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Show that $R = \mathbb{C}[x,y]/(y^2-x^3-1)$ is not a PID.

My idea is to find an ideal of $\mathbb{C}[x,y]$ containing $(y^2-x^3-1)$ and show that its image is not principal.

So I have $J = (x,y+1)$, then suppose it is generated by $\alpha(x,y)$. Since we can reduce the degree of $y$ of $\alpha$, we can assume it is of the form $p(x)y+q(x)$. Then there exists polynomials $f(x,y),g(x,y),h(x,y),k(x,y)$ such that

$$x = \alpha(x,y) f(x,y) + h(x,y)(y^2-x^3-1)$$ $$y+1 = \alpha(x,y) g(x,y) + k(x,y)(y^2-x^3-1)$$

But then I am kind of stuck because I have no idea how to continue. I m not even sure if I picked the right ideal to work with

user26857
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Phantom
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    You could try to show it's not a UFD. How does $x^3+1$ factor? – rogerl Mar 06 '19 at 01:30
  • Actually I have $y^2-1 = (y+1)(y-1) = x^3$, but the problem is how do I prove that $y\pm 1$ and $x$ are primes? – Phantom Mar 06 '19 at 01:35
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    @Phantom Actually none of them is prime. Instead they are irreducible (which is not easily seen!). For a method to prove the irreducibility see here. – user26857 Mar 06 '19 at 08:13

1 Answers1

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Suppose it were an UFD, then primes are irreducibles in that case. It's easier to see that $x$ and $y-1$ are irreducibles in that case and non units so that $y^2 -1=(y-1)(y+1)=x^3$, that is it wouldn't have a unique factorization. Thus $R$ is not an UFD which means that $R$ is not a PID.

Edit: As it has been said on the comments. Proving they are irreducibles is much more difficult than I thought and that this post makes it seem. Maybe there are other ways of doing this?

hedphelym
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    how to you show that $x,y-1$ are irreducibles under this quotient? – Phantom Mar 06 '19 at 02:45
  • Informally, this is what I was thinking. You can clearly see that they are irreducibles over $\Bbb C[x,y]$, for example by noticing they are irreducibles over $\Bbb C[x][y]$ since it is primitive and irreducible as a one variable polynomial. The extra conditions imposed by the quotient don't affect neither $x$ nor $y-1$ but higher degree polynomials. They also can't be units in this quotient as they are not units in $\Bbb C[x,y]$. Now it's getting very late for me, but tomorrow I can try to write this more rigorously if you need it. – hedphelym Mar 06 '19 at 05:16