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I have to show that the ring $R=K[t^2,t^3]$ is not a PID, where $K$ is a field.

Consider the ideal $I=(t^2,t^3)$. If $R$ is a PID then there exist $f(t^2,t^3)\in R$ such that $(t^2,t^3)= (f(t^2,t^3))$. Since, $t^2\in I\implies t^2=f(t^2,t^3)g(t^2,t^3) \implies $ either $f$ is constant polynomial or $f$ is of degree 2 polynomial. If it is degree 2 polynomial then it can't give $t^3$ but what if it is constant?

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  • $\begingroup$ A nonzero constant is a unit. $\endgroup$ – quasi Sep 1 '17 at 0:10
  • $\begingroup$ Sorry, but I did not get it. $\endgroup$ – XYZABC Sep 1 '17 at 0:18
  • $\begingroup$ An ideal containing a unit (an invertible element) also contains $1$, hence is the full ring. $\endgroup$ – quasi Sep 1 '17 at 0:26
  • $\begingroup$ Alternatively, any element of the ideal $(t^2,t^3)$ is a multiple of $t^2$ in $K[t]$, hence is not constant. $\endgroup$ – quasi Sep 1 '17 at 0:28
  • $\begingroup$ Okay, I got it. Thanks :) $\endgroup$ – XYZABC Sep 1 '17 at 0:43
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Here is another take.

Every PID is a UFD.

$a=t^2$ and $b=t^3$ are irreducible elements of $R$ but $a^3=b^2$ contradicts unique factorization.

Therefore, $R$ is not a UFD and so cannot be a PID.

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