The ring $K[t^2,t^3]$ is not a PID

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?

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

$$a=t^2$$ 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.