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?

Solution 1:

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.

Solution 2:

Suppose that it is a PID, then it is also an UFD and every UFD is normal, (i.e.) integrally closed. But we can clearly see that the element $t\in k[t]$ is integrally closed over $k[t^2,t^3]$, for instance via the polynomial $x^2-t^2$, but $t\notin k[t^2,t^3]$.