# $(1-x,y)$ is not principal in $\Bbb Q[x,y]/(x^2＋y^2-1)$

I want to show $$(1-x,y)$$ is not principal in $$\Bbb Q[x,y]/(x^2＋y^2-1)$$,

I challenged this problem like trying to show there is no $$f(x,y),g(x,y)$$ such that $$1-x=f(x,y)g(x,y)+(x^2+y^2-1).$$

But I cannot go further.My attempt may be fruitless. I would be appreciated if you solve it in the way above or in completely another way.

• Do you wanna show that $(1-x,y)$ is not principal or its image in $\mathbb{Q}[x,y]/(x^2+y^2-1)$, i.e. $\overline{(1-x,y)}$, is not principal? – Jaca Mar 8 at 19:16
• I wanna show the latter . – buoyant Mar 9 at 4:04

Take $$A = \mathbb{Q}[x,y]/(x^2 + y^2 - 1)$$.

First note that there's a natural injection of rings $$\mathbb{Q}[x] \hookrightarrow A$$ by $$f(x) \mapsto f(x) + (x^2 + y^2 - 1)$$ (that it's an injection follows from comparing the degree wrt $$y$$).

Second, any element of $$A$$ is of the form $$s = a(x) + b(x)y + (x^2 + y^2 - 1)$$ with $$a,b \in \mathbb{Q}[x]$$ and $$s = 0$$ iff $$a = 0$$ and $$b = 0$$. This is again easy to verify by comparing the degrees wrt $$y$$.

Third, the units of $$A$$ are precisely $$\mathbb{Q}^{*} = \mathbb{Q} \setminus \{ 0 \}$$, the image of the natural embedding $$\mathbb{Q}^* \to A$$ via $$\alpha \mapsto \alpha + (x^2 + y^2 - 1)$$. This is where we need that $$\mathbb{Q}$$ is a subfield of $$\mathbb{R}$$ (I'll make a note later that $$(x-1,y)$$ becomes principal if we extend the scalars from $$\mathbb{Q}$$ to a field containing $$i$$).

To check this, suppose $$st = 1$$ in $$A$$, with $$t = c(x) + d(x)y + (x^2 + y^2 + 1)$$ and with $$s = a(x) + b(x)y + (x^2 + y^2 + 1)$$ as above. Then $$ac + bd(1 - x^2) + (ad + bc)y + (x^2 + y^2 - 1) = 1$$ in $$A$$, so by the second note we must have $$ac + bd(1 - x^2) = 1$$ and $$ad + bc = 0$$ in $$\mathbb{Q}[x]$$. Multiplying the first equation by $$b$$ and using the second relation, we have $$b = bac + b^2 d(1 - x^2) = -a^2 d + b^2 d (1 -x ^2)$$, so that $$d$$ divides $$b$$ in $$\mathbb{Q}[x]$$. Likewise, $$b$$ divides $$d$$, so $$d = \alpha b$$ for some $$\alpha \in \mathbb{Q}^*$$. If $$b = 0$$ then $$d = 0$$ so $$s$$ is a unit in $$\mathbb{Q}[x]$$, hence the claim. So assume that $$b \neq 0$$. Then from $$ad = -bc$$ we obtain $$c = -\alpha a$$. So $$st = 1$$ writes as $$-\alpha (a^2 + b^2 x^2 - b^2) = 1$$ in $$\mathbb{Q}[x]$$. But note that $$a^2 + b^2 x^2$$ must have a positive leading term (sum of squares is nonnegative in $$\mathbb{R}$$) and its degree is strictly bigger than that of $$b^2$$. Therefore the equation $$-\alpha (a^2 + b^2 x^2 - b^2) = 1$$ is impossible in $$\mathbb{Q}[x]$$.

Fourth, calculating inside the ring $$A$$ (so slight abuse of notation in what ensues), we have that the square of the ideal in question, $$(x-1,y)^2 = ((x-1)^2 , (x-1)y, y^2) = ((x-1)^2 , (x-1)y, 1 - x^2) = (x-1, (x - 1)y) = (x-1)$$. That is, the square of the ideal in $$A$$ is principal and is generated by $$x-1$$. Suppose, for contradiction, that $$s = a(x) + b(x)y + (x^2 + y^2 - 1) \in A$$ generates $$(x-1,y)$$ in $$A$$. Then $$(s^2) = (x - 1)$$ in $$A$$, so $$a^2 + b^2 (1 - x^2) + 2aby + (x^2 + y^2 - 1) = \alpha (x - 1) + (x^2 + y^2 - 1)$$ for some $$\alpha \in \mathbb{Q}^*$$ by the third note, so $$a^2 + b^2 (1 - x^2) = \alpha (x - 1)$$ and $$2ab = 0$$. So $$a = 0$$ or $$b = 0$$. Either case, we have impossible equations (if $$a = 0$$ then $$b^2 ( 1 - x^2) = \alpha (x-1)$$ and if $$b = 0$$ then $$a^2 = \alpha (x-1)$$, and $$\alpha$$ is scalar so these are impossible equations in $$\mathbb{Q}[x]$$).

Note, however, that your ideal becomes principal when we extend the scalars to a field containing $$i$$. $$(x-1, y) = (y - i(x-1))$$ inside the ring $$B = \mathbb{C}[x,y]/(x^2 + y^2 - 1$$). The RHS is clearly contained in the LHS. For the other inclusion, $$(y - i(x-1))(y + i(x-1)) = y^2 + (x-1)^2 = 1 - x^2 + x^2 - 2x + 1 = -2(x - 1)$$, calculating in $$B$$. So the RHS contains $$x-1$$ and so it must contain $$y$$ as well.

There's a more geometric way to understand this. That $$(x-1,y)$$ is principal in $$A$$ means that $$(x-1,y) = (f(x,y) , x^2 + y^2 - 1)$$ in $$\mathbb{Q}[x,y]$$, for some $$f \in \mathbb{Q}[x,y]$$ (we may as well assume that this is nonzero). This equation must hold then after extending scalar to $$\mathbb{C}$$ (or just to algebraic closure of $$\mathbb{Q}$$ is fine). Projectifying and working over $$\mathbb{P}_{\mathbb{C}}^2$$, what this says is that the projective curve defined by $$F(X,Y,Z) = Z^{d}f(X/Z, Y/Z)$$, with $$d$$ the total degree of $$f$$, and $$X^2 + Y^2 - Z^2$$ meet at intersection multiplicity exactly $$1$$ at the point $$(1:0:1)$$ and meet at no other point in the affine chart $$Z \neq 0$$. The other points of potential intersections in $$\mathbb{P}^2$$ are easy to calculate (setting $$Z = 0$$): They are $$(\pm i : 1 : 0)$$. By Bezout's, we have $$2d = 1 + r + s$$ where $$r,s$$ are intersection multiplicities at $$(i : 1: 0)$$ and $$(-i : 1 : 0)$$, respectively. Since $$F$$ is a rational polynomial, the "complex conjugation" is an involution of $$\mathbb{P}_{\mathbb{C}}^2$$ taking the projective varieties $$X^2 + Y^2 - Z^2 = 0$$ and $$F = 0$$ to themselves and mapping $$(i: 0 : 1)$$ to $$(-i : 0 : 1)$$, so $$r = s$$ and $$2d = 1 + 2r$$, a contradiction (even $$\neq$$ odd).

• Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Mar 16 at 21:03