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.
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 theorem, we have $2d = 1 + r + s$ wherep $r,s$ 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).