Lemma 54.12.2. Let $\kappa$ be a field. Let $I \subset \kappa [x, y]$ be an ideal. Let

$a + b x + c y + d x^2 + exy + f y^2 \in I^2$

for some $a, b, c, d, e, f \in k$ not all zero. If the colength of $I$ in $\kappa [x, y]$ is $> 1$, then $a + b x + c y + d x^2 + exy + f y^2 = j(g + hx + iy)^2$ for some $j, g, h, i \in \kappa$.

Proof. Consider the partial derivatives $b + 2dx + ey$ and $c + ex + 2fy$. By the Leibniz rules these are contained in $I$. If one of these is nonzero, then after a linear change of coordinates, i.e., of the form $x \mapsto \alpha + \beta x + \gamma y$ and $y \mapsto \delta + \epsilon x + \zeta y$, we may assume that $x \in I$. Then we see that $I = (x)$ or $I = (x, F)$ with $F$ a monic polynomial of degree $\geq 2$ in $y$. In the first case the statement is clear. In the second case observe that we can write any element in $I^2$ in the form

$A(x, y) x^2 + B(y) x F + C(y) F^2$

for some $A(x, y) \in \kappa [x, y]$ and $B, C \in \kappa [y]$. Thus

$a + b x + c y + d x^2 + exy + f y^2 = A(x, y) x^2 + B(y) x F + C(y) F^2$

and by degree reasons we see that $B = C = 0$ and $A$ is a constant.

To finish the proof we need to deal with the case that both partial derivatives are zero. This can only happen in characteristic $2$ and then we get

$a + d x^2 + f y^2 \in I^2$

We may assume $f$ is nonzero (if not, then switch the roles of $x$ and $y$). After dividing by $f$ we obtain the case where the characteristic of $\kappa$ is $2$ and

$a + d x^2 + y^2 \in I^2$

If $a$ and $d$ are squares in $\kappa$, then we are done. If not, then there exists a derivation $\theta : \kappa \to \kappa$ with $\theta (a) \not= 0$ or $\theta (d) \not= 0$, see Algebra, Lemma 10.158.2. We can extend this to a derivation of $\kappa [x, y]$ by setting $\theta (x) = \theta (y) = 0$. Then we find that

$\theta (a) + \theta (d) x^2 \in I$

The case $\theta (d) = 0$ is absurd. Thus we may assume that $\alpha + x^2 \in I$ for some $\alpha \in \kappa$. Combining with the above we find that $a + \alpha d + y^2 \in I$. Hence

$J = (\alpha + x^2, a + \alpha d + y^2) \subset I$

with codimension at most $2$. Observe that $J/J^2$ is free over $\kappa [x, y]/J$ with basis $\alpha + x^2$ and $a + \alpha d + y^2$. Thus $a + d x^2 + y^2 = 1 \cdot (a + \alpha d + y^2) + d \cdot (\alpha + x^2) \in I^2$ implies that the inclusion $J \subset I$ is strict. Thus we find a nonzero element of the form $g + hx + iy + jxy$ in $I$. If $j = 0$, then $I$ contains a linear form and we can conclude as in the first paragraph. Thus $j \not= 0$ and $\dim _\kappa (I/J) = 1$ (otherwise we could find an element as above in $I$ with $j = 0$). We conclude that $I$ has the form $(\alpha + x^2, \beta + y^2, g + hx + iy + jxy)$ with $j \not= 0$ and has colength $3$. In this case $a + dx^2 + y^2 \in I^2$ is impossible. This can be shown by a direct computation, but we prefer to argue as follows. Namely, to prove this statement we may assume that $\kappa$ is algebraically closed. Then we can do a coordinate change $x \mapsto \sqrt{\alpha } + x$ and $y \mapsto \sqrt{\beta } + y$ and assume that $I = (x^2, y^2, g' + h'x + i'y + jxy)$ with the same $j$. Then $g' = h' = i' = 0$ otherwise the colength of $I$ is not $3$. Thus we get $I = (x^2, y^2, xy)$ and the result is clear. $\square$

There are also:

• 4 comment(s) on Section 54.12: Rational double points

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).