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

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!

  • $\begingroup$ $x^2+x^{-2}-1$ is not a polynomial, so it's not even an element of $R$, let alone an element of $I$. $\endgroup$ Commented Dec 29, 2019 at 2:44
  • $\begingroup$ @MichaelBurr Oh, that makes sense. Would you be able to explain to me part (a) then? $\endgroup$ Commented Dec 29, 2019 at 2:48

1 Answer 1


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 $x^4-x^2+1$.

  • $\begingroup$ Why is it valid to multiply the first equation by $x^2$? It seems like that should not be allowed because we are artificially adding zeros on the $y$-axis? $\endgroup$ Commented Dec 29, 2019 at 3:04
  • $\begingroup$ @tekay-squared Ideals are closed under multiplication by ring elements. Therefore, multiplying by $x$ results in another element of the ideal. The goal isn't necessarily to find the smallest element of the ideal, just some element of the ideal (although we do find the smallest such element). Also, you can see that you're not adding roots since when $x=0$, the second equation has no solutions since $(0)y-1=-1$. $\endgroup$ Commented Dec 29, 2019 at 3:08
  • $\begingroup$ Okay, I understand the staying within the ideal. But when we are just trying to eliminate $y$ are we implicitly working in an ideal? I was reading the question as "do high school algebra to get rid of $y$", but it sounds like I am missing something important. $\endgroup$ Commented Dec 29, 2019 at 3:12
  • $\begingroup$ The distinction is between working with equations and working with elements of an ideal. The approaches are similar, but slightly different due to the types of objects under consideration. $\endgroup$ Commented Dec 29, 2019 at 3:22

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