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I want to show that the ideal $I = (x^2, y)$ in the ring $R = \mathbb{Z}[x,y]$ is not a principal. Any ideas?

AMD
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    If $I$ were equal to $(f)$, the polynomial $f$ would have to divide both $x^2$ and $y$. – user49640 Jun 10 '17 at 03:16
  • @user49640 sorry you are right. – Vim Jun 10 '17 at 03:24
  • @user49640 To finish the argument, $f$ dividing both $x^2$ and $y$ forces $f = 1$ or $-1 $ in which case $(f) = R$. Is this correct? – AMD Jun 10 '17 at 03:27
  • Yes. I mean, to justify your argument, there's a little bit less to prove if you already know that $Z[x,y]$ is a unique factorization domain. (Though it's easy either way.) After that, you still also need to prove that $(x^2,y)$ is not equal to $R$. – user49640 Jun 10 '17 at 03:31

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Hint $ $ If $\,(x^2,y) = (f)\,$ then $\,f = \underbrace{ \gcd(x^2,y) = 1}_{\large {\rm by\ prime}\ x\,\nmid\, y}$ so $1 = x^2 g + y h\,$ $\Rightarrow1 = 0$ by eval at $x,y = 0$

Bill Dubuque
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