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I know that $\mathbb Z$ is not a field so this doesn't rule out non-principal ideals. I don't know how to find them though besides with guessing, which could take forever. As for $\mathbb Q[x,y]$ I know $\mathbb Q$ is a field which would mean $\mathbb Q[x]$ is a principal ideal domain, but does this still apply for $\mathbb Q[x,y]$ ?

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  • $\begingroup$ $\Bbb Z$ is a principle ideal domain. I don't know about $\Bbb Z[x]$. $\endgroup$ – dfeuer Nov 12 '13 at 12:42
  • $\begingroup$ Have you tried guessing yet? Yes, sometimes guessing takes forever. However, a lot of the time (especially with homework questions), it solves the problem very quickly. And frequently when it doesn't, working through your guesses help you understand the problem better which lets you make your next guess even better, or even gives you ideas for solving the problem! $\endgroup$ – user14972 Nov 12 '13 at 12:52
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Here is a general result:

If $D$ is a domain, then $D[X]$ is a PID iff $D$ is a field.

One direction is a classic result. For the other direction, take $a\in D$, consider the ideal $(a,X)$, and prove that it is principal iff $a$ is a unit.

This immediately answers both questions: $(2,X)$ is not principal in $\mathbb Z[X]$ and $(X,Y)$ is not principal in $\mathbb Q[X,Y]=\mathbb Q[X][Y]$.

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Hints:

  • Consider the polynomials $a_nx^n+\dots+a_1x+a_0\in\mathbb{Z}[x]$ where $a_0$ is even.

  • For $\mathbb{Q}[x,y]$: you have two indeterminates.

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