1
$\begingroup$

The exericse is as follows: Let $A, B \in \overline{K}$. Characterize the values of A and B for which the following variety is singular. $$ V: Y^2 Z+ AXYZ + BYZ^2=X^3 $$ First assume that $B=0$. Then we find that the point $[0:0:1]$ is singular; it obviously lies on the curve and the jacobian on the $Z=1$ patch gives the conditions $$ AY-3X^2=0 \text{ and } 2Y+AX=0 $$ which also obviously hold.

Now assume $B\neq 0$. I would proced as follows. Assume there is a point $P\in V$ which is singular. We have to extract some relation between $A$ and $B$. First let's check the $Z=1$ piece again. $$ \begin{align*} Y^2+AXY+BY-X^3&=0\\ AY-3X^2&=0\\ 2Y+AX+B &= 0 \end{align*} $$

We notice that we can assume $X\neq 0$ and $Y\neq 0$. Because if one is zero, the other has to be, because of the second equation. However both being zero violates the third equation.

Considering the first two equations we find the linear system $$ \begin{pmatrix} XY & Y\\ Y & 0 \end{pmatrix}\begin{pmatrix} A\\ B \end{pmatrix} = \begin{pmatrix} X^3-Y^2 \\ -3X^2 \end{pmatrix} $$ So we obtain $BY=-2X^3-Y^2$ and $AY=3X^2$.

And this is basically where I am stuck. I can now play around with this and the thrid equation and get some expressions. Of course I tried some variations already as well, but from the solution given in the back of the book I should arrive at $B(A^3-27B)=0$ and I don't see how to get there.

$\endgroup$
1
$\begingroup$

Solving for $Y$ from the last equation gives $$ Y=-\frac12(B+AX).\qquad(*) $$ Plugging $(*)$ into the first equation gives $$ f(X):=-\frac{B^2}4-\frac{ABX}2-\frac{A^2X^2}4-X^3=0. $$ Plugging $(*)$ into the second equation gives $$ g(X):=-\frac{AB}2-\frac{A^2X}2-3X^2. $$ With $Y$ thus eliminated we need to check when $f(X)$ and $g(X)$ might have a common zero. The tool for that is their resultant. They have a common zero if and only if the resultant $R(f,g)$ vanishes. Also sprach Mathematica $$ R(f,g)=\frac1{16}(A^3-27B)B^3. $$ This is zero iff $B(A^3-27B)=0$ confirming the answer in the back of your book.

$\endgroup$
  • 1
    $\begingroup$ If you don't want to compute the resultant you can proceed and produce a sequence of polynomials in $X$. Progressively eliminating higher degree terms using suitable linear combinations of $f$ and $g$ (and the intermediate linear combinations). Any common zero of $f(X)$ and $g(X)$ is also a zero of all such linear combinations. That process may introduce extra factors depending on how careful you are. I got an extra $A^2$ in my first attempt :-) $\endgroup$ – Jyrki Lahtonen Oct 14 '17 at 21:12
  • $\begingroup$ Just to make sure I understand: in order to fully answer the question, one should also look at the $Y=1$ patch (by symmetry, this gives the same set of solutions as the $Z=1$ patch), and also notice that the only point not covered by those two patches, $[1\,:\,0\,:\,0]$, does not lie on the curve. Right? $\endgroup$ – rogerl Mar 4 '18 at 15:49

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.