0
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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? $\endgroup$ – Jaca Mar 8 at 19:16
  • $\begingroup$ I wanna show the latter . $\endgroup$ – buoyant Mar 9 at 4:04
2
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Mar 16 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.