Quadratics in square ideals of colength greater than one

The two partial derivatives of $q$ lie in $I$ by the Leibniz rule, because $q\in I^2$ and derivations of $\kappa [x,y]$ carry $I^2$ into $I$; if a nonzero constant were among these derivatives, then $I$ would contain a unit and the hypothesis of colength greater than one would fail.

Suppose a nonconstant linear derivative is nonzero; after an affine change of coordinates one has $x\in I$, and then either $I=(x)$ or $I=(x,F(y))$ with $F$ monic of degree at least two. Reducing an element of $I^2$ modulo $x^2$ shows that the coefficient of $x$ is divisible by $F$ and the part constant in $x$ is divisible by $F^2$; the degree bound on $q$ then forces both parts to vanish, leaving $q=$ a constant times $x^2$.

Remaining case: both partial derivatives vanish, so the characteristic is two and $q=a+d{x}^2+f{y}^2$. A nonzero constant cannot lie in a proper $I^2$. If only one variable occurs, normalize its quadratic coefficient to $1$; differentiating the remaining scalar coefficient gives a nonzero constant in $I$ unless its ratio to the quadratic coefficient is a square, in which case $q$ is already a scalar multiple of a square. Now suppose both variables occur and normalize $f=1$. If $a$ and $d$ are both squares, then $q$ is a square. If $d=s^2$ and $a$ is nonsquare, the invertible coordinate change $z=y+sx$ gives $q=a+z^2$ in characteristic two, reducing to the one-variable case just treated and contradicting its nonsquare alternative. Thus the remaining case has $d$ nonsquare.

Since $d$ is nonsquare, choose by [F5] a field derivation $\theta$ with $\theta (d)\mathrm{\ne }0$, and extend it to $\kappa [x,y]$ by $\theta (x)=\theta (y)=0$. Because derivations carry $I^2$ into $I$, applying $\theta$ to $q$ gives $\theta (a)+\theta (d)x^2\in I$; put $\alpha =\theta (a)\mathrm{/}\theta (d)$, so $x^2+\alpha \in I$. Also $q+d(x^2+\alpha )=y^2+a+d\alpha \in I$, hence $J=(x^2+\alpha ,y^2+a+d\alpha )\subseteq I$. The quotient $\kappa [x,y]\mathrm{/}J$ has basis $1,x,y,xy$. As $q$ is a nonzero linear combination of these two regular-sequence generators, it is not in $J^2$, so $I\mathrm{/}J\mathrm{\ne }0$.

If $I\mathrm{/}J$ contains a nonzero vector with zero $xy$-coefficient, then $I$ contains a nonzero affine linear polynomial, and the case of step 2.1 applies. Otherwise $I\mathrm{/}J$ is one-dimensional and $I$ has colength three; extending scalars to an algebraic closure and translating makes $J=(x^2,y^2)$, and the one-dimensional added ideal is spanned by $g+hx+iy+jxy$ with $j\mathrm{\ne }0$.

In the situation of step 5.1, $g\mathrm{\ne }0$ makes that element a unit in the four-dimensional local algebra $\kappa [x,y]\mathrm{/}(x^2,y^2)$, while $h\mathrm{\ne }0$ or $i\mathrm{\ne }0$ makes its ideal contain two independent vectors, so the quotient would have length at most two; both contradict length three. Hence $I=(x^2,xy,y^2)$ after the extension, whose square contains no nonzero polynomial of degree at most two, contradicting $q\in I^2$.

All cases are excluded except the one where $q$ is a scalar times the square of an affine linear polynomial, which proves the claim; infinite colength is allowed throughout because the arguments use only the two generators of the relevant ideals.