \langle x^2 + y^2 - 1 \rangle$, $g$ would have to be a unit times $x+iy-i$. The lemma below gives a characterization of the group of units of this ring. Combining these gives that a unit times $x+iy-i$ is never a real polynomial, contradicting the assumption that $g \in A$ generates the ideal.

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Full proof: First, observe that if we use base extension from $\mathbb{R}$ to $\mathbb{C}$, then the ideal becomes principal with generator $x+iy-i$. Namely, $y-1 = -\frac{1}{2} [x^2 + (y-1)^2] = -\frac{1}{2} (x-iy+i) (x+iy-i)$, and then $x = (x+iy-i) - i (y - 1)$. Therefore, if $g$ were a generator of $\langle x, y-1 \rangle$ in $A$, then $g$ would have to be equal to a unit of $A \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathbb{C}[x,y] / \langle x^2 + y^2 - 1 \rangle$ times $x+iy-i$. From here, it will be useful to see what the units of $\mathbb{C}[x,y] / \langle x^2 + y^2 - 1 \rangle$ actually are.

Lemma: Every unit of $\mathbb{C}[x,y]/\langle x^2+y^2-1 \rangle$ is equal to $\lambda (x+iy)^n$ for some $\lambda \in \mathbb{C}^*$, $n \in \mathbb{Z}$ (where we use $(x+iy)^{-1} = x-iy$ to calculate negative powers).

Proof: Every element of $\mathbb{C}[x,y]/\langle x^2 + y^2 - 1 \rangle$ can be uniquely written as $a(x) + b(x) y$ for some $a, b \in \mathbb{C}[x]$. Now suppose $a(x) + b(x) y$ is a unit with inverse $c(x) + d(x) y$. Then by writing out the multiplication, we get: $$a(x) c(x) + b(x) d(x) (1-x^2) = 1,\\ a(x) d(x) + b(x) c(x) = 0.$$ Now if $b = 0$ or $d = 0$, then $a(x)$ is a nonzero constant polynomial, so $a(x) + y b(x) = a (x+iy)^0$. Otherwise, if $b$ and $d$ are both nonzero, then the first equation implies $a$ and $c$ are also both nonzero. We now prove this case by induction on $\deg b$.

Looking at the degrees in the above equations, we see $\deg a + \deg c = \deg b + \deg d + 2$ and $\deg a + \deg d = \deg b + \deg c$. Solving, we find $\deg a = \deg b + 1$, $\deg c = \deg d + 1$. Letting $A := a(x) / x^{\deg a}$ and similarly for $B, C, D$, we get polynomials in $x^{-1}$ with $AC - BD (1-x^{-2}) = 0$ and $AD + BC = 0$. Truncating terms of $x^{-2}, x^{-3}$ and so on (i.e. considering images in $\mathbb{C}[x^{-1}] / \langle x^{-2} \rangle$), we see $A C \equiv B D \pmod{x^{-2}}$ and $A D = - B C$, from which we can conclude $B \equiv \pm i A \pmod{x^{-2}}$ (since $A,B,C,D$ are all units in this quotient ring, we get $A^2 \equiv - B^2 \pmod{x^{-2}}$; and also, either $B - iA$ or $B + iA$ must be a unit).

Now, if $B \equiv i A \pmod{x^{-2}}$, then $$(a(x) + b(x) y) (x - iy) = (x a(x) + i (x^2 - 1) b(x)) + (x b(x) - i a(x)) y.$$ By the assumption, $x b(x) - i a(x)$ has degree strictly smaller than $\deg b$ since the two leading terms cancel; therefore, by inductive hypothesis $(a(x) + b(x) y) (x - iy) = \lambda (x + iy)^n$ for some $\lambda \in \mathbb{C}^*, n \in \mathbb{Z}$. It follows that $a(x) + b(x) y = \lambda (x + iy)^{n+1}$. (Either that, or $x b(x) - i a(x) = 0$ in which case $(a(x) + b(x) y) (x - iy) = \lambda (x + iy)^0$ by the special case we considered first.) Similarly, if $B \equiv -i A \pmod{x^{-2}}$, by inductive hypothesis we get $(a(x) + b(x) y) (x + iy) = \lambda (x + iy)^n$ for some $\lambda \in \mathbb{C}^*, n \in \mathbb{Z}$ so $a(x) + b(x) y = \lambda (x + iy)^{n-1}$. $\Box$

Now, observe that $(x + iy)^n = p_n(x) + i q_n(x) y$, where $p_n$ and $q_n$ are the Chebyshev polynomials such that $p_n(\cos \theta) = \cos (n \theta)$ and $q_n(\cos \theta) \sin \theta = \sin (n \theta)$. We know that $\deg p_n = |n|$ and $\deg q_n = |n|-1$ (except that for $n=0$, $q_n(x) = 0$); and also that $p_n, q_n$ are real polynomials. Now, suppose $\lambda (x + iy)^n (x+iy - i)$ were a real polynomial. Then it would be equal to $\lambda (p_{n+1}(x) - i p_n(x)) + \lambda i (q_{n+1}(x) - i q_n(x)) y$. If $n \ge 0$, then the leading coefficient of $p_{n+1}(x) - i p_n(x)$ is real, so $\lambda$ would have to be real. But then $i p_n(x)$ would also have to be real, giving a contradiction. Similarly, if $n < 0$, then the leading term of $p_{n+1}(x) - i p_n(x)$ is pure imaginary, so $\lambda$ would have to be pure imaginary, implying $i p_{n+1}(x)$ is real, giving a contradiction.

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Note: I came here from trying to find an example of a f.g. module over an integral domain which is projective (of constant rank) but not free. The ideal is a projective module since localizing by $y-1$, the ideal is generated by $y-1$ since $x = \frac{x}{y-1} (y-1)$; while localizing by $y+1$, the ideal is generated by $x$ since $y-1 = -\frac{x}{y+1} \cdot x$; and $y-1, y+1$ generate the unit ideal. I was expecting to find the Picard group of the circle $\operatorname{Spec} \mathbb{C}[x,y] / \langle x^2 + y^2 - 1 \rangle$ to be nontrivial since the divisor $(0,1)$ has odd rank, while in the projective closure $\operatorname{Proj} \mathbb{C}[x,y,z] / \langle x^2 + y^2 - z^2 \rangle$ every principal divisor must have even rank by Bezout's theorem. But as I found out, it's actually possible to have principal divisors with odd rank on the affine circle (and therefore odd rank intersected with the points at infinity $[1 : i : 0], [1 : -i : 0]$). However, as this shows, restricting the scalar field to $\mathbb{R}$ disallows that case (though then it's not algebraically complete so certain algebraic geometry techniques won't necessarily work) and we do get the example I was searching for.

It's also interesting to note that $I \otimes_A I \simeq I^2 \simeq A$ (geometrically, $2 (0,1)$ is the principal divisor cut out by $y-1$). I also seem to have found that $I \oplus I \simeq A \oplus A$ by considering a free resolution of $I$.