I have just started reading Cox, Little, and O'Shea's book Ideals, Varieties, and Algorithms, and I am working through exercise 1.4.1 and am confused. I am hoping someone here will be able to spot where I have gone wrong.
The exercise says to "[c]ondsider the following equations: $$x^2+y^2-1=0\\xy-1=0,$$ which describe the intersection of a circle and a hyperbola.
a. Use algebra to eliminate $y$ from the above equations
b. Show how the polynomial found in part (a) lies in $\langle x^2+y^2-1, xy-1\rangle$. Your answer should be similar to what we did in (1). Hint: Multiply the second equation by $xy+1$"
So my first point of confusion is that when I use SAGE to plot the two varieties (using implicit_plot) they do not intersect at all. To me this made sense because the closest the curve $xy-1$ will be to the origin is at $x=y=1$, but that is distance $\sqrt{2}$ from the origin so it is outside of the unit circle defined by $x^2+y^2-1=0$.
Ok now taking the intersection issue at faith, then I can do part (a), to write $x^2+\frac{1}{x^2}-1=0$. I couldn't figure out part (b), so I decided to run the following SAGE code:
sage: R.<x,y> = RR['x','y']
sage: R
Multivariate Polynomial Ring in x, y over Real Field with 53 bits of precision
sage: f = x^2+y^2-1
sage: g = x*y-1
sage: I = (f,g)*R
sage: I
Ideal (x^2 + y^2 - 1.00000000000000, x*y - 1.00000000000000) of Multivariate Polynomial Ring in x, y over Real Field with 53 bits of precision
sage: x^2+x^(-2)-1 in I
False
This seems to indicate that the exercise is wrong, but it seems more likely that I am wrong in my understanding of part (a).
I also followed the hint and saw that multiplying the first equation by $x^2$ and the second equation by $xy+1$ we get $$ x^4 + x^2y^2 -x^2 = 0\\ x^2y^2-1 = 0$$ and so $x^2(x^2+y^2-1)-(xy+1)(xy-1) = x^4+1-x^2 \in \langle x^2+y^2-1, xy-1\rangle$. Then we can notice that $x^4+1-x^2 = x^2(x^2+\frac{1}{x^2} -1)$ which is what we wanted with an extra factor of $x^2$, but that doesn't seem to help. If it does help could someone please explain why/how?
Thank you!
The first problem is that $x^2+x^{-2}-1$ is not a polynomial since the exponent on $x$ is negative. Therefore, this Laurent polynomial is not in $R$ (and, therefore, not in the ideal $I$).
Starting with $x^2+y^2-1=0$ and $xy-1=0$, you could multiply the first equation by $x^2$ to get $x^4+x^2y^2-x^2=0$. Since $xy=1$, $x^2y^2=1$, so we can substitute to get $x^4+1-x^2=0$. Therefore, $x^4-x^2+1\in I$.
For the second part of the problem, one could, instead multiply $x^2+y^2-1$ by $x^2$ to get $x^4+x^2y^2-x^2$ and multiply $xy-1$ by $xy+1$ to get $x^2y^2-1$. Subtracting these from each other gives the desired result. In other words, $$ x^2(x^2+y^2-1)-(xy+1)(xy-1)=x^4-x^2+1. $$
Once you get a bit further in the book, you'll find that the set of polynomials of $I$ which only involve $x$ are all multiples of $x^4-x^2+1$.