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Let $$ F(x,y) = y^2 + x^2 - y - x \\ G(x,y) = y^2 - x^2 - 6xy + 11y + 7x - 12$$ both on $\Bbb{C}[x,y]$ . Compute $\mathcal{Z}(F,G) = \{ (x,y) \in \Bbb{C}^2 : F(x,y) = 0 \, \land \, G(x,y) = 0 \} $

I am trying to do this excercise from my Algebraic curves book and I am getting confused since I have realised that it is not as easy as it seems. On many videos and websites, when showing how to find the intersection of two conics, it is always "perfect" in the sense that it is easy to find a two grade equation for $x$ or for $y$ or directly to find the values of one of them. This is not the case and I have search on this website for a Method to compute this but I am getting more confused.

On this post, the author says:

You can work by considering the pencil containing the two conics ($\lambda f(x,y)+(1-\lambda) g(x,y)=0$) and find its degenerate element, which is formed of two straight lines. Then intersecting the straight lines with one of the conics is straightforward.

And I do not understund it. What does mean to find "its degenerate element" and which are the straight lines that are mentioned? Any possible help or explanations would be very appreciated.

Superdivinidad
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    From $F(x,y)=0∧G(x,y)=0$ you can deduce that also $F(x,y)-G(x,y)=0$ and that one then eliminates $y^2$. Applying the midnight formula results in $x=f(y)$. - Does that help? – Dr. Richard Klitzing Sep 24 '24 at 17:00
  • @Dr.RichardKlitzing No much because what remains after applicate the Midnight formula is a complicated expression that only leads to many calculations when substituting on one of the expressions of $F$ or $G$ – Superdivinidad Sep 24 '24 at 17:18
  • Given the very specific choice of integers appearing as coefficients, I suspect that they wanted you to "guess" that $(1,1)$ and $(0,1)$ are solutions. – Ted Shifrin Sep 24 '24 at 17:40
  • @TedShifrin Yes, I note it, but there can be 2 more points since the Peano Theorem and that we are working on $\Bbb{C}$, am I right? – Superdivinidad Sep 24 '24 at 17:45
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    Sure, you're right. But perhaps if you substitute $z=y-1$ and rewrite the equations, you'll end up with a system of equations you can easily solve. – Ted Shifrin Sep 24 '24 at 18:06
  • In particular, using Ted's $y = 1 + z$ (I used letter $t$ ) makes the sum of the two equations come out very nicely – Will Jagy Sep 24 '24 at 18:12
  • Thanks both for your help, but this is a very particular way to solve it to this specific curves and, as I asked, I was searching for a Method that can be applied more generally, I do not have faith on my capacity to intuit a propper substitution in other cases. Either it is applying what I linked or other method it is fine. – Superdivinidad Sep 24 '24 at 18:42
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    I think what you're looking for can be found in my answer to this question –  Sep 24 '24 at 21:21
  • @HosamHajeer This is exactly what I needed, thank you so much! – Superdivinidad Sep 25 '24 at 09:12

1 Answers1

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Just as already mentioned in my first comment, but apply the same idea instead onto $F(x,y)+G(x,y)=0$. This directly results in $2y=-5+3x\pm(7-3x)$, i.e. either in $y=1$ or in $y=3x-6$.

Inserting the case $y=1$ into $F$ or $G$ results in $x^2-x=0$ which is $x=0$ or $x=1$, thus being the solutions found in the comment by Ted Shifrin.

Inserting instead the other $y=3x-6$ into $F$ or $G$ both results in $10x^2-40x+42=0$ with solutions $x=2\pm i/\sqrt5$, then being the remaining 2 searched for solutions.

Addition: So far as to a particular solution of the provided example conics. - The more general approach runs very close to those very lines. Instead of using the minus sign of my trial in the comment or the plus sign of the above final solution, you could use a one-parametric combination approach, for instance similar to your cited snippet. Again you try to solve that single combined equation accordingly via the midnight formula, thereby simply choosing the free parameter in such a way, that the radicant can be extracted, as was achieved in the second line of this answer.

Dr. Richard Klitzing
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